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NCERT Exemplar Class 12 Chemistry Solutions Chapter 4 Chemical Kinetics

NCERT Exemplar Class 12 Chemistry Solutions Chapter 4 Chemical Kinetics

Edited By Sumit Saini | Updated on Sep 16, 2022 05:03 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Chemistry solutions chapter 4 discuss questions related to a branch of chemistry that studies the rate and time of occurrence of any reaction. NCERT exemplar Class 12 Chemistry chapter 4 solutions deals with the concepts of reaction rates of different chemical compounds based on different concentrations of the reactants and how these factors of chemical reaction could be altered through temperature, pressure, catalyst and concentration in bringing changes in its chemical properties, and its reaction time. The NCERT lesson also describes the method for elementary and complex reactions and provides a formulation for the rate constant, order, molecularity, and its influence in the change of properties of chemical reactions. Class 12 Chemistry NCERT exemplar solutions chapter 4 also covers the effect of concentration and rate constant on zero-order, and the first-order reaction through the integration of the differential rate equation. NCERT exemplar Class 12 Chemistry solutions chapter 4 PDF download is also available for students.

This Story also Contains
  1. NCERT Exemplar Class 12 Chemistry Solutions Chapter 4: MCQ (Type 1)
  2. Question:1
  3. NCERT Exemplar Class 12 Chemistry Solutions Chapter 4: MCQ (Type 2)
  4. NCERT Exemplar Class 12 Chemistry Solutions Chapter 4: Short Answer Type
  5. NCERT Exemplar Class 12 Chemistry Solutions Chapter 4: Matching Type
  6. NCERT Exemplar Class 12 Chemistry Solutions Chapter 4: Assertion and Reason Type
  7. NCERT Exemplar Class 12 Chemistry Solutions Chapter 4: Long Answer Type
  8. NCERT Exemplar Class 12 Chemistry Solutions Chapter 4 Chemical Kinetics -Main Subtopics
  9. What will the students learn through NCERT Exemplar Class 12 Chemistry Solutions Chapter 4 Chemical Kinetics?
  10. NCERT Exemplar Class 12 Chemistry Solutions
  11. Important Topics To Cover For Exams From Chemical Kinetics

NCERT Exemplar Class 12 Chemistry Solutions Chapter 4: MCQ (Type 1)

Question:1

The role of a catalyst is to change ______________.
(i) gibbs energy of reaction.
(ii) enthalpy of reaction.
(iii) activation energy of reaction.
(iv) equilibrium constant.
Answer:

The answer is the option (iii). Activation energy for the reaction is lowered by a catalyst.

Question:2

In the presence of a catalyst, the heat evolved or absorbed during the reaction ___________.
(i) increases.
(ii) decreases.
(iii) remains unchanged.
(iv) may increase or decrease.
Answer:

The answer is the option (iii). Catalyst only eases the threshold for a reaction without participating directly in it. So, it has no effect on heat evolved or absorbed during reaction.

Question:3

Activation energy of a chemical reaction can be determined by _____________.
(i) determining the rate constant at standard temperature.
(ii) determining the rate constants at two temperatures.
(iii) determining the probability of collision.
(iv) using catalyst.
Answer:

The answer is the option (ii). According to Arrhenius equation, activation energy can be calculated by measuring the rate constants at two different temperatures.
log\left ( \frac{k_{2}}{k_{1}} \right )=\frac{E_{a}}{2.303R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})
Where, E_{a}=\text {Activation energy}
T_{1}=\text {lower temperature}
T_{2}=\text {Higher temperature}
k_{1}=\text {rate constant at temperature }T_{1}
k_{2}=\text {rate constant at temperature }T_{2}

Question:4

Consider Fig. 4.1 and mark the correct option.
(i) The activation energy of forward reaction is E1 + E2 and the product is less stable than reactant.
(ii) The activation energy of forward reaction is E1 + E2 and product is more stable than reactant.
(iii) The activation energy of both forward and backward reaction is E1 + E2 and reactant is more stable than the product.
(iv) The activation energy of the backward reaction is E1 and the product is more stable than reactant.


Answer:

The answer is the option (i) E_{a}\; \text {forward}=E_{1}+E_{2}
As reactants have lower energy than the products, the products are less stable.

Question:5

Consider a first order gas phase decomposition reaction given below :
A(g)\rightarrow B(g)+C(g)
The initial pressure of the system before decomposition of A was pi. After lapse of time ‘t’, total pressure of the system increased by x units and became ‘pt'. The rate constant k for the reaction is given as _________.
(i) k=\frac{2.303}{t} log\;\frac{P_{i}}{P_{i}-x}
(ii) k=\frac{2.303}{t} log\;\frac{P_{i}}{2P_{i}-P_{t}}
(iii) k=\frac{2.303}{t} log\;\frac{P_{i}}{2P_{i}+P_{t}}
(iv) k=\frac{2.303}{t} log\;\frac{P_{i}}{P_{i}+x}
Answer:

The answer is the option (ii)


A(g)

B(g)

C(g)

Initial Pressure:

Pi

0

0

Pressure after time (t):

(Pi - x)

x

x

P_{i}=P_{i}+x
x=P_{t}-P_{i}
P_{A}= \text {Pressure of A after time t}= \text {P}_{i}-\text {x}=\text {P}_{i}-(\text {P}_{i}-\text {P}_{i})=2\text {P}_{i}-\text {P}_{t}
k=\frac{2.303}{t}log\frac{[A]_{0}}{[A]}=\frac{2.303}{t}log\frac{P_{i}}{2P_{i}-P_{t}}

Question:6

According to Arrhenius equation rate constant k is equal to A \; e^{-E_{a} /RT}. Which of the following options represents the graph of ln k vs \frac{1}{T}?

Answer:

The answer is the option (i) According to Arrhenius equation:
k=Ae^{-\frac{E_{a}}{RT}}
ln (k)=ln\left ( Ae^{-\frac{E_{a}}{RT}} \right )
ln\; k =ln\; A-\frac{E_{a}}{RT}

Question:7

Consider the Arrhenius equation given below and mark the correct option. k = A \; e^{-E_{a} /RT}
(i) Rate constant increases exponentially with increasing activation energy and decreasing temperature.
(ii) Rate constant decreases exponentially with increasing activation energy and decreasing temperature.
(iii) Rate constant increases exponentially with decreasing activation energy and decreasing temperature.
(iv) Rate constant increases exponentially with decreasing activation energy and increasing temperature
Answer:

The answer is the option (iv). Increase in T leads to an increase in -\frac{E_{a}}{RT} and thus, an increase in k. Similarly, a decrease in E_{a} leads to an increase in -\frac{E_{a}}{RT} and thus, an increase in k.

Question:8

A graph of volume of hydrogen released vs time for the reaction between zinc and dil.HCl is given in Fig. 4.2. On the basis of this mark the correct option.

(i) Average rate up to 40 s is \frac{V_{3}-V_{2}}{40}
(ii) Average rate up to 40 seconds is \frac{V_{}3- V_{}2}{40-30}
(ii) Average rate up to 40 seconds is \frac{V_{}3}{40}
(iv) Average rate up to 40 seconds is \frac{V_{}3- V_{}1}{40-20}

Answer:

The answer is the option (iii) Average rate of the reaction can be determined by change in concentration of H_{2} upon time from (0,0) to (40, V_{3})=\frac{V_{3}-0}{40-0}=\frac{V_{3}}{40}

Question:9

Which of the following statements is not correct about an order of a reaction.
(i) The order of a reaction can be a fractional number.
(ii) Order of a reaction is experimentally determined quantity.
(iii) The order of a reaction is always equal to the sum of the stoichiometric coefficients of reactants in the balanced chemical equation for a reaction.
(iv) The order of a reaction is the sum of the powers of the molar concentration of the reactants in the rate law expression.
Answer:

The answer is the option (iii). The summation of powers of the reactant’s concentration is used for determining the order of reaction. The order of a reaction is always equal to the sum of the stoichiometric coefficients of reacting species in the balanced chemical equation.

Question:10

Consider the graph given in Fig. 4.2. Which of the following options does not show the instantaneous rate of reaction at 40th second?
(i)\; \frac{V_{5}-V_{2}}{50-30}
(ii)\; \frac{V_{4}-V_{2}}{50-30}
(iii)\; \frac{V_{3}-V_{2}}{40-30}
(iv)\; \frac{V_{3}-V_{1}}{40-20}
Answer:

The answer is the option (ii). Instantaneous rate is the rate of a reaction at a particular moment of time. Option (ii) does not show instantaneous rate of reaction.

Question:11

Which of the following statements is correct?
(i) The rate of a reaction decreases with passage of time as the concentration of reactants decreases.
(ii) The rate of a reaction is the same at any time during the reaction.
(iii) The rate of a reaction is independent of temperature change.
(iv) The rate of a reaction decreases with increase in the concentration of the reactant(s).
Answer:

The answer is the option (i). The rate of a reaction is defined as the change in concentration of the reactant per unit time. Therefore, it is dependent upon the concentration of reactants.

Question:12

Which of the following expressions is correct for the rate of reaction given below?
5Br^{}-(aq)+BrO_{3}^{}-(aq)+6H^{}+(aq)\rightarrow 3Br_{2}(aq)+3H_{2}O(l)
(i) \frac{\Delta [Br^{-}]}{\Delta t}=5\frac{\Delta [H^{+}]}{\Delta t}
(ii) \frac{\Delta [Br^{-}]}{\Delta t}=\frac{6}{5}\frac{\Delta [H^{+}]}{\Delta t}
(iii) \frac{\Delta [Br^{-}]}{\Delta t}=\frac{5}{6}\frac{\Delta [H^{+}]}{\Delta t}
(iv) \frac{\Delta [Br^{-}]}{\Delta t}=6\frac{\Delta [H^{+}]}{\Delta t}

Answer:

We can rewrite the rate law expression for the given equation
5Br^{}-(aq)+BrO_{3}^{}-(aq)+6H^{}+(aq)\rightarrow 3Br_{2}(aq)+3H_{2}O(l)
-\frac{1}{5}\frac{\Delta [Br^{-}]}{\Delta t}=-\frac{\Delta [BrO_{3}^{-}]}{\Delta t}=-\frac{1}{6}\frac{\Delta [H^{+}]}{\Delta t}=\frac{1}{3}\frac{\Delta [Br_{2}]}{\Delta t}
-\frac{\Delta [Br^{-}]}{\Delta t}=-\frac{\Delta [BrO_{3}^{-}]}{\Delta t}=-\frac{5}{6}\frac{\Delta [H^{+}]}{\Delta t}
\frac{\Delta [Br^{-}]}{\Delta t}=\frac{5}{6}\frac{\Delta [H^{+}]}{\Delta t}
(iii) \frac{\Delta [Br^{-}]}{\Delta t}=\frac{5}{6}\frac{\Delta [H^{+}]}{\Delta t}

Question:13

Which of the following graph represents the exothermic reaction?

(i) (a) only
(ii) (b) only
(iii) (c) only
(iv) (a) and (b)
Answer:

The answer is the option (i). In an exothermic reaction the reaction activation energy of product should be higher than that of reactant whereas the enthalpy of reactant should be greater than that of products.

Question:14

The rate law for the reaction A+2B\rightarrow C is found to be
Rate =k[A][B]
The concentration of reactant ‘B’ is doubled, keeping the concentration of ‘A’ constant, the value of rate constant will be______.
(i) the same
(ii) doubled
(iii) quadrupled
(iv) halved
Answer:

The answer is the option (i). Rate constant of a reaction is independent of the concentrations of the reactants.

Question:15

Which of the following statements is incorrect about the collision theory of chemical reaction?
(i) It considers reacting molecules or atoms to be hard spheres and ignores their structural features.
(ii) Number of effective collisions determine the rate of reaction.
(iii) Collision of atoms or molecules possessing sufficient threshold energy results into the product formation.
(iv) Molecules should collide with sufficient threshold energy and proper orientation for the collision to be effective.
Answer:

The answer is the option (iii) For a reaction to occur, collision theory mandates several conditions that are:
(i) Collision of molecules must take place with sufficient threshold energy and should be effective.
(ii) They should have proper orientation.

Question:16

A first-order reaction is 50% completed in 1.26 \times 10^{14} \; s. How much time would it take for 100% completion?
(i)\; 1.26\times 10^{15}\; s
(ii)\; 2.52\times 10^{14}\; s
(iii)\; 2.52\times 10^{28}\; s
(iv) \text {infinite}
Answer:

The answer is the option (iv). Since whole of the substance never reacts, reaction is considered 100% complete only after an infinite time that cannot be calculated.

Question:17

Compounds ‘A’ and ‘B’ react according to the following chemical equation.
A(g)+2B(g)\rightarrow 2C(g)
The concentration of either ‘A’ or ‘B’ was changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.

Experiment

Initial Concetration of [A]/molL-1

Initial Concentration of [B]/molL-1

Initial rate concentration of [C]/molL-1 S-1

1

0.30

0.30

0.10

2

0.30

0.60

0.40

3

0.60

0.30

0.20

(i) Rate = k[A]^{2}[B]
(ii) Rate =k[A][B]^{2}
(iii) Rate = k[A][B]
(iv) Rate = k[A]^{2}[B]^{0}

Answer:

The answer is the option (ii)
We know that Rate of a reaction is =k[A]^{x}[B]^{y}
On doubling the concentration of B, keeping the concentration of A constant, the rate of formation of C increases by a factor of four. Thus, the rate of reaction is directly proportional to the square of concentration of B. However, on doubling the concentration of A, the rate of formation of C becomes the double of the initial value. Thus the rate is directly proportional to the first power of concentration of A. Hence Rate = =K[A][B]^{2}

Question:18

Which of the following statement is not correct for the catalyst?
(i) It catalyses the forward and backward reaction to the same extent.
(ii) It alters \Delta G of the reaction.
(iii) It is a substance that does not change the equilibrium constant of a reaction.
(iv) It provides an alternate mechanism by reducing activation energy between reactants and products.
Answer:

The answer is the option (ii)
A catalyst increases the rate of both the forward and backward reaction to the same extent. Hence, option a is correct.
\Delta G= \text {-RT ln Q where Q is the reaction quotient}
Reaction quotient is the relation between the concentration of reactants and products. Thus the gibbs free energy is unaffected by the addition of catalyst. It does not affect the equilibrium of the reaction since equilibrium constant depends only on the concentration of reactants and products.

Question:19

The value of rate constant of a pseudo first order reaction ____________.
(i) depends on the concentration of reactants present in a small amount.
(ii) depends on the concentration of reactants present in excess.
(iii) is independent of the concentration of reactants.
(iv) depends only on temperature.
Answer:

The answer is the option (ii). The value of rate constant reaction is dependent on the concentration of reactants present in excess in a pseudo first order reaction. This is because the excess reactants do not get altered much while during the reaction

Question:20

Consider the reaction A \rightleftharpoons B. The concentration of both the reactants and the products varies exponentially with time. Which of the following figures correctly describes the change in concentration of reactants and products with time ?

Answer:

The answer is the option (ii) A->B

With time, there is an exponential decrease in the concentration of reactants and an increase in the concentration of products.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 4: MCQ (Type 2)

Question:21

In the following questions two or more options may be correct.
Rate law cannot be determined from the balanced chemical equation if _______.
(i) the reverse reaction is involved.
(ii) it is an elementary reaction.
(iii) it is a sequence of elementary reactions.
(iv) any of the reactants is in excess
Answer:

The answer is the option (i, iii, iv). From a balanced chemical equation rate law can only be determined if it is an elementary reaction.

Question:22

In the following questions two or more options may be correct.
Which of the following statements are applicable to a balanced chemical equation of an elementary reaction?
(i) Order is the same as molecularity.
(ii) Order is less than the molecularity.
(iii) Order is greater than the molecularity.
(iv) Molecularity can never be zero.
Answer:

The answer is the option (i, iv). The order and molecularity are same for an elementary or single step reaction with molecularity of a reaction never equal to zero.

Question:23

In the following questions two or more options may be correct.
In any unimolecular reaction ______________.
(i) only one reacting species is involved in the rate-determining step.
(ii) the order and the molecularity of the slowest step are equal to one.
(iii) the molecularity of the reaction is one and order is zero.
(iv) both molecularity and order of the reaction are one.
Answer:

The answer is the option (i, ii). Only one reactant is involved in rate determining step in an unimolecular reaction and the order and molecularity of rate determining or slowest step are equal to one.
A\rightarrow B
Rate = k[A]

Question:24

In the following questions two or more options may be correct.
For a complex reaction ______________.
(i) order of an overall reaction is the same as molecularity of the slowest step.
(ii) order of an overall reaction is less than the molecularity of the slowest step.
(iii) order of an overall reaction is greater than molecularity of the slowest step.
(iv) molecularity of the slowest step is never zero or non-integer.
Answer:

The answer is the option (i, iv).
The answer is the option (i) Order of overall reaction is same as the molecularity of slowest step for a complex reaction.
The rate of overall reaction is dependent on total number of molecules involved in slowest step of the reaction making molecularity of the slowest step equal to order of overall reaction.

Question:25

In the following questions two or more options may be correct.
At high pressure, the following reaction is zero order.
2NH_{3}(g)\overset{1130 K \; \text {Platinum catalyst}}{\rightarrow} \; N_{2}(g)+3H_{2}(g)
Which of the following options are correct for this reaction?
(i) Rate of reaction = Rate constant
(ii) Rate of the reaction depends on concentration of ammonia.
(iii) Rate of decomposition of ammonia will remain constant until ammonia
disappears completely.
(iv) Further increase in pressure will change the rate of reaction.
Answer:

The answer is the option (i, iii, iv) The rate of reaction is not dependent on the concentration of ammonia in a zero-order reaction

Question:26

In the following questions two or more options may be correct.
During decomposition of an activated complex
(i) energy is always released
(ii) energy is always absorbed
(iii) energy does not change
(iv) reactants may be formed
Answer:

The answer is the option (i, iv) Activated complex is an intermediate formed at the highest level of energy (E_{a}) of the system. Some energy is released with the decomposition of this unstable complex to give products. If the reaction is reversible, reactants might be formed.

Question:27

In the following questions two or more options may be correct.
According to Maxwell Boltzmann distribution of energy, __________.
(i) the fraction of molecules with most probable kinetic energy decreases at higher temperatures.
(ii) the fraction of molecules with most probable kinetic energy increases at higher temperatures.
(iii) most probable kinetic energy increases at higher temperatures.
(iv) most probable kinetic energy decreases at higher temperatures.
Answer:

The answer is the option (i, iii) Maxwell Boltzmann distribution of kinetic energy can be described by a graph plotted with fraction of molecules on y-axis versus kinetic energy on x-axis.

Most probable kinetic energy is the kinetic energy of maximum fraction of molecule. The peak of the graph shifts forward and downward with an increase in temperature.
This implies that the most probable kinetic energy increases and fraction of molecules with most probable K.E decreases with an increase in temperature.

Question:28

In the following questions two or more options may be correct.
In the graph showing Maxwell Boltzman distribution of energy, ___________.
(i) area under the curve must not change with an increase in temperature.
(ii) area under the curve increases with increase in temperature.
(iii) area under the curve decreases with increase in temperature.
(iv) with the increase in the temperature, curve broadens and shifts to the right-hand side.

Answer:

The answer is the option (i, iv) Area under the curve should remain constant with an increase in temperature as the probability must be 1 for all the time. As the temperature rises, the curve broadens out and shifts to the right-hand higher energy value side.

Question:29

Which of the following statements are in accordance with the Arrhenius equation?
(i) Rate of a reaction increases with increase in temperature.
(ii) Rate of a reaction increases with decrease in activation energy.
(iii) Rate constant decreases exponentially with an increase in temperature.
(iv) Rate of reaction decreases with decrease in activation energy.
Answer:

The answer is the option (i, iv). According to Arrhenius equation, the rate of reaction increases with increase in temperature and decrease in the activation energy. It gets double with every 10° change. As the rate of reaction increases, there is an exponential increase in the rate constant of the reaction.

Question:30

Mark the incorrect statements.
(i) Catalyst provides an alternative pathway to a reaction mechanism.
(ii) Catalyst raises the activation energy.
(iii) A catalyst lowers the activation energy.
(iv) Catalyst alters enthalpy change of the reaction.
Answer:

The answer is the option (ii, iv)
(i) On adding the catalyst to a reaction, medium rate of reaction increases with decreasing activation energy of molecule.
(ii) The enthapy change of reaction remains unchanged due to a catalyst. In both the catalysed and uncatalysed reaction, energy of reactants and products are alike.

Question:31

Which of the following graphs is correct for a zero-order reaction?

Answer:

The answer is the option (i, iv). On plotting [R] against t, a straight line is obtained with slope =-k and intercept equal to [R]_{0}from the equation
[R]=-kt+[R]_{0}
The rate is directly proportional to the zero power of concentration in a Zero-order reaction.
Rate = -\frac{d[R]}{dt}=k[R]^{0}

Question:32

Which of the following graphs is correct for a first-order reaction?

Answer:

The answer is the option (i, iv) For a first order reaction
t_{\frac{1}{2}}=\frac{0.693}{k} and rate =-\frac{d[R]}{dt}=k[R]
Since \; k=\frac{2.303}{t}\; log\; [\frac{R^{0}}{R}] the slope for a graph between log log [\frac{R^{0}}{R}] versus t is \frac{k}{2.303}

NCERT Exemplar Class 12 Chemistry Solutions Chapter 4: Short Answer Type

Question:33

State a condition under which a bimolecular reaction is a kinetically first-order reaction.
Answer:

In a biomolecular reaction:
A+B\rightarrow Product
where the rate is given by k[A] [B] …………. (i)
If the value of [B] is very high, rate law can be represented as Rate =k'[A]
where, k'=k[B]
The order of reaction will thus be one.

Question:34

Write the rate equation for the reaction 2A+B\rightarrow C if the order of the reaction is zero.
Answer:

2A+B \rightarrow C
If the order of reaction is zero, the rate equation can be written as
Rate = k[A]^{o} [B]^{o}

Question:35

How can you determine the rate law of the following reaction?
2NO (g) + O_{2} (g) \rightarrow 2NO_{2} (g)
Answer:

The rate of this reaction is measured as a function of initial concentration for the determination of the rate law. This is done either by keeping the concentration of one of the reactants constant while changing the other or by changing the concentration of both the reactants. The rate law can be determined from the concentration dependence of rate.

Question:36

For which type of reactions, order and molecularity have the same value?
Answer:

In elementary reactions the value of order and molecularity is equal.

Question:37

In a reaction if the concentration of reactant A is tripled, the rate of reaction becomes twenty-seven times. What is the order of the reaction?
Answer:

\text {Rate}=[\text {A}]^{\alpha }
27*\text {Rate}=\text {k}[\text {3A}]^{\alpha }
Dividing both the equations we get \alpha =3
Therefore the order of reaction is 3

Question:38

Derive an expression to calculate the time required for completion of the zero-order reaction.
Answer:

In case of a zero order reaction
R=[R]_{0}-kt
For the reaction to achieve completion [R]=0
kt=[R]_{0}\; or\; t=\frac{[R]_{0}}{k}

Question:39

For a reaction A+B\rightarrow products, the rate law is — Rate = k [A][B]^{3/2}. Can the reaction be an elementary reaction? Explain.
Answer:

Given that the rate of reaction is k[A][B]^{\frac{3}{2}}
Order =1+\frac{3}{2}=\frac{5}{2}
The order of elementary reaction cannot be fractional.
Hence, This indicates that the reaction is not an elementary
reaction.

Question:40

For a certain reaction large fraction of molecules has energy more than the threshold energy, yet the rate of reaction is very slow. Why?
Answer:

For a certain reaction, large fraction of molecules has energy more than the threshold energy, yet the rate of reaction is very slow due to lack of proper orientation

Question:41

For a zero-order reaction will the molecularity be equal to zero? Explain.
Answer:

Molecularity of a reaction cannot be equal to zero in any case.

Question:42

For a general reaction A\rightarrow B, plot of concentration of A vs time is given in Fig. 4.3. Answer the following question based on this graph.
(i) What is the order of the reaction?
(ii) What is the slope of the curve?
(iii) What are the units of the rate constant?

Answer:

(i) The reaction is of zero order.
(ii) [R]=[R]_{0}-kt
Thus, the slope of the curve is -k
(iii) \frac{molL^{-1}}{s}=k[A]^{0}\; or\; k=molL^{-1}s^{-1}

Question:43

The reaction between H_{2}(g) and O_{2}(g) is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of water. Explain.
Answer:

2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)
The reaction between H2 and O2 to form water does not take place at room temperature because the activation energy of the reaction is very high.

Question:44

Why does the rate of a reaction increase with a rise in temperature?
Answer:

The rate of a reaction increases with a rise in temperature as larger fraction of colliding particles can cross the energy barrier (i.e. the activation energy).

Question:45

Oxygen is available in plenty in the air yet fuels do not burn by themselves at room temperature. Explain.

Answer:

Oxygen does not burn by themselves in spite of being available in plenty in air as the activation energy or E_{a} for combustion reactions of fuels is extremely high at room temperature.

Question:46

Why is the probability of reaction with molecularity higher than three very rare?
Answer:

The low possibility of more than three molecules colliding simultaneously makes the probability of reaction with molecularity higher than three very rare.

Question:47

Why does the rate of any reaction generally decrease during the reaction?
Answer:

We know that the rate of a reaction is dependent on the concentration of reactants. During the course of the reaction, the concentration of reactants keeps decreasing because the reactants keep getting converted to products.

Question:48

Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.

Answer:

Gibbs free energy determines the thermodynamic feasibility of a reaction. For spontaneous reaction, \Delta G should be negative whereas
Activation energy determines the kinetic feasibility of the reaction.
\text {Diamond}\rightarrow \text {Graphite}\; \text {where}\;\Delta G=-ve

Question:49

Why in the redox titration of KMnO_{4} vs oxalic acid, we heat oxalic acid solution before starting the titration?
Answer:

In the redox titration of KMnO_{4} and oxalic acid we heat the oxalic acid solution before starting the titration to increase the rate of reaction as it is very slow.

Question:50

Why can’t molecularity of any reaction be equal to zero?
Answer:

Molecularity can be defined as the number of molecules involved in the process. It cannot be less than one.

Question:51

Why molecularity is applicable only for elementary reactions and order is applicable for elementary as well as complex reactions?
Answer:

A complex reaction involves many elementary reactions. Numbers of molecules in each elementary reaction can vary leading to different molecularity at each step. Therefore, the molecularity is not applicable in case of overall complex reaction. Whereas the slowest step is responsible for the experimental determination of the order of a complex and is therefore, applicable even in the case of complex reactions.

Question:52

Why can we not determine the order of a reaction by taking into consideration the balanced chemical equation?
Answer:

As reactions can be multi-stepped, the rate of reaction is determined by slowest step
KClO_3+6FeSO_4+3H_2 SO_4\rightarrow KCl+3H_2 O+3Fe_2 (SO_4 )_3
Given the above equation, it may seem like a reaction with an order of ten, but in reality this is a second order reaction. Order is determined via experiments and expresses the dependence of the rate of reaction on the concentration of reactants.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 4: Matching Type

Question:53

Match the graph given in Column I with the order of reaction given in Column II. More than one item in Column I may link to the same item of Column II.


Column I

Column II

(i)

(a) 1 st order

(ii)

(iii)

(b) Zero order

(iv)


Answer:

i\rightarrow (a), ii\rightarrow (b), iii\rightarrow (b), iv\rightarrow (a)
For zero order reaction rate equation may be written as [R]=-kt+[R_{0}]
which denotes a straight line equation similar to y=mx+c
on transforming \frac{[R]-[R_{0}]}{t}=-k
k=\frac{[R]-[R_{0}]}{T}
k=rate
Rate = k[t]^{0}
rate \; \alpha\; [t]^{0}
For a first order reaction \frac{dx}{dt}\; \alpha \; [\text {concentration}]
Graph between rate and concentration may be drawn as k =\frac{2.303}{t}\; log\frac{[R]_{0}}{[R]}
\frac{kt}{2.303}=log \frac{[R]_{0}}{[R]}
\frac{kt}{2.303}=log [R]_{0}-log [R]
log[R]=\left ( -\frac{k}{2.303} \right )t+log[R]_{0}

Question:54

Match the statements given in column I and Column II.

Column I

Column II

(i)

Catalyst alters the rate of reaction

(a)

Cannot be fraction or zero

(ii)

Molecularity

(b)

Proper orientation is not there always

(iii)

Second half life of first order reaction

(c)

by lowering the activation energy

(iv)

e^{-E_{a}/RT}

(d)

is same as the first

(v)

Energetically favourable reactions are sometimes slow

(e)

total probability is one

(vi)

Area under the Maxwell- Boltzmann curve is constant

(f)

refers to the fraction of molecules with energy equal to or greater than activation energy.

Answer:

(i) → (c) (ii) → (a) (iii) → (d) (iv) → (f) (v) → (b) (vi) → (e)
(i) Catalyst can change the rate of reaction by decreasing the (E_{a}) or activation energy.
(ii) Molecularity cannot be fraction or zero. No reaction is possible if molecularity is zero.
(iii) The first and second half-life of first order reaction are same since half-life is temperature dependent. ‘
(iv) In Arrhenius equation, e-\frac{E_{a}}{RT} denotes the fraction of molecules with K.E equal to or greater than activation energy.
(v) The activation energy and proper orientation of the molecules together determine the criteria for effective collision in collision theory.
(vi) Since total probability of molecules must be equal to one at all times, area under the Maxwell-Boltzmann curve remains constant.

Question:55

Match the items of Column I and Column II.

Column I

Column II

(i)

Diamond

(a)

Short interval of time

(ii)

Instantaneous rate

(b)

Ordibarily rate of conversion is imperceptible

(iii)

Average rate

(c)

long duration of time

Answer:

(i ->b), (ii-> a), (iii-> c)
(i) The conversion of diamond to graphite does not take place under ordinary condition.
(ii) Instantaneous rate of reaction is calculated for a very short time interval.
(iii) Average rate of reaction is calculated for a long duration of time.

Question:56

Match the items of Column I and Column II.

Column I

Column II

(i)

Mathematical expression for rate of reaction

(a)

Rate constant

(ii)

Rate of reaction for zero order reaction is equal to

(b)

rate law

(iii)

Units of rate constant for zero order reaction is same as that of

(c)

order of slowest step

(iv)

Order of a complex reaction is determined by

(d)

rate of a reaction

Answer:

(i -> b), (ii -> a), (iii —> d), (iv-> c)
(i) Mathematical expression for rate of reaction is termed as rate law.
(ii) Rate for zero order reaction is equal to rate constant r=k[A]
(iii) Unit of rate constant for zero order reaction is same as that of rate of reaction.
(iv)The slowest step or the rate determining step helps in finding the order of a complex reaction.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 4: Assertion and Reason Type

Question:57

In the following question, a statement of Assertion The answer is the option (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion The answer is the option (i): Order of the reaction can be zero or fractional.
Reason (R): We cannot determine order from balanced chemical equation.
(i) Both Assertion and Reason are correct and the reason is the correct explanation of Assertion.
(ii) Both assertion and reason are correct but Reason is not the correct explanation of Assertion
(iii) Assertion is correct but Reason is incorrect.
(iv) Both Assertion and Reason are incorrect.
(v) Assertion is incorrect but Reason is correct.
Answer:

The answer is the option (ii) Order of a reaction may be zero or fractional and rate law expression can be used to determine it.

Question:58

In the following question, a statement of Assertion The answer is the option (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion The answer is the option (i): Order and molecularity are same.
Reason (R): Order is determined experimentally, and molecularity is the sum of the stoichiometric coefficient of rate determining elementary step.
(i) Both Assertion and Reason are correct and the reason is the correct explanation of Assertion.
(ii) Both assertion and reason are correct but Reason is not the correct explanation of Assertion
(iii) Assertion is correct but Reason is incorrect.
(iv) Both Assertion and Reason are incorrect.
(v) Assertion is incorrect but Reason is correct.
Answer:

The answer is the option (v). While balanced stoichiometric equation is used to calculate molecularity order can only be measured through experiments.

Question:59

In the following question, a statement of Assertion The answer is the option (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion The answer is the option (i): The enthapy of reaction remains constant in the presence of a catalyst.
Reason (R): A catalyst participating in the reaction, forms different activated complex and lowers down the activation energy but the difference in energy of reactant and product remains the same.
(i) Both Assertion and Reason are correct and the reason is the correct explanation of Assertion.
(ii) Both assertion and reason are correct but Reason is not the correct explanation of Assertion
(iii) Assertion is correct but Reason is incorrect.
(iv) Both Assertion and Reason are incorrect.
(v) Assertion is incorrect but Reason is correct.
Answer:

The answer is the option (i) \Delta H = Activation Energy of forward reaction – Activation Energy of reverse reaction.
The heat of reaction remains unaltered by the catalysts because it does not involve directly in the reaction and only affects the activation energy reactions. The enthalpy of reaction remains unchanged due to the addition of catalyst.

Question:60

In the following question, a statement of Assertion The answer is the option (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion The answer is the option (i): All collision of reactant molecules lead to product formation.
Reason (R): Only those collisions in which molecules have correct orientation and sufficient kinetic energy lead to compound formation.
(i) Both Assertion and Reason are correct and the reason is the correct explanation of Assertion.
(ii) Both assertion and reason are correct but Reason is not the correct explanation of Assertion
(iii) Assertion is correct but Reason is incorrect.
(iv) Both Assertion and Reason are incorrect.
(v) Assertion is incorrect but Reason is correct.
Answer:

The answer is the option (v). Only the molecular collisions with sufficient kinetic energy and proper orientation bring out the formation of product.

Question:61

In the following question, a statement of Assertion The answer is the option (i) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
Assertion The answer is the option (i): Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules.
Reason (R): Reactant molecules undergo chemical change irrespective of their orientation during collision.
(i) Both Assertion and Reason are correct and the reason is the correct explanation of Assertion.
(ii) Both assertion and reason are correct but Reason is not the correct explanation of Assertion
(iii) Assertion is correct but Reason is incorrect.
(iv) Both Assertion and Reason are incorrect.
(v) Assertion is incorrect but Reason is correct.
Answer:

The answer is the option (iii). The criteria for an effective collision and rate of chemical reaction is determined together by activation energy and proper orientation of molecules in the collision theory.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 4: Long Answer Type

Question:62

All energetically effective collisions do not result in a chemical change. Explain with the help of an example.
Answer:

Products are only formed via effective collisions, which means molecules need to collide with sufficient kinetic energy and should have proper orientation. It is a required condition to break molecular bonds, which give way to formation of new bonds i.e. – products.
Taking the example of reaction of bromomethane with hydroxyl ion, we can see that improper orientation does not let the reaction complete.

Question:63

What happens to most probable kinetic energy and the energy of activation with increase in temperature?
Answer:


On increasing the temperature, the curve flattens out as there are more molecules with higher kinetic energies. Because of this shift, the most probable kinetic energy (the modal value) also increases slightly.

Question:64

Describe how does the enthalpy of reaction remain unchanged when a catalyst is used in the reaction?

Answer:

Catalyst doesn’t participate in the reaction but increases the speed of reaction. The “Intermediate complex formation theory” claims that the reactants combine with the catalyst to form highly active radical complex. These complex compounds then decompose to create the product and the catalyst.
The intermediate formed with catalyst possesses much lower potential energy than the intermediate formed in the normal reaction, lowering the activation energy required. The following diagram shows the impact of using the catalyst on the potential energy and thus, the activation energy.

Question:65

Explain the difference between instantaneous rate of a reaction and average rate of a reaction.
Answer:

The change in concentration of reactants or products per unit time is the average rate of reaction.
\text {Average rate}=-\frac{\Delta [R]}{\Delta T}=\frac{\Delta [P]}{\Delta T}
It is calculated over a large time frame and, in most scenarios, will not give the rate of reaction for future. Instantaneous rate is the actual rate of reaction at a given instance and can be represented as follows
\text {Instantaneous rate}= \lim_{\Delta T\rightarrow 0}-\frac{\Delta [R]}{\Delta T}=\lim_{\Delta T\rightarrow 0}\frac{\Delta [P]}{\Delta T}=-\frac{d[R]}{dT}=\frac{d[P]}{dT}
While average rate can be measured for multi-step and elementary reaction, instantaneous rate can’t be measured.

Question:66

With the help of an example explain what is meant by pseudo first order reaction?
Answer:

Pseudo-first order reaction is a reaction where one reactant has a very high concentration and during the course of the reaction, it doesn’t change significantly. The reaction proceeds as a first order reaction in such a case. Take for instance the hydrolysis of ethyl acetate and inversion of cane sugar
Hydrolysis of ethyl acetate:
CH_{3}COOC_{2}H_{5}\; \; \; \; \; \; +\; \; \; \; \; H_{2}O\; \; \; \; \; \; \; \overset{H^{+}}{\rightarrow}\; \; \; \; \; CH_{3}COOH\; \; \; \; \; \; \; \; +\; \; \; \; \; C_{2}H_{5}OH
t=0 0.01\; mol 10\; mol 0\; mol 0\; mol
t=t 0\; mol 9.99 \; mol 0.01 \; mol 0.01 \; mol
\text {Rate of reaction}=k[CH_{3}COOC_{2}H_{5}]
Where, k=k'[H_{2}O]
Inversion of cane sugar:
C_{12}H_{22}O_{11}\; \; \; \; \; \; \; +\; \; \; \; \; H_{2}O\; \; \; \; \; \; \overset{H^{+}}{\rightarrow}\; \; \; \; \; \; \; C_{6}H_{12}O_{6}\; \; \; \; \; \; +\; \; \; \; \; \; C_{6}H_{12}O_{6}
\text {Rate of reaction}=k[C_{12}H_{22}O_{11}]
Where, k=k'[H_{2}O]

NCERT Exemplar Class 12 Chemistry Solutions Chapter 4 Chemical Kinetics -Main Subtopics

  • The rate of a Chemical Reaction
  • Factors Influencing the Rate of a Reaction
  • Dependence of Rate on Concentration
  • Rate Expression and Rate Constant
  • Order of a Reaction
  • Molecularity of a Reaction
  • Integrated Rate Equations
  • Zero Order Reactions
  • First-Order Reactions
  • Half-Life of a Reaction
  • Temperature Dependence of the Rate of a Reaction
  • Effect of Catalyst
  • Collision Theory of Chemical Reactions
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What will the students learn through NCERT Exemplar Class 12 Chemistry Solutions Chapter 4 Chemical Kinetics?

Class 12 Chemistry NCERT exemplar solutions chapter 4 will help students to understand the process of chemical reactions, their rate and time and also about various other factors that will accelerate or hinder such chemical reactions. The NCERT Solutions for Class 12 Chemistry will also teach students about various different catalysts used to increase the rate of reactions and how temperature change can lead or affect the rate of any reaction. By use of NCERT exemplar Class 12 Chemistry solutions chapter 4, the learners will be able to calculate the rate and time of any given reaction and can also alter the rate and time of any reaction with a change in temperature, pressure, concentration and also through the use of different catalysts which will help them to understand how preservatives are used for the long life of food which will help them to connect these theories to their application in real lives.

NCERT Exemplar Class 12 Chemistry Solutions

Important Topics To Cover For Exams From Chemical Kinetics

· NCERT exemplar solutions for Class 12 Chemistry chapter 4 reflects upon temperature dependence of chemical reactions effectively through the use of the Arrhenius Equation, which establishes a relationship of temperature to activation energy and kinetic energy of molecules with temperature change.

· NCERT exemplar Class 12 Chemistry solutions Chapter 4 also deals with the impact of a catalyst on any chemical reaction without hindering the Gibbs energy and equilibrium constant of a reaction.

· NCERT exemplar Class 12 Chemistry chapter 4 solutions concludes with the explanation of the collision theory of chemical reactions providing a wider insight into the energetic and mechanistic aspect of any given reaction.

NCERT Exemplar Class 12 Solutions

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Frequently Asked Questions (FAQs)

1. Q: How many questions are there in this chapter?

Ans: NCERT exemplar Class 12 Chemistry solutions chapter 4 6 exercies in total with 66 questions that are classified into 2 sets of MCQs, short answer, long answer, matching type and assertion & reason questions.

2. Q: Are these solutions helpful for competitive exams?

Ans: Yes, NCERT exemplar solutions for class 12 chemistry chapter 4 are designed to help students study for boards as well as competitive exams like NEET or JEE Main.

3. Q: How to read this chapter more effectively?

Ans: To read NCERT exemplar Class 12 Chemistry solutions chapter 4 more effectively you need to understand the context well. Once you recognize the important points you can use markers like highlighters or underline them for quick go through later.

4. Q: Who has prepared the solutions?

Ans: The solutions are prepared by us. We are an expert team who prepare solutions by referring to many scholar books.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read More

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read More

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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