NCERT Exemplar Class 11 Chemistry Solutions Chapter 6 Thermodynamics

NCERT Exemplar Class 11 Chemistry Solutions Chapter 6: MCQ (Type 1)

Question:1

1) Thermodynamics is not concerned about.........
(i) energy changes involved in a chemical reaction.
(ii) the extent to which a chemical reaction proceeds.
(iii) the rate at which a reaction proceeds.
(iv) the feasibility of chemical reaction.
Answer:

The answer is option (iii) the rate at which a reaction proceeds.
Explanation: Thermodynamics is actually concerned with the initial and final states of a system and not with the rate at which the transformations take place. The laws can be applied to a system only when the system is in equilibrium.

Question:2

Which of the following statements is correct ?
(i) The presence of reacting species in a covered beaker is an example of open system.
(ii) There is an exchange of energy as well as matter between the system and the surroundings in a closed system.
(iii) The presence of reactants in a closed vessel made up of copper is an example of a closed system.
(iv) The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system.
Answer:

The answer is an option (iii) the presence of reactants in a closed vessel made up of copper is an example of a closed system.
Explanation: When the reactants are in a closed vessel, the matter cannot be exchanged. However, the energy may be exchanged through walls. So the reactants in a copper or a steel vessel are an example of a closed system.

Question:3

The state of gas can be described by quoting the relationship between _________ .
(i) pressure, volume, temperature
(ii) temperature, amount, pressure
(iii) amount, volume, temperature
(iv) pressure, volume, temperature, amount
Answer:

The answer is option (iv) pressure, volume, temperature, amount.
Explanation: The states of a gas can be described by the relationship between the variables such as p, V, T, and the amount of gas. These variables are called state variables. Their values are only concerned with the current state of the system.

Question:4

The volume of gas is reduced to half from its original volume. The specific heat will ________ .
(i) reduce to half
(ii) be doubled
(iii) remain constant
(iv) increase four times
Answer:

The answer is that option (iii) remains constant.
Explanation: The property of Specific heat capacity is an intensive property since it refers to the quantity of heat needed to elevate the temperature of one unit mass of a substance by one degree Celsius. It depends on the volume of the substance.

Question:5

During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is
(i)
$2C_4{H}_{10}(g)+13O_2(g)\to 8C{O}_2(g)+10H_2{O}_{(l)}{\mathrm{\Delta }}_{c}H=-2658.0kJmo{l}^{-1}$
(ii)
$C_4{H}_{10}(g)+\frac{13}{2}{O}_2(g)\to 4C{O}_2(g)+5H_{2}O(g){\mathrm{\Delta }}_{c}H=-1329.0kJmo{l}^{-1}$
(iii)
$C_4{H}_{10}(g)+\frac{13}{2}{O}_2(g)\to 4C{O}_2(g)+5H_{2}O(l){\mathrm{\Delta }}_{c}H=-2658.0kJmo{l}^{-1}$
(iv)
$C_4{H}_{10}(g)+\frac{13}{2}{O}_2(g)\to 4C{O}_2(g)+5H_{2}O(g){\mathrm{\Delta }}_{c}H=+2658.0kJmo{l}^{-1}$
Answer:

The answer is the option (iii).
Explanation: The definition of the enthalpy of combustion is when the substance undergoes combustion and the enthalpy change that happens per mole in that substance. All other reactants are assumed to be in their standard states at that time.

Question:6

${\mathrm{\Delta }}_f{U}^o$ of formation of $C{H}_4(g)$ at a certain temperature is –393 kJ mol^–1. The value of ${\mathrm{\Delta }}_f{H}^{\ominus }$ is
(i) zero
(ii) $<{\mathrm{\Delta }}_f{U}^o$
(iii) $>{\mathrm{\Delta }}_f{U}^o$
(iv) equal to ${\mathrm{\Delta }}_f{U}^o$
Answer:

The answer is option (ii) $<{\mathrm{\Delta }}_f{U}^o$
Explanation: The reaction for the formation of $C{H}_4$ is:
$C{H}_4+2O_2\to C{O}_2+2H_{2}O$
$\mathrm{\Delta }{n}_g=1-3=-2$
$\mathrm{\Delta }{H}^o=\mathrm{\Delta }{U}^o+\mathrm{\Delta }{n}_{g}RT$
$\mathrm{\Delta }{n}_g=-2$
Hence, $\mathrm{\Delta }{H}^o<{\mathrm{\Delta }}_f{U}^o$

Question:7

In an adiabatic process, no transfer of heat takes place between system and surroundings.Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.
(i) $q=0,\mathrm{\Delta }T\ne 0,w=0$
(ii) $q\ne 0,\mathrm{\Delta }T=0,w=0$
(iii) $q=0,\mathrm{\Delta }T=0,w=0$
(iv) $q=0,\mathrm{\Delta }T<0,w\ne 0$
Answer:

The answer is option (iii) $q=0,\mathrm{\Delta }T=0,w=0$
Explanation: w =0 in the situation of free expansion since the volume is also constant. Als0, q=0 considering the process is adiabatic. The first law of thermodynamics states that $\mathrm{\Delta }U=q+w=0$ which is a constant.

Question:8

The pressure-volume work for an ideal gas can be calculated by using the expression
$w=-{\int }_{v_i}^{v_f}{P}_{ex}dV$
The work can also be calculated from the pV - plot by using the area under the curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from V_i to V_f, choose the correct option.
(i) w (reversible) = w (irreversible)
(ii) w (reversible) < w (irreversible)
(iii) w (reversible) > w (irreversible)
(iv) w (reversible) = w (irreversible) + p_ex. $\mathrm{\Delta }V$
Answer:

The answer is option (ii) w (reversible) < w (irreversible).
Explanation: In the situation of irreversible compression, the area under the curve is always larger compared

Question:9

The entropy change can be calculated by using the expression $\mathrm{\Delta }S=(q_{rev}/T)$.
When water freezes in a glass beaker, choose the correct statement amongst the following :
(i) $\mathrm{\Delta }S$ (system) decreases, but $\mathrm{\Delta }S$ (surroundings) remains the same.
(ii) $\mathrm{\Delta }S$ (system) increases but $\mathrm{\Delta }S$ (surroundings) decreases.
(iii) $\mathrm{\Delta }S$ (system) decreases, but $\mathrm{\Delta }S$ (surroundings) increases.
(iv) $\mathrm{\Delta }S$ (system) decreases, and $\mathrm{\Delta }S$ (surroundings) also decreases.
Answer:

The answer is option (iii) $\mathrm{\Delta }S$ (system) decreases, but $\mathrm{\Delta }S$ (surroundings) increases.
Explanation: As we all know in the case of an exothermic process, heat is released, which leads to a decrease in the surrounding entropy and as a result, the system entropy increases. As freezing is an exothermic process, we choose this option.

Question:10

On the basis of thermochemical equations (a), (b) and (c), find out which of the algebraic relationships given in options (i) to (iv) is correct.
(a) $C(graphite)+O_2(g)\to C{O}_2(g);{\mathrm{\Delta }}_{r}H=xkJmo{l}^{-1}$
(b) $C(graphite)+\frac{1}{2}{O}_2(g)\to CO(g);{\mathrm{\Delta }}_{r}H=ykJmo{l}^{-1}$
(c) $C{O}_{(g)}+\frac{1}{2}{O}_{2(g)}\to C{O}_{2(g)};{\mathrm{\Delta }}_{r}H=zkJmo{l}^{-1}$
(i) z = x + y
(ii) x = y – z
(iii) x = y + z
(iv) y = 2z – x

Answer:

The answer is option (iii) x = y + z
x =y + z is the correct option as the Algebraic sum of y and z will give x.
Explanation: When y is subtracted from x, we get z. Explained below:
$C(graphite)+O_2(g)\to C{O}_2(g);{\mathrm{\Delta }}_{r}H=xkJmo{l}^{-1}$________(i)
$C(graphite)+\frac{1}{2}{O}_2(g)\to CO(g);{\mathrm{\Delta }}_{r}H=ykJmo{l}^{-1}$________(ii)
When subtracted, we receive:
$C{O}_{(g)}+\frac{1}{2}{O}_{2(g)}\to C{O}_{2(g)};{\mathrm{\Delta }}_{r}H=zkJmo{l}^{-1}$________(iii)
So, z= x-y, or, easily put, x=y+z. Hence, option c.

Question:11

Consider the reactions given below. On the basis of these reactions find out which of the algebraic relations given in options (i) to (iv) is correct?
(i) $C(g)+4H(g)\to C{H}_4(g);{\mathrm{\Delta }}_{r}H=xkJmo{l}^{-1}$
(ii)
$C(graphite,s)+2H_2(g)\to C{H}_4(g);{\mathrm{\Delta }}_{r}H=ykJmo{l}^{-1}$
(i) x = y
(ii) x = 2y
(iii) x >y
(iv) x< y
Answer:

The answer is option (iii).
Explanation: In the case of reaction b, the bonds are broken, but in reaction a, the bonds are not broken. However, the same bonds are formed in both cases.

Question:12

The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound
(i) is always negative
(ii) is always positive
(iii) may be positive or negative
(iv) is never negative
Answer:

The answer is option (iii) maybe positive or negative
Explanation: It can be positive or negative as the reaction could be exothermic or endothermic.

Question:13

Enthalpy of sublimation of a substance is equal to
(i) enthalpy of fusion + enthalpy of vaporization
(ii) enthalpy of fusion
(iii) enthalpy of vaporization
(iv) twice the enthalpy of vapourization.
Answer:

The answer is an option (i) enthalpy of fusion + enthalpy of vaporization
Explanation: In the case of enthalpy of sublimation, a solid converts to liquid and then to gas. In this process, the conversion to liquid requires enthalpy of fusion and the conversion to vapor requires the enthalpy of vapourisation.

Question:14

Which of the following is not correct?
(i) $\mathrm{\Delta }G$ is zero for a reversible reaction.
(ii) $\mathrm{\Delta }G$ is positive for a spontaneous reaction
(iii) $\mathrm{\Delta }G$ is negative for a spontaneous reaction
(iv) $\mathrm{\Delta }G$ is positive for a non-spontaneous reaction.
Answer:

The answer is option (ii) $\mathrm{\Delta }G$ is positive for a spontaneous reaction
Explanation: As we all know for a spontaneous reaction, $\mathrm{\Delta }G$ is always negative.

NCERT Exemplar Class 11 Chemistry Solutions chapter 6: MCQ (Type 2)

Question:15

Thermodynamics mainly deals with
(i) interrelation of various forms of energy and their transformation from one form to another.
(ii) energy changes in the processes which depend only on the initial and final states of the microscopic system containing a few molecules.
(iii) how and at what rate these energy transformations are carried out.
(iv) the system in an equilibrium state or moving from one equilibrium state to another equilibrium state.
Answer:

The answer is option (i) and (iv)
Explanation: The prerequisite conditions required for the thermodynamics laws to be applicable is for the system to be in equilibrium. Also, they are relevant to the macroscopic systems and not to the microscopic systems which have very few numbers of molecules.

Question:16

In an exothermic reaction, heat is evolved, and the system loses heat to the surroundings. For such system
(i) $q_p$ will be negative
(ii) ${\mathrm{\Delta }}_{r}H$ will be negative
(iii) $q_p$ will be positive
(iv) ${\mathrm{\Delta }}_{r}H$ will be positive.
Answer:

The answer is options (i) and (ii)
Explanation: In an exothermic reaction, the heat is released to the surroundings of the system. So,$q_p$ and ${\mathrm{\Delta }}_{r}H$ will be negative.

Question:17

The spontaneity means, having the potential to proceed without the assistance of an external agency. The processes which occur spontaneously are
(i) flow of heat from colder to warmer body.
(ii) gas in a container contracting into one comer.
(iii) gas expanding to fill the available volume.
(iv) burning carbon in oxygen to give carbon dioxide.
Answer:

The answer is options (iii) and (iv).
Explanation: Both the processes, namely, the filling of empty space by gas, as well as the burning of carbon which leads to the formation of carbon dioxide, are spontaneous processes.

Question:18

For an ideal gas, the work of reversible expansion under isothermal conditions can be calculated by using the expression $w=-nRT$ In $V_f/V_i$
A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option.
(i) Work done at 600 K is 20 times the work done at 300 K.
(ii) Work done at 300 K is twice the work done at 600 K
(iii) Work done at 600 K is twice the work done at 300 K.
(iv) $\mathrm{\Delta }U=0$ in both cases.
Answer:

The answer is (iii) and (iv).
Explanation: Now, q = -w, in the case of an isothermal reversible change
$Q=-w=nRTIn\frac{vf}{vi}$
$\frac{w(600k)}{w(300k)}=\frac{1\times R\times 600kIn\frac{10}{1}}{1\times R\times 300KIn\frac{10}{1}}=2$
In case of ideal gases, $\mathrm{\Delta }U=0$ for the expansion, which is isothermal. There is no change in the internal energy as the temperature is constant, and, hence $\mathrm{\Delta }U=0$.

Question:19

Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below:
$2Zn(s)+O_2(g)\to 2ZnO(s);\mathrm{\Delta }H=-693.8kJmo{l}^{-1}$
(i) The enthalpy of two moles $ZnO$ is less than the total enthalpy of two moles of $Zn$ and one mole of oxygen by 693.8 kJ.
(ii) The enthalpy of two moles of $ZnO$ is more than the total enthalpy of two moles of $Zn$ and one mole of oxygen by 693.8 kJ.
(iii) 8 kJ mol ^-1 energy is evolved in the reaction.
(iv) 693.8 kJ mol^-1 energy is absorbed in the reaction.
Answer:

The answers are options (i) and (iii).
Explanation:
${\mathrm{\Delta }}_{f}H=\sum _i{a}_i{\mathrm{\Delta }}_{f}H-\sum _i{b}_i{\mathrm{\Delta }}_{f}H$
The reaction depicted above is exothermic in nature, and, hence, the reactant enthalpy is more than the product enthalpy.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 6: Short Answer Type

Question:20

18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol^-1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourization for water?
Answer:

1 mol of water or 18g of water has an enthalpy of change for vaporization. Now it is given that the change of enthalpy required for 1 mole of water is 40.79 kJ/mol.
For 2 moles of water, the enthalpy change will be 2 x 40.79 = 81.58 KJ. Therefore the standard enthalpy of vaporisation of water is ${\mathrm{\Delta }}_{vap}H=40.79kJ/mol$

Question:21

One mole of acetone requires less heat to vaporise than 1 mol of water. Which of the two liquids has a higher enthalpy of vapourization?
Answer:

Water has a greater enthalpy of vaporization than acetone. Acetone molecules have a weak force of attraction between them, and hence they require less heat to break the molecular bonds.

Question:22

Standard molar enthalpy of formation, ${\mathrm{\Delta }}_f{H}^{\ominus }$ is just a special case of enthalpy of reaction, ${\mathrm{\Delta }}_r{H}^{\ominus }$ . Is the ${\mathrm{\Delta }}_r{H}^{\ominus }$ for the following reaction same as ${\mathrm{\Delta }}_f{H}^{\ominus }$ ? Give a reason for your answer.
$CaO(s)+C{O}_2(g)\to CaC{O}_3(s);{\mathrm{\Delta }}_f{H}^{\ominus }=-178.3kJmo{l}^{-1}$
Answer:

No, since CaCO₃ has been formed from other compounds and not from its constituent elements.

Question:23

The value of ${\mathrm{\Delta }}_f{H}^{\ominus }$ for $N{H}_3$ is – 91.8 kJ mol^–1. Calculate the enthalpy change for the following reaction :
$2N{H}_3(g)\to N_2(g)+3H_2(g)$
Answer:

The enthalpy change for the given reaction can be calculated as: -

$N_2(g)+3H_2(g)\to 2N{H}_3(g)$

${\mathrm{\Delta }}_{f}H=-91.8kg/mol$
The enthalpy of the reaction is +91.8 kg/mol. This happens as with the reversal of reaction, the value of${\mathrm{\Delta }}_{r}H$ also gets reversed.

Question:24

Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A→B along one route is ${\mathrm{\Delta }}_{r}H$and ${\mathrm{\Delta }}_r{H}_1$, ${\mathrm{\Delta }}_r{H}_2$,${\mathrm{\Delta }}_r{H}_3$….. represent enthalpies of intermediate reactions leading to product B. What will be the relation between ${\mathrm{\Delta }}_{r}H$ for overall reaction and ${\mathrm{\Delta }}_r{H}_1$,${\mathrm{\Delta }}_r{H}_2$ etc. for intermediate reactions.
Answer:

According to Hess’s law, ${\mathrm{\Delta }}_{r}H={\mathrm{\Delta }}_r{H}_1+{\mathrm{\Delta }}_r{H}_2+{\mathrm{\Delta }}_r{H}_3.....$
This is so because, during the reaction $A\to B$, B’s formation undergoes various intermediate reactions, with the overall value of the enthalpy being ${\mathrm{\Delta }}_{r}H$.

Question:25

The enthalpy of atomisation for the reaction $C{H}_4(g)\to C(g)+4H(g)is1665kJmo{l}^{-1}$. What is the bond energy of $C-H$ bond?
Answer:

The reaction presented in the question is
$C{H}_4(g)\to C(g)+4H(g)$
Now, ${\mathrm{\Delta }}_{a}H=1665kJ/mo{l}^{-1}$
The mean bond enthalpy of the C-H bond should be used here. For the atomization of 4 moles of C-H bonds, the value is 1665 kJ/mol-1 So, per mole energy = 1665/4 = 416.2 kJ / mol^-1

Question:26

Use the following data to calculate ${\mathrm{\Delta }}_{lattice}$ $H^{\ominus }$ for NaBr.
${\mathrm{\Delta }}_{sub}$ $H^{\ominus }$ for sodium metal = 108.4 kJ mol^–1
Ionization enthalpy of sodium = 496 kJ mol^-1
Electron gain enthalpy of bromine = – 325 kJ mol^-1
Bond dissociation enthalpy of bromine = 192 kJ mol^-1
${\mathrm{\Delta }}_f{H}^{\ominus }$ for NaBr (s) = – 360.1 kJ mol^-1
Answer:

In order to calculate the lattice enthalpy of NaBr,
(i)$Na(s)\to Na(g);{\mathrm{\Delta }}_{sub}{H}^{\circ }=108.4kJmo{l}^{-1}$
(ii) $Na\to N{a}^{+}+e^{-};{\mathrm{\Delta }}_i{H}^{\circ }=496kJmo{l}^{-1}$
(iii) $\frac{1}{2}B{r}_2\to Br,\frac{1}{2}{\mathrm{\Delta }}_{diss}{H}^{\circ }=96kJmo{l}^{-1}$
(iv) $Br+e^{-}\to B{r}^{-};\mathrm{\Delta }eg{H}^{\circ }=-325kJmo{l}^{-1}$
${\mathrm{\Delta }}_f{H}^{\circ }={\mathrm{\Delta }}_{sub}{H}^{\circ }+\frac{1}{2}{\mathrm{\Delta }}_{diss}{H}^{\circ }+{\mathrm{\Delta }}_i{H}^{\circ }+{\mathrm{\Delta }}_{eg}{H}^{\circ }+{\mathrm{\Delta }}_{latttice}{H}^{\circ }$
${\mathrm{\Delta }}_{latttice}{H}^{\circ }=-360.1-108.4-96-496+325=-735.5KJmo{l}^{-1}$

Question:27

Given that $\mathrm{\Delta }H=0$ for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
Answer:

Entropy in an isolated system increases with the increase in the disorder in the system. As diffusion is a spontaneous process, the process also will be a spontaneous one.

Question:28

Heat has a randomizing influence on a system, and the temperature is the measure of the average chaotic motion of particles in the system. Write the mathematical relation that relates these three parameters.
Answer:

Temperature and the entropy change are both inversely proportional quantities. Q and T are related to the entropy in the case of reversible reaction in the following manner:
$\mathrm{\Delta }S=q_{rev}/T$

Question:29

An increase in the enthalpy of the surroundings is equal to a decrease in the enthalpy of the system. Will the temperature of the system and surroundings be the same when they are in thermal equilibrium?
Answer:

Yes. The temperature of the surroundings and the system will be identical when we look at the condition of thermal equilibrium. The enthalpy varies in an inversely proportionate manner in the case of system and surroundings. If the enthalpy of the system increases, the enthalpy of the surroundings deceases.

Question:30

At 298 K, K_p for the reaction $N_2{O}_4(g)\rightleftharpoons 2N{O}_2(g)$ is 0.98. Predict whether the reaction is spontaneous or not.
Answer:

As per the information given in the question, K_p for the above reaction is 0.98
${\mathrm{\Delta }}_{y}G={\mathrm{\Delta }}_{y}H-T{\mathrm{\Delta }}_{y}G=-RTlnK$
For a process to be of spontaneous nature, $\mathrm{\Delta }G$ should be negative
As ${\mathrm{\Delta }}_{y}G=-RTln{K}_p$
Hence ${\mathrm{\Delta }}_{y}G$ has a negative connotation, and the reaction given is of a spontaneous nature

Question:31

A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression, as shown in the figure. What will be the value of $\mathrm{\Delta }H$ for the cycle as a whole?

Answer:

In case of a cyclic change, the value of $\mathrm{\Delta }H=0$ i.e. there is not going to be any change in the enthalpy.

Question:32

The standard molar entropy of $H_{2}O(l)$ is 70 J K^-1 mol^-1. Will the standard molar entropy of $H_{2}O$(s) be more, or less than 70 J K^-1 mol^-1?
Answer:

The standard molar entropy of H₂O (s) will be less than 70 JK^-1mol^-1. This is because entropy decreases when ice is formed from water as molecules reduce their movement and are arranged in an orderly fashion.

Question:33

Identify the state functions and path functions out of the following: enthalpy, entropy, heat, temperature, work, fand ree energy.
Answer:

Out of the options mentioned, the state functions are: Temperature, Enthalpy, Entropy, and Free energy and the path functions are Work, Heat, etc.

Question:34

The molar enthalpy of vapourization of acetone is less than that of water. Why?
Answer:

As the intermolecular forces of attraction is lower in acetone than in water, the enthalpy of vaporization of water is higher than that of acetone.

Question:35

Which quantity out of ${\mathrm{\Delta }}_{r}G$ and ${\mathrm{\Delta }}_r{G}^{⊝}$ will be zero at equilibrium?
Answer:

${\mathrm{\Delta }}_{r}G$ will always be zero.
${\mathrm{\Delta }}_r{G}^{⊝}$ is zero for K = 1 because ${\mathrm{\Delta }}_r{G}^{⊝}$= – RT lnK, ${\mathrm{\Delta }}_r{G}^{⊝}$ will be non zero for other values of K.

Question:36

Predict the change in internal energy for an isolated system at constant volume.
Answer:

There is no change in internal energy for an isolated system at a constant volume if there is no transfer of heat as work which are the conditions of an isolated system. Hence, $\mathrm{\Delta }H=0$.

Question:37

Although heat is a path function heat absorbed by the system under certain specific conditions is independent of the path. What are those conditions? Explain.
Answer:

The conditions in which heat becomes path-independent are:
a) when the volume remains the same
$\mathrm{\Delta }V=0,so\mathrm{\Delta }U=q+p\mathrm{\Delta }V=q$
As $\mathrm{\Delta }U$ is a state function, q is also a state function.
b) when the pressure remains the same
$\mathrm{\Delta }U=q-p\mathrm{\Delta }V$
So, $q=p\mathrm{\Delta }V+\mathrm{\Delta }U=\mathrm{\Delta }H$ which is a state function
So, at a constant volume and at constant pressure heat change is a state function because it is equal to a change in internal energy and a change in enthalpy respectively which are state functions.

Question:38

The expansion of a gas in a vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 liter of an ideal gas expands isothermally into a vacuum until its total volume is 5 liter?
Answer:

A free expansion is known as the expansion of gas in a vacuum. The equation can be written as
$(-w)=p(V_2-V_1)=0\times (5-1)=0$
q =0, when we consider isothermal expansion.
Referring to the first law of thermodynamics we have
$q=\mathrm{\Delta }U+(-w)$
$0=\mathrm{\Delta }U+0$
Hence, $\mathrm{\Delta }U=0$

Question:39

Heat capacity $(C_p)$ is an extensive property but specific heat (c) is an intensive property. What will be the relation between $(C_p)$ and c for 1 mol of water?
Answer:

The quantity of heat needed to raise the temperature of one mole of the substance by one degree Celsius is known as its heat capacity.
For the molar heat capacity calculation, we multiply 18 and the value of the specific heat
$C_p=18\times c$
We know that, C (specific heat) = $4.18J{g}^{-1}{K}^{-1}$ Heat capacity,
Therefore, $C_p=18\times 4.18J{K}^1=75.24J{K}^{-1}$

Question:40

The difference between $C_{p}and{C}_v$ can be derived using the empirical relation $H=U+pV$. Calculate the difference between $C_{p}and{C}_v$ for 10 moles of an ideal gas.
Answer:

In the case of an ideal gas, the difference between $C_{p}and{C}_v$ is nR, where where n represents the number of moles and R is the universal gas constant.
Therefore, for 10 moles of an ideal gas-
$C_p-C_v=10R$
$C_p-C_v=10\times 4.18J$
= 41.84 J

Question:41

If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be the molar enthalpy change? Give the significance of the sign also.
Answer:

We know that, 1 mole of carbon = 12g
To calculate the molar enthalpy, change of graphite, we need to multiply enthalpy change for 1g carbon along with its molar mass. Which as per the information provided in the question can be calculated as,
$-20.7\times 12=-248.4kJmo{l}^{-1}$ (negative sign depicts exothermic reaction as there is heath involved)

Question:42

The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus the amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction?
$H_2(g)+B{r}_2(g)\to 2HBr(g)$
Given that Bond energy of $H_2$, $B{r}_2$ and HBr is 435 kJ mol^-1 , 192 kJ mol^-1 and 368 kJ mol^–1 respectively.
Answer:

We know that to calculate Enthalpy change, the formula is
= (Bond energy of H-H bond + Br-Br bond) – (2 × bond energy of H-Br)
In our case, we can calculate enthalpy change by substituting the values.
= $(435+192)kJmo{l}^{-1}-(2\times 368)kJmo{l}^{-1}$
= -109 kJ mol^-1

Question:43

The enthalpy of vapourization of $CC{l}_4$ is 30.5 kJ mol^–1. Calculate the heat required for the vapourization of 284 g of $CC{l}_4$ at constant pressure. (Molar mass of $CC{l}_4$ = 154 g mol^–1).
Answer:

As per the information provided in the question, for one mole of $CC{l}_4(154g)$, the heat of vaporization required is 30.5 kJ/mol.
Hence for the vaporisation of 284 g of $CC{l}_4$, we require:
$=\frac{284g}{154gmo{l}^{-1}}\times 30.5kJmo{l}^{-1}$
$=56.2kJ$

Question:44

The enthalpy of reaction for the reaction:
$2H_2(g)+O_2(g)\to 2H_{2}O(l)$ is ${\mathrm{\Delta }}_r{H}^{⊝}=-572kJmo{l}^{-1}$. What will be the standard enthalpy of the formation of $H_{2}O(l)$?
Answer:

Standard molar enthalpy of formation is the enthalpy change for the formation of one mole of a compound from its most stable states or reference states. As per the given information in the question, the standard enthalpy for the given equation is – 572 kJ mol^–1
Now the enthalpy of formation for $H_{2}O$ will be half the enthalpy of the value in the given equation. So, now we can calculate that:
${\mathrm{\Delta }}_{f}H=\frac{1}{2}\times {\mathrm{\Delta }}_{y}H=-\frac{52}{2}=-286kJmo{l}^{-1}$

Question:45

What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, $p_{ext}$ in a single step as shown in Fig. 6.2. Explain graphically.

Answer:

Assumption: The cylinder is filled with one-mole gas, and the piston is frictionless. Let the pressure of the gas inside be p and the volume of gas be $V_I$.
A piston is moved towards the inside to make the external pressure $(P_{ext})$ equal to p. Now, let us assume that this change takes place in a single step, hence, V is the final volume. The work done by the piston is depicted in the graph shown below by shading the area.
$P_{ext}\mathrm{\Delta }V=A{V}_I(orB{V}_{II})\times (V_I-V_{II})$

Question:46

How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps?
Answer:

When a process can be reversed by bringing an extremely small change in it, we call it a reversible process. The pressure-volume graph can be used to calculate the work done. The pressure is not constant, and changes in infinitesimal amounts as compression happens from initial volume V_i to the final volume V_f. The below graph depicts the work done with the shaded area.

Question:47

Represent the potential energy/enthalpy change in the following processes graphically.
(a) Throwing a stone from the ground to roof.
(b)
$\frac{1}{2}{H}_2(g)+\frac{1}{2}C{l}_2(g)\rightleftharpoons HCl(g){\mathrm{\Delta }}_r{H}^{\ominus }=-92.32kJmo{l}^{-1}$
In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?
Answer:

(a) Throwing a stone from the ground to the roof

b) the reaction involved is a process where the energy decreases after the reaction. It can be represented as:

In process b), potential energy/enthalpy change is contributing factor to the spontaneity.

Question:48

Enthalpy diagram for a particular reaction is given in Fig. 6.3. Is it possible to decide the spontaneity of a reaction from given diagram? Explain.

Answer:

No, for the state of spontaneity, the enthalpy change is not the only criteria. Entropy also needs to be taken into account here.

Question:49

1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in figure. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K.

Answer:

We can conclude from the figure that this change is a reversible change.
Now,
$W=-2.303nRTlogp1/p2$

$=-2.303\times 1\times 8.314\times 298\times log2$

$=-2.303\times 8.314\times 298\times 0.3010J$

$=-1717.46J$

Question:50

An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case?
(Given that 1 L bar = 100 J)
Answer:

We know that the amount of work done =$-p_{ext}\mathrm{\Delta }V$
On substituting the values in the formula, we get,
$-2bar\times (50-10)L=-80Lbar$
According to the described problem,$1LBar=100J$

Therefore, $-80Lbar=(-80\times 100)=-8000J$

$=-8kJ,$ which is the amount of work done
The significance of the negative sign states that the work is done on the surroundings of the system. In the case of reversible expansion, the work done will be more.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 6: Matching Type

Question:51

Match the following:

Answer:

(i) - (e); (ii)- (d); (iii)- (f); (iv)- (a); (v)- (g), (k), (l); (vi)- (b); (vii)- (c); (viii)- (j); (ix) (h); (x)- (i); (xi)- (a), (l), (m); (xii)- (g), (k)

Question:52

Match the following processes with entropy change:

Answer:

(i) - b
(ii) - c
(iii) -a
Explanation:

Question:53

Match the following :

Answer:

(i) - c
(ii) - a
(iii) -b
Explanation:

Question:54

Match the following :

Answer:

(i) - b, d
(ii) - b
(iii) -c
(iv) -a
Explanation:

NCERT Exemplar Class 11 Chemistry Solutions Chapter 6: Assertion and Reason Type

Question:55

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Combustion of all organic compounds is an exothermic reaction.
Reason (R) : The enthalpies of all elements in their standard state are zero.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.
Answer:

The answer is option (ii) Both A and R are true but R is not the correct explanation of A.
Explanation: the reactant enthalpies are always greater than the product enthalpies when it comes to the combustion reaction.

Question:56

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : Spontaneous process is an irreversible process and may be reversed by some external agency.
Reason (R) : Decrease in enthalpy is a contributory factor for spontaneity.
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Answer:

The answer is option (ii) Both A and R are true, but R is not the correct explanation of A.
Explanation: In the case of a spontaneous process, the randomness should be positive, and ΔU should be negative.

Question:57

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): A liquid crystallizes into a solid and is accompanied by a decrease in entropy.
Reason (R): In crystals, molecules organize in an ordered manner.
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Answer:

The answer is option (i) Both A and R are true, and R is the correct explanation of A.
Explanation: As the liquid starts to crystallize the randomness in the molecules decreases as they get arranged in an orderly fashion. Hence, the entropy decreases.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 6: Long Answer Type

Question:58

Derive the relationship between $\mathrm{\Delta }H$ and $\mathrm{\Delta }U$ for an ideal gas. Explain each term involved in the equation.
Answer:

Now, the heat absorbed at constant volume is equal to the change in the internal Energy.
$\mathrm{\Delta }U=q_p-p\mathrm{\Delta }V$
qp is the heat absorbed.
p $\mathrm{\Delta }V$ is the expansion work done by the system
the equation can be represented as $U_2-U_1=p(V_2-V_1)$
so, we get,
$q_p=(U_2+P{v}_2)-(U_1+P{v}_1)-----------------(1)$
now we take into consideration, the enthalpy
$H=U+pV--------------------------(2)$
Hence, we can rewrite the first equation as,
$q_p=H_2-H_1=\mathrm{\Delta }H$
now, in the case of the state functions.
H is a state function because it depends on U, p, and V, all of which are state functions.
So, $\mathrm{\Delta }H$, as well as qp, is path independent.
We can rewrite equation 2 in case of finite changes as,
$\mathrm{\Delta }H=\mathrm{\Delta }U+p\mathrm{\Delta }V---------------------(3)$
We measure changes in enthalpy, when heat is being absorbed at a constant pressure.
ΔV = 0, for constant volume. So, equation 3 becomes ΔH = ΔU = qv
Let Va be the total volume of gaseous reactants and Vb be the total volume of gaseous products. Let nA be the number of moles of gas reactants and nB be the number of moles for the gaseous products. The ideal gas law is:
pV_a= n_A R T
pV_b= n_B R T
now,
pV_b - pV_a = (n_B - n_A) RT
hence,
$p\mathrm{\Delta }V=\mathrm{\Delta }{n}_{g}RT-------------------(4)$
now, by using equation 3 and equation 4,
we can write
ΔH = ΔU + Δn_g RT

Question:59

Extensive properties depend on the quantity of matter, but intensive properties do not. Explain whether the following properties are extensive or intensive.
Mass, internal energy, pressure, heat capacity, molar heat capacity, density, mole fraction, specific heat, temperature and molarity.
Answer:

Extensive properties Mass, internal energy, heat capacity.
Intensive properties Pressure, molar heat capacity, density, mole fraction, specific heat, temperature and molarity.
Ratio of two extensive properties is always intensive.
Extensive/Extensive = Intensive
So, mole fraction and molarity are intensive properties.

Question:60

The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compound present in its gaseous state, dissociates into its ions. It is impossible to determine it directly by experiment. Suggest and explain an indirect method to measure lattice enthalpy of $NaCl(s).$
Answer:

When one mole of an ionic compound dissociates into its ions, the enthalpy change related to this process is called lattice enthalpy.
$N{a}^{+}C{l}^{-}\to N{a}^{+}+C{l}^{-}$
$\mathrm{\Delta }{H}_{lattice}=788kJ/mol$
Lattice enthalpy for NaCl is calculated as follows:

1) Sublimation of sodium metal
$Na(s)\to Na(g)$
2) Ionization enthalpy: ionization of sodium atoms
$Na(g)\to N{a}^{+}(g)+e^{-1}(g)$
3) Bond dissociation enthalpy: dissociation of chlorine
$\frac{1}{2}C{l}_2(g)\to C{l}^{-}(g)$
4) Electron gain enthalpy: electron gain by chlorine
$C{l}^{-}(g)+e^{-1}(g)\to Cl(g)$
$5)N{a}^{+}(g)+C{l}^{-}(g)\to N{a}^{+}C{l}^{-}(s)$
These steps collectively form the BORN-HABER Cycle, wherein the sum total of enthalpy changes around a cycle is equal to 0.

Question:61

$\mathrm{\Delta }G$ is net energy available to do useful work and is thus a measure of “free energy”. Show mathematically that $\mathrm{\Delta }G$ is a measure of free energy. Find the unit of $\mathrm{\Delta }G$. If a reaction has positive enthalpy change and positive entropy change, under what condition will the reaction be spontaneous?
Answer:

We are aware of the equation that,
$\mathrm{\Delta }{S}_{tot}=\mathrm{\Delta }{S}_{sys}+\mathrm{\Delta }{S}_{surr}$
The temperature of the system and surroundings will be equal if both are in equilibrium. Also, the enthalpy changes are inversely proportional in both of them.
We can represent the entropy changes in the surroundings as:
$\mathrm{\Delta }{S}_{Surr}=\mathrm{\Delta }{H}_{surr}/T=-(\mathrm{\Delta }{H}_{sys}/T)$
$\mathrm{\Delta }{S}_{tot}=\mathrm{\Delta }{S}_{sys}-(\mathrm{\Delta }{H}_{sys}/T)$
Hence, we can rewrite this as:
$T\mathrm{\Delta }{S}_{tot}=T\mathrm{\Delta }{S}_{sys}-\mathrm{\Delta }{H}_{sys}$
In case of a spontaneous process:
$-(\mathrm{\Delta }{H}_{sys}-T\mathrm{\Delta }{S}_{sys})>0,since\mathrm{\Delta }{S}_{tot}>0$
We can write the above equation as:
$-\mathrm{\Delta }G>0$
$\mathrm{\Delta }G=\mathrm{\Delta }H-T\mathrm{\Delta }S>0$
ΔG is the measure of free energy as it is the net available energy for performing some useful work.
Criteria for spontaneity:
$\mathrm{\Delta }G<0$: spontaneous process
$\mathrm{\Delta }G>0$: non-spontaneous process

Question:62

Graphically show the total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from $(p_i,V_i)$ to $(p_f,V_f)$. With the help of a pV plot compare the work done in the above case with that carried out against a constant external pressure $P_f$.
Answer:

Let us assume that a cylinder has 1-mole gas without any weight or friction having area of cross-section A. Hence, the total volume is depicted by V_i and the initial pressure is assumed to be p.
Let us assume that pext is the external pressure.
Now, if p_ext> p, the piston moves down until p_ext = p,
Hence, the final volume is V_f and the distance moved by piston be ?l
Thus, ?V = ?l × A (eq-1)
?V = V_f - Vi
We know that, force = pressure ×area
Therefore, F = p_ext× A (eq-2)
If w is work done on the system
W = force × displacement
= p_ext × A ×?l
From eq-1
W = p_ext×(-?V)
W = -p_ext?V
W = -p_ext( vf – vi)
If v_f> v_i work is done by the system and w is negative. If vf< v_i work is on the system and w is positive