NCERT Exemplar Class 11 Chemistry Solutions Chapter 7 Equilibrium 2021

Q: Does NCERT exemplar class 11 Chemistry solutions chapter 7 carry reasonable mark distribution in the exams?

Though the individual topic and chapter of equilibrium altogether might not garner the highest number of marks in the final examination it is significant for the laying down the basic concept foundation for many more complex chapters and topics to come so it essential to be well versed with this chapter for ease in the long run

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Generally speaking, the questions asked from this topic are frequently observed to be practical numerical based where we have to apply the principles and rules of equilibrium.

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Yes, the solutions are largely beneficial for students who desire to extend their practical knowledge by solving problems relating to the topic which are followed by specifically designed solutions which are easily understandable and suggest the best approach to solve the question.

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 7 Equilibrium

Edited By Sumit Saini | Updated on Sep 10, 2022 05:24 PM IST

NCERT exemplar Class 11 Chemistry solutions chapter 7 is curated by experts who have a clear understanding of chemistry. Making the optimum use of NCERT exemplar Class 11 Chemistry chapter 7 solutions allows the students to extend the horizon of their problem-solving skills by putting their theoretical knowledge to use in solving the provided questions and seek assistance from the answers whenever they are struck. NCERT exemplar class 11 Chemistry solutions chapter 7 revolves solely upon the introducing and strengthening the concept of chemical equilibrium and the principal and rules instated regarding the same which are used while balancing chemical equations.

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: MCQ (Type 1)
NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: MCQ (Type 2)
NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: Short Answer Type
NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: Matching Type
NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: Assertion and Reason Type
NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: Long Answer Type
Topics and Subtopics in NCERT Exemplar Class 11 Chemistry Solutions Chapter 7 Equilibrium
What the students will Learn from NCERT Exemplar Class 11 Chemistry Solutions Chapter 7 Equilibrium?
NCERT Exemplar Class 11 Chemistry Solutions Chapter-Wise
Important Topics To Cover From NCERT Exemplar Class 11 Chemistry Solutions Chapter 7 Equilibrium

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: MCQ (Type 1)

Question:1

We know that the relationship between $K_{c}$ and $K_{p}$ is
$K_{p}$ $= K_{c}(RT)^{\Delta n}$
What would be the value of $\Delta n$ for the reaction $NH_{4}Cl(s) \leftrightharpoons NH_{3}(g) + HCl(g)$
(i) 1
(ii) 0.5
(iii) 1.5
(iv) 2

Answer:

The answer is the option (iv) 2
Explanation: Because,
$\Delta n$ = Number of moles of product - Number of moles of reactant.
$=2-0$
$=2$

Question:2

For the reaction $H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g)$ the standard free energy is $\Delta G^{\circleddash }>0.$ The equilibrium constant (K) would be,
(i) K = 0
(ii) K > l
(iii) K = 1
(iv) K < 1
Answer:

The answer is the option (iv) K < 1
Explanation: As $\Delta G^{\Theta }=-RT$ lnK and so if $\Delta G^{\Theta }>0$ , then $-\Delta G^{\Theta }/RT$ is negative, which means K<1. This certainly implies that a non-spontaneous reaction or a reaction proceeds in the forward direction to such a small degree that only a very small quantity of product is formed.

Question:3

Which of the following is not a general characteristic of equilibria involving physical processes?
(i) Equilibrium is possible only in a closed system at a given temperature.
(ii) All measurable properties of the system remain constant.
(iii) All the physical processes stop at equilibrium.
(iv) The opposing processes occur at the same rate, and there is a dynamic but stable condition.
Answer:

The answer is the option (iii) All the physical processes stop at equilibrium.
Explanation: What happens at equilibrium is that both the processes occur at the same rate, and thus there is a dynamic but a pretty much stable condition. Thus, no further physical processes occur.

Question:4

$PCl_{5}$ , $PCl_{3}$ and $Cl_{2}$ are at equilibrium at 500 K in a closed container and their concentrations are $0.8\times 10^{-3}\; mol\; L^{-1}$ , $1.2 \times 10^{-3} mol L^{-1}$ and $1.2 \times 10^{-3} mol L^{-1}$ respectively. The value of $Kc$ for the reaction PCl₅(g) $\rightleftharpoons$ PCl₃(g) + Cl₂ (g) will be
(i) $1.8\times 10^{3}mol\; L^{-1}$
(ii) $1.8\times 10^{-3}$ mol L^-1
(iii) $1.8\times 10^{-3}\; Lmol^{-1}$
(iv) $0.55\times 10^{4}$

Answer:

The answer is the option (ii) $1.8\times 10^{-3}$ mol L^-1
Explanation :
$K_{c}=\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}=\frac{1.2\times 10^{-3}\times 1.2\times 10^{-3}}{0.8\times 10^{-3}}$
$=1.8\times 10^{-3}\; mol\; L^{-1}$

Question:5

Which of the following statements is incorrect?
(i) In equilibrium mixture of ice and water kept in perfectly insulated flask mass of ice and water does not change with time.
(ii) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate.
(iii) On addition of a catalyst the equilibrium constant value is not affected,
(iv) Equilibrium constant for a reaction with negative ΔH value decreases as the temperature increases.
Answer:

Ans. (ii) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate.
Explanation: Oxalic Acid which is $H_{2}C_{2}O_{4}$ reacts with the $Fe^{3+}$ ions forming a stable complexion that is $[Fe(C_{2}O_{4})_{3}]^{3-}$ thereby decreasing the concentration of free $Fe^{3+}$ ions and because of which concentration of $[Fe(SCN)]^{2+}$ decreases and thus the intensity of red colour decreases.

Question:6

When hydrochloric acid is added to cobalt nitrate solution at room temperature, the following reaction takes place, and the reaction mixture becomes blue. On cooling the mixture, it becomes pink. On the basis of this information mark the correct answer.
$[Co(H_{2}O)_{6}]^{3+} (aq) + 4Cl^{-}(aq) \rightleftharpoons [CoCl_{4}]^{2-}(aq) + 6H_{2}O(l)$
(i) ΔH > 0 for the reaction
(ii) ΔH < 0 for the reaction
(iii) ΔH = 0 for the reaction
(iv) The sign of ΔH cannot be predicted on the basis of this information.
Answer:

(i) $\Delta H > 0$ for the reaction
Explanation: On the cooling of the mixture, the reaction tends to move towards the backward direction, which is an endothermic reaction, and, hence, $\Delta H > 0$ .

Question:7

The pH of neutral water at 25°C is 7.0. As the temperature increases, ionisation of water increases, however, the concentration of H⁺ ions and OH^-ions are equal. What will be the pH of pure water at 60°C?
(i) Equal to 7.0
(ii) Greater than 7.0
(iii) Less than 7.0
(iv) Equal to zero
Answer:

The answer is the option (iii) Less than 7.0
Explanation: Kw increases $[H+ ][OH- ]>10^{-14}$
$As [H+ ]=[OH- ]$
$or [H+ ]^{2} =10^{-14}$
$or [H+ ]>10^{-7} M$
$pH< 7$

Question:8

The ionisation constant of an acid, K_a, is the measure of the strength of an acid. The K_a values of acetic acid, hypochlorous acid and formic acid are $1.74\times 10^{-5}, 3.0\times 10^{-8}$ and $1.8\times 10^{-4}$ respectively. Which of the following orders of pH of 0.1 mol dm^-3 solutions of these acids is correct?
(i) acetic acid > hypochlorous acid > formic acid
(ii) hypochlorous acid > acetic acid > formic acid
(iii) formic acid > hypochlorous acid > acetic acid
(iv) formic acid > acetic acid > hypochlorous acid
Answer:

The answer is the option (iv) formic acid > acetic acid > hypochlorous acid
Explanation: Higher the value of Ka higher will be the acidic strength.
$1.8\times 10^{-4} > 1.74\times 10^{-5} > 3.0\times 10^{-8}$

Question:9

$K_{a1},$ , $K_{a2},$ $K_{a3}$ are the respective ionisation constants for the following reactions.
$H_{2}S \rightleftharpoons H^{+} + HS^{-}$
$HS^{-} \rightleftharpoons H^{+} + S^{2-}$
$H_{2}S \rightleftharpoons 2H^{+} + S^{2-}$
The correct relationship between $K_{a1},$ , $K_{a2},$ $K_{a3}$ is
(i) $K_{a3} = K_{a1} \times K_{a2}$
(ii) $K_{a3} = K_{a1} + K_{a2}$
(iii) $K_{a3} = K_{a1} - K_{a2}$
(iv) $K_{a3} = \frac{K_{a1} }{K_{a2}}$
Answer:

The answer is the option (i) $K_{a3} = K_{a1} \times K_{a2}$
Explanation:
$K_{a1}=\frac{[H^{+}][HS^{-}]}{[H_{2}S]}.....(1)$
$K_{a2}=\frac{[H^{+}]+[S^{2-}]}{[HS]}.....(2)$
On adding equation (i.e. multiplication of equilibrium constants) 1 and 2, we get,
$K_{a3}=\frac{[H^{+}]^{2}[S^{2-}]}{[H_{2}S]}$
Which proves that for a dibasic acid $K_{a3} = K_{a1} \times K_{a2}$

Question:10

The acidity of BF₃ can be explained on the basis of which of the following concepts?
(i) Arrhenius concept
(ii) Bronsted Lowry concept
(iii) Lewis concept
(iv) Bronsted Lowry as well as Lewis concept.
Answer:

The answer is the option (iii) Lewis concept
Explanation: All Lewis Acids can accept a pair of electrons and hence they can be called be electron-deficient species and so BF3 is a Lewis acid.

Question:11

Which of the following will produce a buffer solution when mixed in equal volumes?
(i) $0.1 mol\; dm^{-3} NH_{4}OH \; and\; 0.1\; mol \; dm^{-3} HCl$
(ii) $0.05 \; mol\; dm^{-3} NH_{4}OH \; and\; 0.1\; mol \; dm^{-3} HCl$
(iii) $0.1 \; mol\; dm^{-3} NH_{4}OH \; and\; 0.05\; mol \; dm^{-3} HCl$
(iv) $0.1 \; mol\; dm^{-3} \;CH_{3}COONa \; and\; 0.1\; mol \; dm^{-3} NaOH$
Answer:

The answer is the option (iii) $0.1 \; mol\; dm^{-3} NH_{4}OH \; and\; 0.05\; mol \; dm^{-3} HCl$
Explanation: A mixture of ammonium chloride and ammonium hydroxide acts as a buffer.

Question:12

In which of the following solvents is silver chloride most soluble?
(i) $0.1\; mol\; dm^{-3}\; AgNO3\; solution$
(ii) $0.1 \; mol \; dm^{-3} HCl \: solution$
(iii) $H_{2}O$
(iv) Aqueous ammonia
Answer:

The answer is the option (iv) Aqueous ammonia
Explanation: Aqueous ammonia absorbs chloride ions, and thus the equilibria will have to shift in the forward direction and thus the solubility of silver chloride increases.

Question:13

What will be the value of pH of 0.01 mol dm^-3 $CH_{3}COOH(K_{a}=1.74\times 10^{-5})$
(i) 3.4
(ii) 3.6
(iii) 3.9
(iv) 3.0
Answer:

The answer is the option (i) 3.4
Explanation: We know that the ionization of $CH_{3}COOH$ occurs as follows: -
$CH_{3}COOH+H_{2}O\leftrightharpoons H_{3}O^{+}+CH_{3}COO^{-}$
Now, we know that the initial concentration: 0.01
We also know that the euilibrium concentration: 0.01 – x x x
Now, the ionization constant $K_{a}=\frac{[H_{3}O^{+}][CH^{3}COO^{-}]}{[CH_{3}COOH]}=\frac{x^{2}}{0.01-x}$
$As, x>>0.01 \; Therefore, 0.01 - \times \sim 0.01 .$
$X^{2}=1.74\times 10^{-5}\times 0.01$
$X = 4.2 \times 10^{-4}$
Which can also be written as, $pH = -log (4.2 \times 10^{-4} ) = 3.4$

Question:14

K_a for $CH_{3}COOH$ is $1.8\times 10^{-5}$ and $K_{b}$ , for $NH_{4}OH$ is $1.8\times 10^{-5}$ . The pH of ammonium acetate will be
(i) 7.005
(ii) 4.75
(iii) 7.0
(iv) Between 6 and 7
Answer:

The answer is the option (iii) 7.0
Explanation: The formation of ammonium acetate is formed by a weak base $NH_{4}OH \; and \; CH_{3}COOH.$
We know that in case of salts with weak acids,

$pH = \frac{1}{2} [pKw + pKa - pKb ]$

In this case,

$= \frac{1}{2} [14 - log (1.8 \times 10-5) + log (1.8 \times 10-5 )]$

$=7$

Question:15

Which of the following options will be correct for the stage of half completion of the reaction $A\rightleftharpoons B$ ?
(i) $\Delta G^{\Theta } =0$
(ii) $\Delta G^{\Theta } >0$
(iii) $\Delta G^{\Theta } <0$
(iv) $\Delta G^{\Theta } =-RT\; In\; 2$
Answer:

The answer is the option (i) $\Delta G^{\Theta } =0$
Explanation: At the stage of half completion of reaction the two concentrations rae equal, [A] = [B]. Therefore, K = 1.

Question:16

On increasing the pressure, in which direction will the gas-phase reaction proceed to re-establish equilibrium, is predicted by applying Le Chatelier’s principle. Consider the reaction.
$N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)$
Which of the following is correct, if the total pressure at which the equilibrium is established, is increased without changing the temperature?
(i) K will remain the same
(ii) K will decrease
(iii) K will increase
(iv) K will increase initially and decrease when pressure is very high
Answer:

The answer is the option (i) K will remain the same
Explanation: If the temperature is not changed, equilibrium constant K will remain the same.

Question:17

What will be the correct order of vapour pressure of water, acetone and ether at 30°C? Given that among these compounds, water has a maximum boiling point and ether has a minimum boiling point?
(i) Water < ether < acetone
(ii) Water < acetone < ether
(iii) Ether < acetone < water
(iv) Acetone < ether < water
Answer:

The answer is the option (ii) Water < acetone < ether
Explanation: More will be the boiling point, and less will be the vapour pressure.

Question:18

At 500 K, the equilibrium constant, K_c, for the following reaction is 5.
$\frac{1}{2} H_{2} (g) + \frac{1}{2} I_{2} (g) \rightleftharpoons HI (g)$
What would be the equilibrium constant $K_{c}$ for the reaction
$2HI (g)\rightleftharpoons H_{2} (g) + I_{2} (g)$
(i) 0.04
(ii) 0.4
(iii) 25
(iv) 2.5
Answer:

The answer is the option (i) 0.04
Explanation: If the given equation is multiplied by 2, the equilibrium constant for the new equation will be squared, and on reversing the reaction the value of the equilibrium constant is reciprocated. Thus
$K=5_{2}=25$
For the required reaction equation $K=\frac{1}{25}=0.04$

Question:19

In which of the following reactions, the equilibrium remains unaffected on the addition of a small amount of argon at constant volume?
(i) $H_{2}(g)+I_{2}(g)\leftrightharpoons 2HI(g)$
(ii) $PCl_{5}(g)\leftrightharpoons PCl_{3}(g)+ Cl_{2}(g)$
(iii) $N_{2}(g)+3H_{2}(g)\leftrightharpoons 2NH_{3}(g)$
(iv) The equilibrium will remain unaffected in all three cases.
Answer:

The answer is the option (iv) The equilibrium will remain unaffected in all three cases.
Explanation: At constant volume, equilibrium remains unaffected even on the addition of inert gas.
In each of the other cases,
(i) $H_{2}(g)+I_{2}(g)\leftrightharpoons 2HI(g)$
(ii) $PCl_{5}(g)\leftrightharpoons PCl_{3}(g)+ Cl_{2}(g)$
(iii) $N_{2}(g)+3H_{2}(g)\leftrightharpoons 2NH_{3}(g)$ the Equilibrium will remain unaffected on addition of Argon which is an inert gas.
Thus, the correct answer has to be the option (iv).

NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: MCQ (Type 2)

Question:20

For the reaction $N_{2}O_{4}(g)\leftrightharpoons 2NO_{2}(g),$ , the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is correct?
(i) The reaction is endothermic.
(ii) The reaction is exothermic.
(iii) If $NO_{2}$ (g) and $N_{2}O_{4}(g)$ are mixed at 400 K at partial pressures, 20 bar and 2 bar respectively, more $N_{2}O_{4}(g)$ will be formed.
(iv) The entropy of the system increases.
Answer:

The answer is the option (i), (iii) and (iv)
Explanation: K increases with the increase in temperature, and when K increases, the reaction must be endothermic. Thus, the number of molecules increases, and there is an increase in entropy.

Question:21

At a particular temperature and atmospheric pressure, the solid and liquid phases of a pure substance can exist in equilibrium. Which of the following term defines this temperature?
(i) Normal melting point
(ii) Equilibrium temperature
(iii) Boiling point
(iv) Freezing point
Answer:

The answer is the option (i) and (iv)
Explanation: Ice and water are in equilibrium only at a certain temperature and pressure. For any pure substance at atmospheric pressure, the temperature at which the solid and liquid phases are at equilibrium is called the normal melting point or normal freezing point of the substance.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: Short Answer Type

Question:22

The ionisation of hydrochloric acid in water is given below:
$HCl(aq) + H_{2}O (l )\rightleftharpoons H_{3}O^{+} (aq) + Cl^{-} (aq)$
Label two conjugate acid-base pairs in this ionisation
Answer:

The two conjugate acid-base pairs in the reaction for the ionisation of Hydrochloric acid in water as per the given equation are
(1) $(HCl-Cl^{-})$ – Herein, $Cl^{-}$ is the conjugate base and HCl is the conjugate acid.
(2) $(H_{2}O-H_{3}O^{+})$ – Herein $H_{2}O$ is the conjugate base and $H_{3}O^{+}$ is the conjugate acid.
$HCl(aq)+H_{2}O\leftrightharpoons H_{3}O^{+}(aq)+Cl^{-}(aq)$

Question:23

The aqueous solution of sugar does not conduct electricity. However, when sodium chloride is added to water, it conducts electricity. How will you explain this statement on the basis of ionisation, and how is it affected by the concentration of sodium chloride?
Answer:

The conductance of any substance depends on the ions present in the solution since sodium chloride (NaCl) is an ionic compound which ionises completely in solution (NaCl has a dissociation rate of 100% i.e. it ionizes completely in water) forming an aqueous solution and hence the solution conducts electricity. Sodium Chloride ionizes into sodium ion (Na⁺) and chloride ion (Cl^-) in water, which conducts electricity.
Hence, higher would be the number of ions of Socium Chloride in the water, higher would be the conductivity.

Question:24

$BF_{3}$ does not have proton but still acts as an acid and reacts with $NH_{3}$ . Why is it so? What type of bond is formed between the two?
Answer:

$BF_{3}$ does not have a proton but still acts as an acid because it is an electron pair acceptor. This can be represented by
H₃N : → BF₃
Coordinate bond is formed between BF3 and NH3, and here Nitrogen acts as lone pair donator in this combination.

Question:25

Ionisation constant of a weak base MOH is given by the expression
$K_{b}=\frac{[M^{+}][OH^{-}]}{[MOH]}$
Values of ionization constant of some weak bases at a particular temperature are given below:

Base	Dimethylamine	Urea	Pyridine	Ammonia
	$5.4\times 10^{-4}$	$1.3\times 10^{-14}$	$1.77\times 10^{-9}$	$1.77\times 10^{-5}$

Arrange the bases in decreasing order of the extent of their ionization at equilibrium. Which of the above base is the strongest?

Answer:

As per the information available from the question, pK_b values for Di-methylamine, ammonia, pyridine and urea are respectively 3.29, 4.752, 8.752, and 13.8861.
We know that, Higher is the value of Kb stronger will be the base. When we arrange then in decreasing order of extent of their ionization at equilibrium, they can be written as,
$5.4\times 10^{-4} > 1.77\times 10^{-5} > 1.77\times 10^{-9} > 1.3\times 10^{-14}$
So, the answer is Dimethylamine > Ammonia > Pyridine > Urea

Question:26

Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases?
$OH^{-}, RO^{-}, CH_{3}COO^{-} , Cl^{-}$

Answer:

$\begin{aligned} &\text {Conjugate acids of given bases are } \mathrm{H}_{2} \mathrm{O}, \mathrm{ROH}, \mathrm{CH}_{3} \mathrm{COOH}, \mathrm{HCl} \text { . }\\\\ &\text {Their acidic strength is in the order }\\ &\mathrm{HCl}>\mathrm{CH}_{3} \mathrm{COOH}>\mathrm{H}_{2} \mathrm{O}>\mathrm{ROH} \\ \\ &\text {Hence, basic strength is in the order : } \mathrm{RO}^{-}>\mathrm{OH}^{-}>\\ &\mathrm{CH}_{3} \mathrm{COO}^{-}>\mathrm{Cl}^{-} \end{aligned}$

Question:26

Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases?
$OH^{-}, RO^{-}, ch_{3}coo^{-} , cl^{-}$
Answer:

We know that, stronger is the acid, weaker is its conjugate base. Now we know that the conjugate base of a strong acid is weak therefore the decreasing order of basic strength will be; $RO^{-}> OH^{-}> CH_{3}COO^{-} > Cl^{-}$

Question:27

Arrange the following in increasing order of pH.
$KNO_{3}(aq), CH_{3}COONa(aq), NH_{4}Cl(aq), C_{6}H_{5}COONH_{4}(aq)$
Answer:

NH₄Cl<C₆H₅COONH₄<KNO₃<CH₃COONa
Salts of strong acid and strong base do not hydrolyse and form a neutral solution, thus, pH will be nearly 7 of KNO₃. In sodium acetate, acetic acid remains unionised. This results in an increase in OH^- concentration and pH will be more than 7. NH4Cl formed from a weak base, NH4OH and strong acid, HCl, in water dissociate completely, aq. ammonium ions undergo hydrolysis with water to form NH4OH and H⁺ ions resulting in less pH value.

Question:28

The value of K_c for the reaction
$2HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g)$ is $1\times 10^{-4}$ . At a given time, the composition of reaction mixture is $[HI] = 2 \times 10^{-5} mol, [H_{_{2}}] = 1 \times 10^{-5} mol$ and $[I_{2}] = 1 \times 10^{-5} mol$ . In which direction will the reaction proceed?

Answer:

Now, as per the given information in the question
$2HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g)$
It has also been specified that $K_{c} =1 \times 10^{-4}$
Now, when we apply the law of mass action the Equilibrium constant in this given equation we get;
$K_{c} =\frac{[H_{2}][I_{2}]}{[HI]^{2}}$
It is also known to us that the reaction quotient, $Q_{c}$ , specifies the relative ratio of products to reactants at a given point in time.
$Q_{c} =\frac{[H_{2}][I_{2}]}{[HI]^{2}}$
$Q_{c} =\frac{(1\times10^{-5})(1\times10^{-5})}{(2\times10^{-4})}$
$Q_{c} =\frac{1}{4}=0.25$
Here, $Q_{c} >K_{c}$
Therefore, the reaction will proceed in reverse direction.

Question:29

On the basis of the equation pH = - log [H⁺ ], the pH of 10^-8 mol dm^-3 solution of HCl should be 8. However, it is observed to be less than 7.0. Explain the reason.
Answer:

As per the information given in the question, we can determine that the solution will be extremely dilute. It is also known to us that when HCl reacts with water it forms hydronium ion.
Thus, the large concentration of (H⁺) as a result, contributes to a decrease in the pH. Thus, here we cannot afford to ignore the concentration of H₃O⁺ ions in the solution.
So, the total pH will be; $[H_{3}O^{+}] = 10^{-8} + 10^{-7} M \approx 7$
Therefore, the solution will be acidic.

Question:30

pH of a solution of a strong acid is 5.0. What will be the pH of the solution obtained after diluting the given solution 100 times?
Answer:

As per the information given in the question, we know that the pH=5.0
pH = 5 means [H⁺]=10⁻⁵M,
On diluting it 100 times, we get
$[H^{+}]=\frac{10^{-5}}{100}=10^{-7}mol/L$
This should give pH = 7 but it cannot be so because solution is acidic and pH should be less than 7
The reason is that [H⁺] from H₂O cannot be neglected. Thus, total [H⁺]=10⁻⁷M (from HCl) + 10⁻⁷M (from H₂O ) =2×10⁻⁷M

∴pH=−log(2×10⁻⁷)= 7−0.3010=6.699

Question:31

A sparingly soluble salt gets precipitated only when the product of the concentration of its ions in the solution $(Q_{sp})$ becomes greater than its solubility product. If the solubility of $BaSO_{4}$ in water is $8\times 10^{-4} mol \; dm^{-3}$ calculate its solubility in 0.01 mol dm^-3 of $H_{2}SO_{4}.$ .

Answer:

As per the information given in the question, the solubility of $BaSO_{4}$ in water is $8 \times 10^{-4} g/L$
Hence, We can write the equation of disassociation of $BaSO_{4}$ as: -
$BaSO_{4}\rightleftharpoons Ba^{2+}++SO{_{4}}^{2-}$

At t=0	0	0
At t=0	S	S

Now, it is known to us that,

$K_{sp}=S\times S=S^{2}$

$K_{sp}=(8\times 10^{-8})^{2}$

$K_{sp}=64\times 10^{-8}$
Thus, in the presence of 0.01 $H_{2}SO_{4}$ solⁿ

$H_{2}SO_{4}\rightleftharpoons 2H^{+}+SO{_{4}}^{2-}$

Initial	0.01	0	0
Final	0	0.02	0.01

Now we know that,

$BaSO_{4}\rightleftharpoons Ba^{2+}+SO{_{4}}^{2-}$

Final

S+0.01

The expression for K_sp in the presence of sulphuric acid will be
K_sp = (S) (S + 0.01)
Since value of K_sp will not change in the presence of sulphuric acid,
therefore,
(S) (S + 0.01) = 64 × 10^–8
S²+ 0.01 S = 64 × 10^–8
S²+ 0.01 S – 64 × 10^–8 = 0
On solving quadratic equation, we get S= $6\times10^{-4}$ .
Thus the solubility of $BaSO_{4}$ in 0.01 mol dm^-3 of $H_{2}SO_{4}$ is $6\times10^{-4}$ mol dm^-3

Question:32

pH of 0.08 mol dm^-3 $HOCl$ solution is 2.85. Calculate its ionisation constant.
Answer:

As per the information given in the question, C =0.08 mol dm-3 and pH =2.85
We know that the HOCl is a weak acid, therefore its dissociation will be written as-
$HOCl+H_{2}O\leftrightharpoons H_{3}O^{+}+OCl^{-}$
pH = -log [H⁺]
Thus, -2.85 = log [H⁺]
[H⁺] = antilog (-2.85)
[H⁺] = 1.41 × 10^-3
We are aware that in case of a weak mono basic acid-
$H^{+}=\sqrt{K_{a}C}$
$[H^{+}]^{2}=K_{a}C$ (squaring both sides)
$K_{a}=\frac{[H^{+}]^{2}}{C}=\frac{(1.41\times 10^{-3})^{2}}{0.08}$
Thus, $K_{a}= 2.5\times 10^{-5}$
Therefore, we can conclude that the ionization constant of $HOCl$ will be $2.5\times 10^{-5}$

Question:33

Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively.
Answer:

pH of solution A = 6
Therefore, concentration of [H+ ] ion in solution A = 10^-6 mol L^-1
pH of solution B = 4
Therefore, concentration of [H+ ] ion in solution B = 10^-4 mol L^-1 .
On mixing one litre of each solution, total volume = 1L + 1L = 2L.
Amount of H⁺ ions in 1L of solution A= concentration x volume V= 10^-6 mol X 1L
Amount of H⁺ ions in 1L of solution B= 10^-4 mol X 1L
The total amount of H+ ions in the solution formed by mixing solutions A and B is (10^-6 + 10^-4 ) mol X 1L
This amount is present in 2L solution.
$Total\; [H^{+}]in\; 1L\; solution=\frac{10^{-6}+10^{-4}}{2}$
$=10^{-4}(1+\frac{0.01}{2})$
$=10^{-4}\times \frac{0.01}{2}$
$=5\times 10^{-5}mol/L$
Therefore, $pH=-log[H^{+}]=-log[5\times 10^{-5}]$
$=-log(5)+(-5\; log10)$
$=-log(5)+5=4.3$
Hence, the pH will be 4.3

Question:34

The solubility product of $Al(OH)_{3}$ is $2.7\times 10^{-11}$ . Calculate its solubility in gL ^-1 and also find out the pH of this solution. (Atomic mass of $Al=27\; u$ ).
Answer:

As per the information given in the question,
$K_{sp} =2.7\times 10^{-11}$
Hence, we can write the equation of disassociation of $Al(OH)_{3}$ as -
$Al(OH)_{3}\rightleftharpoons Al^{3+}(aq)+3OH^{-}(aq)$

At t=0	1	0	0
At,t=t	1-s	s	3s

It is known to us that,
$K_{sp}= [Al^{3+}] [OH^{-}] ^{3-}$
$=(s)\times (3s)^{3}$
$=27s^{4}$
$S^{4}=\frac{K_{sp}}{27}$
$S^{4}=\frac{2.7\times 10^{-11}}{27}$
$S^{4}=10^{-12}$
$S=(10^{-12})^{\frac{1}{4}}=10^{-3}mol/L$
Now, molar mass of $Al(OH)_{3}=78$
Therefore, solubility = molar mass x s
$=78\times10^{-3}$
$=7.8\times10^{-2}g/L$
Now, it is known to us that,
pH = 14 - pOH
[OH]= 3s = 3 × 10^-3
pOH= 3-log3
pH = 14 – 3 + log3
= 11.4771
Therefore, we can conclude that the solubility in g/L will be 7.8×10^-2 g/L amd the pH of the solution will be 11.47.

Question:35

Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution. (K_sp of PbCl₂ = 3.2 x 10^-8, atomic mass of Pb = 207 u).
Answer:

As per the information available from the question, K_sp of PbCl₂ = 3.2 ×10^-8
Hence, we can write the equation of disassociation of PbCl₂ as follows: -
$PbCl_{2}\rightleftharpoons Pb^{+2}(aq)+2Cl^{-}(aq)$

At t=0	1	0	0
At t=t	1-x	x	2x

$K_{sp}=[Pb^{2+}][Cl^{-}]^{2}$
On substituting the given values, we get
$=(x)\times (2x)^{2}=4x^{3}$
$4x^{3}=3.2\times 10^{-8}$
$x=2\times 10^{-3}mol/L$
We know that solubility is a product of molar mass $(PbCl_{2}) \times 2 \times 10^{-3}$
$=556\times 10^{-3}=0.556g/L$
0.1g of $PbCl_{2}$ will be dissolved in 0.1798 L (0.1/0.0556)
Thus, 0.1798 is the required volume to get a saturated solution of $PbCl_{2}$

Question:36

A reaction between ammonia and boron trifluoride is given below:
$BF_{3}+:NH_{3}\rightarrow BF_{3}:NH_{3}$
Identify the acid and base in this reaction. Which theory explains it? What is the hybridisation of B and N in the reactants?
Answer:

As per the information given in the question, $NH_{3} + BF_{3}\rightarrow H_{3}N: BF_{3}$
$NH_{3} (N=1s^{2}2s^{2}2p^{1})$ is the Lewis acid in this reaction because it has a lone pair of e- to donate in its p-orbital. Moreover, the Lewis base in this case is BF3 as p-orbital of Boron is empty $(B=1s^{2}2s^{2}2p^{1})$ . As such, it will be accepting lone pair from N to form a dative bond, as is clear from the lewis electronic theory. Thus, the Hybridisation of N is sp³ and B is sp².

Question:37

Following data is given for the reaction: $CaCO_{3} (s) \rightarrow CaO (s) + CO_{2} (g)$
$\Delta _{f}H^{\ominus} [CaO (s)] = -635.1 kJ mol^{-1}$
$\Delta _{f}H^{\ominus } [CO_{2} (g)] = -393.5 kJ mol^{-1}$
$\Delta_{f}H^{\ominus } [CaCO_{3} (s)] = -1206.9 kJ mol^{-1}$
Predict the effect of temperature on the Equilibrium constant of the above reaction.
Answer:

We know that,
$\Delta _{r}H^{\Theta } = \Delta _{f}H^{\Theta }[CO_{2}(g)]-\Delta_{f}H^{\Theta }[CaCO_{3}(s)]$
$\Delta _{r}H^{\Theta }= 178.3KJmol^{-1}$
.Thus As per the Le Chatelier’s Principle on increasing the temperature, the reaction will be shifting in the forward direction.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: Matching Type

Question:38

Match the following equilibria with the corresponding condition

Column I		Column II
(i)	$Liquid\rightleftharpoons Vapour$	(a)	Saturated solution
(ii)	$Solid \rightleftharpoons Liquid$	(b)	Boiling point
(iii)	$Solid \rightleftharpoons Vapour$	(c)	Sublimation point
(iv)	$Solute(s) \rightleftharpoons Solute (solution)$	(d)	Melting point ‘
		(e)	Unsaturated solution

Answer:

(i) $\rightarrow$ (b); (ii) $\rightarrow$ (d); (iii) $\rightarrow$ (c); (iv) $\rightarrow$ (a)

Question:39

For the reaction: $N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)$ equilibrium constant,
$K_{c}=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}$
Some reactions are written below in Column I, and their equilibrium constants in terms of Kc are written in Column II. Match the following reactions with the corresponding equilibrium constant.

Column I (Reaction)	Column II (Equilibrium constant)
(i) $2N_{2}(g)+6H_{2}(g)\rightleftharpoons 4NH_{3}(g)$	(a) $2K_{c}$
(ii) $2NH_{3}(g)\rightleftharpoons N_{2}(g)+3H_{2}(g)$	(b) $K{_{c}}^{\frac{1}{2}}$
(iii) $\frac{1}{2}N_{2}(g)+\frac{3}{2}H_{2}(g)\rightleftharpoons NH_{3}(g)$	(c) $\frac{1}{K_{c}}$
	(d) $K{_{c}}^{2}$

Answer:

(i) $\rightarrow$ (d); (ii) $\rightarrow$ (c); (iii) $\rightarrow$ (b)

Question:40

Match standard free energy of the reaction with the corresponding equilibrium constant

(i) $\Delta G^{\ominus }>0$	$K>1$
(ii) $\Delta G^{\ominus }<0$	$K=1$
(iii) $\Delta G^{\ominus }=0$	$K=0$
	$K<1$

Answer:

(i) $\rightarrow$ (d); (ii) $\rightarrow$ (a); (iii) $\rightarrow$ (b)

Question:41

Match the following species with the corresponding conjugate acid.

Species	Conjugate Acid
(i) $NH_{3}$	(a) $CO{_{3}}^{2-}$
(ii) $HCO{_{3}}^{-}$	(b) $NH{_{4}}^{+}$
(iii) $H_{2}O$	(c) $H_{3}O^{+}$
(iv) $HSO{_{4}}^{-}$	(d) $H_{2}SO_{4}$
	(e) $H_{2}CO_{3}$

Answer:

(i) $\rightarrow$ (b); (ii) $\rightarrow$ (e); (iii) $\rightarrow$ (c); (iv) $\rightarrow$ (d)

Question:42

Match the following graphical variation with their description

A	B
(i)	(a) Variation in product concentration with time
(ii)	(b) Reaction at equilibrium
(iii)	(c) Variation in reactant concentration with time

Answer:

(i) $\rightarrow$ (c); (ii) $\rightarrow$ (a); (iii) $\rightarrow$ (b)

Question:43

Match Column (I) with Column (II).

Column I	Column II
(i) Equilibrium	(a) $\Delta G>0,K< 1$
(ii) Spontaneous reaction	(b) $\Delta G=0$
(iii) Non-spontaneous reaction	(c) $\Delta G^{\ominus }=0$
	(d) $\Delta G<0,K> 1$

Answer:

(i) $\rightarrow$ c; (ii) $\rightarrow$ (d); (iii) $\rightarrow$ (a)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: Assertion and Reason Type

Question:44

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): Increasing order of acidity of hydrogen halides is $HF < HCl < HBr < HI$
Reason (R): While comparing acids formed by the elements belonging to the same group of the periodic table, H-A bond strength is a more important factor in determining the acidity of acid than the polar nature of the bond.
(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) Both A and R are false.

Answer:

The answer is the option (i) Both A and R are true, and R is the correct explanation of A
Explanation: H-A bond strength plays a more prominent role in determining the acidity than its polar nature. And as the size of A increases down the group corresponding to that the H-A bond strength decreases.

Question:45

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 solution containing a mixture of acetic acid and sodium acetate maintains a constant value of pH on addition of small amounts of acid or alkali.
Reason (R): A solution containing a mixture of acetic acid and sodium acetate acts as a buffer solution around pH 4.75.
(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.
(iii) Both A and R are false.
Answer:

The answer is the option (i) Both A and R are true, and R is the correct explanation of A.
Explanation: Buffer solutions are such solutions which resist a change in their pH on dilution or with the addition of small amounts of acid or alkali.

Question:46

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): The ionisation of hydrogen sulphide in water is low in the presence of hydrochloric acid.
Reason (R): Hydrogen sulphide is a weak acid.
(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) Both A and R are false.
Answer:

The answer is the option (ii) Both A and R are true, but R is not the correct explanation of A.
Explanation: There is a shift in the backward direction of the equilibrium.

Question:47

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): For any chemical reaction at a particular temperature, the equilibrium constant is fixed and is a characteristic property.
Reason (R): Equilibrium constant is independent of temperature.
(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) Both A and R are false.
Answer:

The answer is the option (iii) A is true, but R is false.
Explanation: The equilibrium constant for an exothermic reaction (negative H) decreases as the temperature increases. The equilibrium constant for an endothermic reaction (positive H) increases as the temperature increases.

Question:48

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): Aqueous solution of ammonium carbonate is basic.
Reason (R): Acidic/basic nature of a salt solution of a salt of a weak acid and weak base depends on K_a and K_b value of the acid and the base forming it.
(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) Both A and R are false.
Answer:

(i) Both A and R are true, and R is the correct explanation of A.
Explanation- Ka and Kb values are the ones responsible for acidic and basic characters of any compound.

Question:49

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): An aqueous solution of ammonium acetate can act as a buffer.
Reason (R): Acetic acid is a weak acid, and $NH_{4}OH$ is a weak base.
(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 false, but R is true.
(iv) Both A and R are false.

Answer:

The answer is the option (iii) A is false, but R is true.
Explanation: Salt of a weak acid and weak base can be used to form a buffer solution.

Question:50

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): In the dissociation of PCI₅ at constant pressure and temperature addition of helium at equilibrium increases the dissociation of PCl₅.
Reason (R): Helium removes Cl₂ from the field of action.
(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) Both A and R are false.
Answer:

(iv) Both A and R are false.
Explanation: If the volume remains constant and an inert gas such as argon is added, the equilibrium remains undisturbed because it does not take part in the reaction.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: Long Answer Type

Question:51

How can you predict the following stages of a reaction by comparing the value of $K_{c} \; and \; Q_{c}$ ?
(i) Net reaction proceeds in the forward direction.
(ii) Net reaction proceeds in the backward direction.
(iii) No net reaction occurs.
Answer:

The values of $K_{c}$ (the equilibrium constant) and $Q_{c}$ (the reaction quotient) less than or greater than one another decides the direction in which reaction will proceed as follows-
(i) As $Q_{c} <K_{c}$ , this implies that in order to reach the equilibrium, the concentration of the products must be increased, hence the reaction will proceed in the forward direction.
(ii) If $Q_{c} >K_{c}$ , this implies that in order to reach the equilibrium, the concentration of the products must be decreased, hence the reaction will proceed in the backward direction.
(iii) If $Q_{c} =K_{c}$ , no net reaction occurs as the equilibrium has been achieved.

Question:52

On the basis of the Le Chatelier principle, explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction:
$N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g) \;\Delta H = -92.38 \; kJ mol^{-1}$
What will be the effect of the addition of argon to the above reaction mixture at constant volume?
Answer:

As per the information given in the question, $N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g) \;\Delta H = -92.38 \; kJ mol^{-1}$
Now, we know that as ΔH is negative, it means that the forward reaction is exothermic in nature.
As per the Le Chatelier principle, when the temperature is reduced the reaction will move in forward direction, hence more yield of the product will be obtained.
We know that when the temperature is increased the reaction will be moving in the backward direction that the yield of product will be reduced.
As per the Le Chatelier principle, when the pressure is increased the equilibrium shifts in the direction where there are lesser numbers of gas molecules. Therefore, when the pressure is increased, the equilibrium will be shifting in the forward direction and consequently the yield of the product will increase.
Thus, number of moles of reactants = 1 + 3 = 4
Number of moles of product = 2
Thus, we can conclude that high pressure and low temperature are the desirable conditions for increasing the yield of the products. Addition of an an inert gas at constant volume does not result in a shift. This is due to the fact that the addition of a non-reactive gas does not cause any change in the partial pressures of the other gases in the container.

Question:53

A sparingly soluble salt having general formula $A{_{x}}^{p+},B{_{y}}^{q-}$ and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt.
Answer:

In order to derive the relationship between the solubility and solubility product for such a salt we will take the following steps,
We will assume the solubility to be S.
Initially, we will assume that 1 mole of A_xB_y was present. Out of it, S moles were dissolved in order to give xS and yS moles of A^p+ and B^q- respectively. We can understand it better through the following equation: -
$A^{p+}\times B^{q-} y \leftrightharpoons xA^{+p} + yB^{-q}$
At t=0, 1 0 0
At Equilibrium, $1-S,xS \; Ys$
We know that, solubility product $(K_{sp}) = [A^{+p}]\times [B^{-q}]^{y}$
= $(xS)\times (yS)^{y}$
= $x^{x}Y^{y}S^{x+y}$

Question:54

Write a relation between AG and Q and define the meaning of each term and answer the following:
(a) Why a reaction proceeds forward when Q < K and no net reaction occurs when Q = K.
(b) Explain the effect of an increase in pressure in terms of reaction quotient Q for the reaction:
CO (g) + 3H₂(g) $\rightleftharpoons$ CH₄(g) + H₂O (g)
Answer:

We can describe the relation between ΔG and Q as
ΔG = ΔG° + RT lnQ ...(1)
Herein, ΔG = Gibb’s energy change, ΔG° = standard Gibb’s energy, T = absolute temperature, Q = reaction quotient, and R = gas constant
We know that at Equilibrium, ΔG° = -RT lnK ...(2) (Wherein K is the Equilibrium constant)
Replacing the values of (2) in (1), we get
ΔG = -RT lnK + RT lnQ
ΔG = RT ln(Q/K) ...(3)
It is known to us that, Kc (which is the Equilibrium constant) is the ratio of concentration of products to that of reactants each raised to their stoichiometric coefficients at Equilibrium.
In contrast, Qc ( which is the reaction quotient) is the ratio of concentration of products to that of reactants each raise to their stoichiometric coefficients at any time during the reaction.
(a) In this case, if Q< K, it implies that the concentration of products has to be increased in order to reach Equilibrium concentration. Hence, the net reaction will proceed in the forward direction.
When Q= K, it implies that the Equilibrium has been achieved and no net reaction occurs. Now lets answer tehs econd part of the question.
(b) CO (g) + 3H₂(g) $\rightleftharpoons$ CH₄(g) + H₂O (g) , As per the Le Chatiliers principle, when the pressure is increased, the Equilibrium will be shifting in that direction where there is a lesser number of gas molecules.
As per the available information,
Number of moles of reactants = 1 + 3 = 4
Number of moles of product = 1 + 1 = 2
Hence, the Equilibrium will shift in forward direction as number of moles of product is less. It is known to us that when Q < K, the Equilibrium will shift in the forward direction. Thus, for the given reaction Q < K

NCERT Exemplar solutions for Class 11 Chemistry chapter 7 lays the foundation for evaluating and deriving information from balanced equations. Thus, students need to be well versed with this chapter for which they can also refer to NCERT exemplar Class 11 Chemistry solutions chapter 7 pdf download as per their convenience.

Also, check NCERT Solutions for Class 11

Topics and Subtopics in NCERT Exemplar Class 11 Chemistry Solutions Chapter 7 Equilibrium

Class 11 Chemistry NCERT Exemplar solutions Chapter 7 include the following topics:

1. Solid-liquid Equilibrium
1.1 Liquid-vapour Equilibrium
1.2 Solid – Vapour Equilibrium
1.3 Equilibrium Involving Dissolution Of Solid Or Gases In Liquids
1.4 General Characteristics Of Equilibria Involving Physical Processes
2. Equilibrium In Chemical Processes – Dynamic Equilibrium
2.1Law Of Chemical Equilibrium And Equilibrium Constant
3. Homogeneous Equilibria
4. 1 Equilibrium Constant In Gaseous Systems
5. Heterogeneous Equilibria
6. Applications Of Equilibrium Constants
6.1 Predicting The Extent Of A Reaction
6.2 Predicting The Direction Of The Reaction
6.3 Calculating Equilibrium Concentrations
7. Relationship Between Equilibrium Constant K, Reaction Quotient Q And Gibbs Energy G
8. Factors Affecting Equilibria
8.1 Effect Of Concentration Change
8.2 Effect Of Pressure Change
8.3 Effect Of Inert Gas Addition
8.4 Effect Of Temperature Change
8.5 Effect Of A Catalyst
9. Ionic Equilibrium In Solution
10. Acids, Bases And Salts
10.1 Arrhenius Concept Of Acids And Bases
10.2 The Bronsted-lowry Acids And Bases
10.3 Lewis Acids And Bases
11. Ionisation Of Acids And Bases
11.1 The Ionization Constant Of Water And It’s Ionic Product
11.2 The Ph Scale
11.3 Ionisation Constants Of Weak Acids
11.4 Ionisation Of Weak Bases
11.5 The Relation Between Ka And Kb
11.6 Di- And Polybasic Acids And Di- And Polyacidic Bases
11.7 Factors Affecting Acid Strength
11.8 Common Ion Effect In The Ionization Of Acids And Bases
11.9 Hydrolysis Of Salts And The Ph Of Their Solutions
12. Buffer Solutions
13. Solubility Equilibria Of Sparingly Soluble Salts
13.1 Solubility Product Constant
13.2 Common Ion Effect on Solubility Of Ionic Salts.

Also read - NCERT Solutions for Class 11 Chemistry Chapter 7

What the students will Learn from NCERT Exemplar Class 11 Chemistry Solutions Chapter 7 Equilibrium?

After completing with the contents of this chapter the students would have an in-depth understanding about the concept of equilibrium in various physical states i.e solid- liquid or gaseous systems etc. Class 11 Chemistry NCERT exemplar solutions chapter 7 primarily focuses upon the state of equilibrium and the factors influencing the same. It also educates the students upon the ionisation of acids and bases. The concept of buffer solutions was also taught to the students for the first time. NCERT exemplar solutions for class 11 Chemistry chapter 7 also include the concept of solubility and its product constant. Through the course of this chapter the students learn several major topics such as equilibrium and factors influencing it, Ph, Solubility and the detailed study of acids, bases and salts.

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NCERT Exemplar Class 11 Chemistry Solutions Chapter-Wise

Chapter 1 Some Basic Concepts of Chemistry

Chapter 2 Structure of Atom

Chapter 3 Classification of Elements and Periodicity in Properties

Chapter 4 Chemical Bonding and Molecular Structure

Chapter 5 States of Matter

Chapter 6 Thermodynamics

Chapter 8 Redox Reactions

Chapter 9 Hydrogen

Chapter 10 The s-Block Elements

Chapter 11 The p-Block Elements

Chapter 12 Organic Chemistry – Some Basic Principles and Techniques

Chapter 13 Hydrocarbons

Chapter 14 Environmental Chemistry

Important Topics To Cover From NCERT Exemplar Class 11 Chemistry Solutions Chapter 7 Equilibrium

Ø Students will take away a new formed understanding of the concept of equilibrium and the factors that bring change or hindrance in the same while also learning the ways in which they can balance the equilibrium in different physical states.

Ø The students will also be educated upon the different types of equilibriums and their different properties such as homogenous, heterogeneous and dynamic equilibrium. They will learn the favourable conditions required to maintain the above.

Ø NCERT exemplar solutions for Class 11 Chemistry chapter 7 also dwells on the extent and direction of a reaction and discusses the relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G after grasping the concept of ionisation of acids and bases.

Also, Check Chapter Wise NCERT Solutions for Class 11 Chemistry

Chapter-1 - Some Basic Concepts of Chemistry

Chapter-2 - Structure of Atom

Chapter-3 - Classification of Elements and Periodicity in Properties

Chapter-4 - Chemical Bonding and Molecular Structure

Chapter-5 - States of Matter

Chapter-6 - Thermodynamics

Chapter-7 - Equilibrium

Chapter-8 - Redox Reaction

Chapter-9 - Hydrogen

Chapter-10 - The S-Block Elements

Chapter-11 - The P-Block Elements

Chapter-12 - Organic chemistry- some basic principles and techniques

Chapter-14 - Hydrocarbons

Chapter-15 - Environmental Chemistry

NCERT Exemplar Class 11 Solutions

NCERT Exemplar Class 11 Biology Solutions

NCERT Exemplar Class 11 Mathematics Solutions

NCERT Exemplar Class 11 Physics Solutions

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Frequently Asked Question (FAQs)

1. Does NCERT exemplar class 11 Chemistry solutions chapter 7 carry reasonable mark distribution in the exams?

Though the individual topic and chapter of equilibrium altogether might not garner the highest number of marks in the final examination it is significant for the laying down the basic concept foundation for many more complex chapters and topics to come so it essential to be well versed with this chapter for ease in the long run

2. What is the general format of the questions asked in this chapter?

Generally speaking, the questions asked from this topic are frequently observed to be practical numerical based where we have to apply the principles and rules of equilibrium.

3. Is NCERT exemplar solutions for class 11 Chemistry chapter 7 a reliable source for studying and practicing the concept?

Yes, the solutions are largely beneficial for students who desire to extend their practical knowledge by solving problems relating to the topic which are followed by specifically designed solutions which are easily understandable and suggest the best approach to solve the question.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 7 Equilibrium

NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: MCQ (Type 1)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: MCQ (Type 2)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: Short Answer Type

NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: Matching Type

NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: Assertion and Reason Type

NCERT Exemplar Class 11 Chemistry Solutions Chapter 7: Long Answer Type

Topics and Subtopics in NCERT Exemplar Class 11 Chemistry Solutions Chapter 7 Equilibrium

What the students will Learn from NCERT Exemplar Class 11 Chemistry Solutions Chapter 7 Equilibrium?

NCERT Exemplar Class 11 Chemistry Solutions Chapter-Wise

Important Topics To Cover From NCERT Exemplar Class 11 Chemistry Solutions Chapter 7 Equilibrium

Also, Check Chapter Wise NCERT Solutions for Class 11 Chemistry

NCERT Exemplar Class 11 Solutions

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