NCERT Solutions Class 11 Chemistry Chapter 7 Equilibrium– Here are all the NCERT solutions for Class 11 Chemistry Chapter 7. This solution contains questions, answers, images, explanations of the complete chapter 1 titled Equilibrium taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Chemistry, then you must come across chapter 7 Equilibrium After you have studied the lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium in one place.
NCERT Solutions Class 11 Chemistry Chapter 7 Equilibrium
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For a better understanding of this chapter, you should also see summary of Chapter 7 Equilibrium , Chemistry, Class 11.
| Class | 11 |
| Subject | Chemistry |
| Book | Chemistry Part I |
| Chapter Number | 7 |
| Chapter Name |
Equilibrium |
NCERT Solutions Class 11 Chemistry chapter 7 Equilibrium
Class 11, Chemistry chapter 7, Equilibrium solutions are given below in PDF format. You can view them online or download PDF file for future use.
Equilibrium
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Question & Answer
Q.1: A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Ans : (a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume. (b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially. (c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.
Q.2: \(\begin{array}{c}{\text { What is } K_{c} \text { for the following equilibrium when the equilibrium concentration of }} \\ {\text { each substance is: }\left[\mathrm{SO}_{2}\right]=0.60 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.82 \mathrm{M} \text { and }\left[\mathrm{SO}_{3}\right]=1.90 \mathrm{M} \text { ? }} \\ {2 \mathrm{SO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})}\end{array}\)
Ans : The equilibrium constant (K,) for the give reaction is: \(\begin{aligned} \mathrm{K}_{c} &=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]} \\ &=\frac{(1.90)^{2} \mathrm{M}^{2}}{(0.60)^{2}(0.821) \mathrm{M}^{3}} \\ &=12.239 \mathrm{M}^{-1}(\text { approximately }) \end{aligned}\) Hence, \(K_{c} \text { for the equilibrium is }^{12.239 \mathrm{M}^{-1}}\)
Q.3: \(\begin{array}{c}{\text { At a certain temperature and total pressure of } 10^{5} \mathrm{Pa}, \text { iodine vapour contains } 40 \%} \\ {\text { by volume of I atoms }} \\ {\quad \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})}\end{array}\)
Calculate Kp for the equilibrium.
Ans : \(\begin{array}{l}{\text { Partial pressure of I atoms, }} \\ {p_{1}=\frac{40}{100} \times p_{\text { lotalal }}} \\ {=\frac{40}{100} \times 10^{5}} \\ {=4 \times 10^{4} \mathrm{Pa}} \\ {\text { Partial pressure of I_ } \mathrm{I}_{2} \text { molecules, }}\end{array}\) \(\begin{array}{l}{p_{1_{2}}=\frac{60}{100} \times p_{\text { total }}} \\ {=\frac{60}{100} \times 10^{5}} \\ {=6 \times 10^{4} \mathrm{Pa}} \\ {\text { Now, for the given reaction, }}\end{array}\) \(\begin{aligned} K_{p} &=\frac{(p \mathrm{I})^{2}}{p_{1_{2}}} \\ &=\frac{\left(4 \times 10^{4}\right)^{2} \mathrm{Pa}^{2}}{6 \times 10^{4} \mathrm{Pa}} \\ &=2.67 \times 10^{4} \mathrm{Pa} \end{aligned}\)
Q.4: Write the expression for the equilibrium constant, Kc for each of the following reactions:
\(\begin{array}{l}{\text { (i) } 2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})} \\ {\text { (ii) } 2 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons 2 \mathrm{CuO}(\mathrm{s})+4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})}\end{array}\)
\(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq})\)
(iv) \(\quad \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\)
(v) \(\quad \mathrm{I}_{2}(\mathrm{s})+5 \mathrm{F}_{2} \rightleftharpoons 2 \mathrm{IF}_{5}\)
Ans : (i) \(\quad K_{e}=\frac{\left[\mathrm{NO}_{(g)}\right]^{2}\left[\mathrm{Cl}_{2(\mathrm{g}\}}\right]}{\left[\mathrm{NOCl}_{(g)}\right]^{2}}\) (ii) \(K_{c}=\frac{\left[\mathrm{CuO}_{(s)}\right]^{2}\left[\mathrm{NO}_{2(\mathrm{g})}\right]^{4}\left[\mathrm{O}_{2(\mathrm{g})}\right]}{\left[\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2(\mathrm{i})}\right]^{2}}\) \(=\left[\mathrm{NO}_{2(\mathrm{g})}\right]^{4}\left[\mathrm{O}_{2(\mathrm{g})}\right]\)
Q.5: Find out the value of Kc for each of the following equilibria from the value of Kp :
\(\begin{array}{ll}{\text { (i) }} & {2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) ; \quad K_{\mathrm{p}}=1.8 \times 10^{-2} \text { at } 500 \mathrm{K}} \\ {\text { (ii) }} & {\mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) ; \quad K_{\mathrm{p}}=167 \text { at } 1073 \mathrm{K}}\end{array}\)
Ans : \(\begin{array}{l}{\text { The relation between } \mathrm{K}_{\mathrm{p}} \text { and } \mathrm{K}_{\mathrm{c}} \text { is given as: }} \\ {\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\Delta \mathrm{n}}} \\ {\text { (a) Here, }} \\ {\Delta \mathrm{n}=3-2=1} \\ {\mathrm{R}=0.0831 \text { barLmol-1 } \mathrm{K}^{-1}} \\ {\mathrm{T}=500 \mathrm{K}} \\ {\mathrm{K}_{\mathrm{p}}=1.8 \times 10^{-2}}\end{array}\) \(\begin{array}{l}{\mathrm{K} \mathrm{ow},} \\ {\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}}} \\ {\Rightarrow 1.8 \times 10^{-2}=K_{c}(0.0831 \times 500)^{1}} \\ {\Rightarrow K_{c}=\frac{1.8 \times 10^{-2}}{0.0831 \times 500}} \\ {=4.33 \times 10^{-4}(\text { approximately })}\end{array}\) \(\begin{array}{l}{\text { (b) Here, }} \\ {\Delta n=2-1=1} \\ {R=0.0831 \text { barLmol }^{-1} K^{-1}} \\ {T=1073 \mathrm{K}} \\ {K_{p}=167} \\ {\text { Now, }} \\ {K_{p}=K_{c}(R T)^{\Delta n}}\end{array}\) \(\begin{aligned} \Rightarrow 167 &=K_{c}(0.0831 \times 1073)^{\Delta n} \\ \Rightarrow K_{c} &=\frac{167}{0.0831 \times 1073} \\ &=1.87(\text { approximately }) \end{aligned}\)
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