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# Chemical Equilibrium | Question Bank for Class 11 Chemistry

Get Chemical Equilibrium important questions for Boards exams. Download or View the Important Question bank for Class 11 Chemistry. These important questions will play significant role in clearing concepts of Chemistry. This question bank is designed by NCERT keeping in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 chemistry chapter wise CBSE. Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, JEE & NEET.  You can access free study material for all three subject’s Physics, Chemistry and Mathematics. Click Here for Detailed Notes of any chapter.

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Q. Give one example of everyday life in which there is gas $\rightleftharpoons$ solution equilibrium.
Ans. Soda-water bottle.
Q. Why gas fizzes out when soda water bottle is opened ?
Ans. The amount of the gas dissolved is very high due to high pressure. On opening the bottle, the pressure tends to decrease to atmospheric pressure. So the solubility decreases, i.e., the dissolved gas escapes out.
Q. The equilibrium constant for gas reaction is $K_{c}=\frac{\left[N H_{3}\right]^{4}\left[O_{2}\right]^{5}}{[N O]^{4}\left[H_{2} O\right]^{6}}$ Write the balance chemical reaction to this expression.
Ans. $4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 4 \mathrm{NH}_{3}+5 \mathrm{O}_{2}$
Q. At equilibrium, the mass of each of the reactants and products remains constant. Does it mean that the reaction has stopped ? Explain.
Ans. No, the reaction does not stop. It continues to take place in the forward as well as backward direction but at equal speed.
Q. Some sugar is added into a saturated solution of sugar in a beaker. What process/processes if any, do you expect to happen with the passage of time? What is this state called?
Ans. Two processes, namely, dissolution and precipitation will continue to take place at equal rates. It is called a state of equilibrium.
Q. Give reasons in brief, indicate whether the following statement is TRUE or FALSE : The rate of an exothermic reaction increases with increasing temperature.
Ans. False because in an exothermic reaction, heat is evolved. Increase of temperature will shift the equilibrium in the backward direction, i.e., the rate of reaction decreases.
Q. What happens to a reversible reaction if a catalyst is added to it?
Ans. The state of equilibrium is not disturbed but it attained quickly because both the rate of forward and backward reaction increases to the same extent.
Q. For the following equilibrium, $K_{c}^{\prime}=6.3 \times 10^{14}$ at $1000 \mathrm{K}$ $N O(g)+O_{3}(g) \rightleftharpoons N O_{2}(g)+O_{2}(g)$ Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is $K_{c}$ for the reverse reaction?
Ans. For the reverse reaction $K_{c}^{\prime}=\frac{1}{K_{c}}=\frac{1}{6.3 \times 10^{14}}$ $=1.59 \times 10^{-15}$
Q. What is meant by reaction quotient ?
Ans. It is defined as product of molar concentration of reaction products each raised to power equal to the respective stoichiometric coefficient in balance chemical equation divided by the product of concentration of the reactants raised to power equal to their individual stoichiometric coefficients at any stage of reaction, e.g., $a A+b B \rightleftharpoons c C+d D$ $Q_{C}=\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$
Q. $\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ What is the relationship between $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{c}} ?$
Ans. $K_{p}=K_{c}$
Q. Predict the effect of compression on the following equilibrium reaction: $\mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(\mathrm{g})$
Ans. The increase of pressure i.e. compression will favour the formation of reactants because reactants have less number of moles than products.
Q. How does a catalyst affect the equilibrium constant ?
Ans. The equilibrium constant is not affected by catalyst.
Q. Vapour pressure of water, acetone and ethanol at 298 K are 2.34, 12.36 and 5.85 kPa respectively. Which of these has the lowest and the highest boiling point ? At 293 K, which of these evaporates least in a sealed container before equilibrium is established?
Ans. Acetone has highest vapour pressure, therefore has lowest boiling point. Water has lowest vapour pressure, therefore has highest boiling point. At $293 K$, water will evaporates least in a sealed container before equilibrium is established.
Q. The concentration quotient of a reversible reaction is $Q$ and the equilibrium constant is $K .$ What do you conclude if $(\text { i }) Q=K$ (ii) $Q>K$ (iii) $Q Ans. (i) If$Q=K,$the reaction is in equilibrium. (ii) If$Q>K, Q$will tend to decrease so as to become equal to$K .$As a result, the reaction will proceed in the backward direction. (iii) If$Q
Q. Explain why pure liquids and solids are ignored while writing the equilibrium constant expression.
Ans. (Pure liquid) or (Pure solid) $=\frac{\text { No. of moles }}{\text { Volume of } \mathrm{L}}=\frac{\text { Mass/mol. mass }}{\text { Volume }}$ $=\frac{\text { Mass }}{\text { Volume }} \times \frac{1}{\text { Mol. mass }}=\frac{\text { Density }}{\text { Mol. mass }}$ As density of a pure liquid or pure solid is constant constant temperature and molecular mass is also. constant, therefore, their molar concentrations are constant and included into the equilibrium constant.
Q. At a certain temperature and a total pressure of $10^{5} \mathrm{Pa}$ iodine vapour contain $40 \%$ by volume of iodine atoms $\left[\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\right]$ Calculate $\mathrm{K}_{\mathrm{p}}$ for the equilibrium.
Ans. Partial pressure of $I$ atoms $\left(p_{1}\right)=\frac{40}{100} \times 10^{5}$ $P a=0.4 \times 10^{5} \mathrm{Pa}$ Partial pressure of $I_{2}\left(p_{I_{2}}\right)=\frac{60}{100} \times 10^{5} P a=0.60 \times 10^{5} P a$ $K_{p}=\frac{p_{I}^{2}}{P_{I_{2}}}=\frac{\left(0.4 \times 10^{5}\right)^{2}}{0.60 \times 10^{5}}=2.67 \times 10^{4}$
Q. The value of for the reaction $3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(\mathrm{g})$ is $2.0 \times 10^{-50}$ at $25^{\circ} \mathrm{C}$. If the equilibrium concentration of $\mathrm{O}_{2}$ in air at $25^{\circ} \mathrm{C}$ is $1.6 \times 10^{-2},$ what is the concentration of $\mathrm{O}_{3} ?$
Ans. $K_{c}=\frac{\left[O_{3}\right]^{2}}{\left[O_{2}\right]^{3}} \therefore 2.0 \times 10^{-50}=\frac{\left[O_{3}\right]^{2}}{\left(1.6 \times 10^{-2}\right)^{3}}$ or $\left[\mathrm{O}_{3}\right]^{2}=\left(2.0 \times 10^{-50}\right)\left(1.6 \times 10^{-2}\right)^{3}=8.192 \times 10^{-56}$ or $\left[O_{3}\right]=2.86 \times 10^{-28} M$
Q. The value $\Delta \mathrm{G}^{\circ}$ of for the phosphorylation of glucose in glycolysis is $13.8 \mathrm{kJ} \mathrm{mol}^{-1} .$ Find the value of $\mathrm{K}_{\mathrm{c}}$ at $298 \mathrm{K}$
Ans. $\Delta G=13.8 \mathrm{kJ} \mathrm{mol}^{-1}=13.8 \times 10^{3} \mathrm{J} \mathrm{mol}^{-1}$ $\Delta G^{\circ}=-2.303 R T \log K$ $\log K=-\frac{\Delta G^{\circ}}{2.303 R T}=-\frac{13.8 \times 10^{3}}{2.303 \times 8.314 \times 298}=-2.418$ $K=3.82 \times 10^{-3}$
Q. Calculate (a) $\Delta \mathrm{G}^{\circ}$ and (b) the equilibrium constant for the formation of $\mathrm{NO}_{2}$ from $\mathrm{NO}$ and $\mathrm{O}_{2}$ at $298 \mathrm{K}$.
Ans. $N O(g)+\frac{1}{2} O_{2}(g) \rightleftharpoons N O_{2}(g)$ $\Delta G^{\circ}=\Delta_{f} G^{\circ}\left(N O_{2}\right)-\left[\Delta_{f} G^{\circ}(N O)+\frac{1}{2} \Delta_{f} G^{\circ}\left(O_{2}\right)\right]$ $=52.0-87.0-\frac{1}{2} \times 0=-35 \mathrm{kJ} \mathrm{mol}^{-1}$ $\mathrm{Now}, \log K=-\frac{\Delta G^{\circ}}{2.303 R T}$ $=-\frac{-35 \times 10^{3} \mathrm{J} \mathrm{mol}^{-1}}{2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \times 298 \mathrm{K}}=6.314$ or $K=1.362 \times 10^{6}$
Q. At $450 K, \mathrm{K}_{\mathrm{p}}=2.0 \times 10^{10} /$ bar for the given reaction at equilibrium $: 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{g})$ What is the $\mathrm{K}_{\mathrm{c}}$ at this temperature?
Ans. $K_{p}=K_{c}(R T)^{\Delta n}$ or $K_{c}=\frac{K_{p}}{(R T)^{\Delta n}}$ $\Delta n=2-(2+1)=-1, T=450 K, R=0.083$ bar $K^{-1} \mathrm{mol}^{-1}$ $\therefore K_{C}=\frac{2.0 \times 10^{10}}{(0.083 \times 450)^{-1}}=2.0 \times 10^{10} \times(0.083 \times 450)$ $=7.47 \times 10^{11} M^{-1}$
Q. Two moles of $\mathrm{NH}_{3}$ are introduced in a previously evacuated container of capacity one litre which is partially dissociated at high temperature as $2 \mathrm{NH}_{3}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}),$ At equilibrium one mole of remains. What is the value of $\mathrm{k}_{\mathrm{C}} ?$
Ans.

Q. 1 mole of $\mathrm{H}_{2} \mathrm{O}$ and 1 mole of $\mathrm{CO}$ are taken in a 10 litre vessel and heated to $725 \mathrm{K}$. At equilibrium, $40 \%$ of $\mathrm{H}_{2} \mathrm{O}$ (by mass) reacts with CO according to the equation: $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})$
Ans.
Q. At $700 K$ the equilibrium constant for the reaction $\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ is $54.8 .$ If $0.5 \mathrm{mol} \mathrm{L}^{-1}$ of $\mathrm{HI}(\mathrm{g})$ is present at equilibrium at $700 K,$ what are the concentrations of $\mathrm{H}_{2}(\mathrm{g})$ and $\mathrm{I}_{2}(\mathrm{g})$ assuming that we initially started with HI (g) and allowed it to reach equilibrium at $700 K ?$
Ans. $\frac{1}{54.8}=\frac{x^{2}}{(0.5)^{2}} \Rightarrow x^{2}=\frac{0.5 \times 0.5}{54.8} \Rightarrow x=0.0675 \mathrm{mol} L^{-1}$
Q. Reaction between nitrogen and oxygen takes place as following $: 2 \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})$ If a mixture of $0.482 \mathrm{mol}$ of $\mathrm{N}_{2}$ and $0.933 \mathrm{mol}$ of $\mathrm{O}_{2}$ is placed in a reaction vessel of volume $10 L$ and allowed to form $\mathrm{N}_{2} \mathrm{O}$ at a temperature for which $\mathrm{K}_{\mathrm{c}}=2.0 \times 10^{-37}$ determine the composition of equilibrium mixture
Ans. $2 N_{2}(g)+O_{2}(g) \rightleftharpoons 2 N_{2} O(g)$ Concentration of $N_{2}=0.482 \mathrm{mol}$ Molar concentration of $N_{2}$ at equilibrium $=\frac{0.482}{10 L}=0.0482 \mathrm{mol} \mathrm{L}^{-1}$ $=\frac{0.482}{10 L}=0.0482 \mathrm{mol} L^{-1}$ Molar concentration of $O_{2}$ at equilibrium $=\frac{0.953}{10 L}=0.0933 \mathrm{mol} L^{-1}$ $K_{c}=\frac{\left[N_{2} \mathrm{O}\right]^{2}}{\left[N_{2}\right]^{2}\left[\mathrm{O}_{2}\right]^{1}} \Rightarrow 2.0 \times 10^{-37}=\frac{\left[N_{2} \mathrm{O}\right]^{2}}{[0.0482]^{2}[0.0933]}$ $\left[N_{2} \mathrm{O}\right]=6.6 \times 10^{-21} \mathrm{mol} L^{-1}$
Q. A sample of pure $\mathrm{PCl}_{5}$ was introduced into an evacuated vessel at $473 \mathrm{K.}$ After equilibrium was attained, concentration of $\mathrm{PCl}_{5}$ was found to be $0.5 \times 10^{-1} .$ If value of $\mathrm{K}$ is $8.3 \times 10^{-3},$ what are the concentrations of $\mathrm{PCl}_{3}$ and $\mathrm{Cl}_{2}$ at equilibrium?
Ans.
Q. For the reaction $C H_{4}(g)+2 H_{2} S(g) \rightleftharpoons C S_{2}(g)+4 H_{2}(g)$ at $1173 K$ the magnitude of the equilibrium constant, $K_{c}$ is $3.6 .$ For each of the following compositions, decide whether reaction mixture is at equilibrium. If it is not, decide which direction the reaction should go: (i) $\left[\mathrm{CH}_{4}\right]=1.07 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{S}\right]=1.20 \mathrm{M},\left[\mathrm{CS}_{2}\right]=0.90 \mathrm{M},\left[\mathrm{H}_{2}\right]=1.78 \mathrm{M}$ (ii) $\left[C H_{4}\right]=1.45 M,\left[H_{2} S\right]=1.29 M,\left[C S_{2}\right]=1.25 M,\left[H_{2}\right]=1.75 M$
Ans. $C H_{4}(g)+2 H_{2} S(g) \rightleftharpoons C S_{2}(g)+4 H_{2}(g)$ (i) $\quad Q_{c}=\frac{\left[C S_{2}\right]\left[H_{2}\right]^{4}}{\left[C H_{4}\right]\left[H_{2} S\right]^{2}}=\frac{[0.90][1.78]^{4}}{[1.07][1.20]^{2}}=\frac{10.03 \times 0.9}{1.44 \times 1.07}$ $K_{c}=3.6 \quad Q_{c}=\frac{9.27}{1.54}=6.02$ since $Q_{c}>K_{c},$ the equilibrium will shift in backward direction. (ii) $\quad Q_{c}=\frac{[1.25][1.75]^{4}}{[1.45][1.29]^{2}}=\frac{9.38 \times 1.25}{1.45 \times 1.66}=\frac{11.725}{2.407}=4.86$ Since , the equilibrium will shift in backward direction.
Q. Consider the following equilibrium: $P C l_{5}(g) \rightleftharpoons P C l_{3}(g)+C l_{2}(g)$ Initially, $0.0646 \mathrm{M} \mathrm{PCl}_{5}$ is introduced into a reaction vessel $350 \mathrm{K} .$ At equilibrium, $0.0444 \mathrm{M} \mathrm{Cl}_{2}$ is found. Calculate $K_{c}$ and $K_{p}$
Ans.
Q. A mixture of $1.57 \mathrm{mol}$ of $N_{2}, 1.92$ mol of $\mathrm{H}_{2}$ and $8.13 \mathrm{mol}$ of $N H_{3}$ is introduced into a $20 L$ reaction vessel at 500 K. At this temperature, the equilibrium constant, $K_{c}$ for the reaction, $N_{2}(g)+3 H_{2}(g) \rightleftharpoons 2 N H_{3}(g)$ is $1.7 \times 10^{2}$ Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction?
Ans.
Q. Equilibrium constant, $K_{c}$ for the reaction $N_{2}(g)+3 H_{2}(g) \rightleftharpoons 2 N H_{3}(g)$ at $500 K$ is 0.061 At a panticular time, the analysis shows that composition of the reaction mixture is $30 \mathrm{mol}^{-1} N_{2}, 20 \mathrm{moL}^{-1} \mathrm{H}_{2}$ and $0.5 \mathrm{mol} L^{-1} \mathrm{NH}_{3} .$ Is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?