Equilibrium
Unit 7
Equilibrium
I. Multiple Choice Questions (Type-I)
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
$NH_4 Cl(s) \rightarrow NH_3(g)+HCl(g)$
(i) 1
(ii) 0.5
(iii) 1.5
(iv) 2
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Answer: (iv)2. For the reaction $H_2(g)+I_2(g) \rightarrow 2 HI(g)$, the standard free energy is $\Delta G^{\ominus}>0$. The equilibrium constant $(K)$ would be ———-.
(i) $K=0$
(ii) $K>1$
(iii) $K=1$
(iv) $K<1$
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Answer: (iv)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 dynamic but stable condition.
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Answer: (iii)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 $K_c$ for the reaction $PCl_5(g) \rightarrow PCl_3(g)+Cl_2(g)$ will be
(i) $1.8 \times 10^{3} mol L^{-1}$
(ii) $1.8 \times 10^{-3}$
(iii) $1.8 \times 10^{-3} L mol^{-1}$
(iv) $0.55 \times 10^{4}$
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Answer: (ii)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 catalyst the equilibrium constant value is not affected.
(iv) Equilibrium constant for a reaction with negative $\Delta H$ value decreases as the temperature increases.
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Answer: (ii)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.
$[Co(H_2 O)_6]^{3+}(a q)+4 Cl^{-}(a q) \rightarrow[CoCl_4]^{2-}(a q)+6 H_2 O(l)$
(pink) (blue)
(i) $\Delta H>0$ for the reaction
(ii) $\Delta H<0$ for the reaction
(iii) $\Delta H=0$ for the reaction
(iv) The sign of $\Delta H$ cannot be predicted on the basis of this information.
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Answer: (i)7. The $pH$ of neutral water at $25^{\circ} C$ 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 $60^{\circ} C$ ?
(i) Equal to 7.0
(ii) Greater than 7.0
(iii) Less than 7.0
(iv) Equal to zero
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Answer: (iii)8. The ionisation constant of an acid, $K_a$, is the measure of 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
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Answer: (iv)9. $K _{a_1}, K _{a_2}$ and $K _{a_3}$ are the respective ionisation constants for the following reactions.
$H_2 S \rightarrow H^{+}+HS$
$HS^{-} \rightarrow H^{+}+S^{2-}$
$H_2 S \rightarrow 2 H^{+}+S^{2-}$
The correct relationship between $K _{a_1}, K _{a_2}$ and $K _{a_3}$ is
(i) $K _{a_3}=K _{a_1} \times K _{a_2}$
(ii) $K _{a_3}=K _{a_1}+K _{a_2}$
(iii) $K _{a_3}=K _{a_1}-K _{a_2}$
(iv) $K _{a_3}=K _{a_1} / K _{a_2}$
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Answer: (i)10. Acidity of $BF_3$ 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.
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Answer: (iii)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) $0.1 mol dm^{-3} CH_4 COONa$ and $0.1 mol dm^{-3} NaOH$
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Answer: (iii)12. In which of the following solvents is silver chloride most soluble?
(i) $0.1 mol dm^{-3} AgNO_3$ solution
(ii) $0.1 mol dm^{-3} HCl$ solution
(iii) $H_2 O$
(iv) Aqueous ammonia
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Answer: (iv)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
(iv) 3.0
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Answer: (i)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
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Answer: (iii)15. Which of the following options will be correct for the stage of half completion of the reaction $A \rightarrow B$.
(i) $\Delta G^{\ominus}=0$
(ii) $\Delta G^{\ominus}>0$
(iii) $\Delta G^{\ominus}<0$
(iv) $\Delta G^{\ominus}=-R T \ln 2$
Show Answer
Answer: (i) $\Delta G^{\ominus}=0$
Justification : $\Delta G^{\ominus}=-R T \ln K$
At the stage of half completion of reaction $[A]=[B]$, Therefore, $K=1$.
Thus, $\Delta G^{\ominus}=0$
16. On increasing the pressure, in which direction will the gas phase reaction proceed to re-establish equilibrium, is predicted by applying the Le Chatelier’s principle. Consider the reaction.
$ N_2(g)+3 H_2(g) \rightarrow 2 NH_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 same
(ii) $K$ will decrease
(iii) $K$ will increase
(iv) $K$ will increase initially and decrease when pressure is very high
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Answer: (i), Justification: According to Le-Chatelier’s principle, at constant temperature, the equilibrium composition will change but $K$ will remain same.17. What will be the correct order of vapour pressure of water, acetone and ether at $30^{\circ} C$. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point?
(i) Water $<$ ether $<$ acetone
(ii) Water $<$ acetone $<$ ether
(iii) Ether $<$ acetone $<$ water
(iv) Acetone $<$ ether $<$ water
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Answer: (ii)18. At $500 K$, equilibrium constant, $K_c$, for the following reaction is 5 .
$\frac{1}{2} H_2(g)+\frac{1}{2} I_2(g) \rightarrow HI(g)$
What would be the equilibrium constant $K_c$ for the reaction
$2 HI(g) \rightarrow H_2(g)+I_2(g)$ (i) 0.04
(ii) 0.4
(iii) 25
(iv) 2.5
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Answer: (i)19. In which of the following reactions, the equilibrium remains unaffected on addition of small amount of argon at constant volume?
(i) $H_2(g)+I_2(g) \rightarrow 2 HI(g)$
(ii) $PCl_5(g) \rightarrow PCl_3(g)+Cl_2(g)$
(iii) $\quad N_2(g)+3 H_2(g) \rightarrow 2 NH_3(g)$
(iv) The equilibrium will remain unaffected in all the three cases.
II. Multiple Choice Questions (Type-II)
In the following questions two or more options may be correct.
Show Answer
Answer: (iv)
II. Multiple Choice Questions (Type-II)
20. For the reaction $N_2 O_4(g) \rightarrow 2 NO_2(g)$, 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.
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Answer: (i), (iii) and (iv)
Justification : (i) Kincreases with increase in temperature.
(iii) $Q>K$, Therefore, reaction proceeds in the backward direction.
(iv) $\Delta n>0$, Therefore, $\Delta S>0$.
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
III. Short Answer Type
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Answer: (i) and (iv)
III. Short Answer Type
22. The ionisation of hydrochloric in water is given below:
$HCl(aq)+H_2 O(l) \rightarrow H_3 O^{+}(aq)+Cl^{-}(aq)$
Label two conjugate acid-base pairs in this ionisation.
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Answer: |$HCl$|$Cl^{-}$| | :—– | :—– | |acid|conjugate base| |$H_2 O$|$H_3 O^{+}$| |base |conjugate acid|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 concentration of sodium chloride?
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Answer: * Sugar does not ionise in water but $NaCl$ ionises completely in water and produces $Na^{+}$and $Cl^{-}$ions.
- Conductance increases with increase in concentration of salt due to release of more ions.
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?
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Answer: $BF_3$ acts as a Lewis acid as it is electron deficient compound and coordinate bond is formed as given below :
$H_3 N: \to BF_3$
25. Ionisation constant of a weak base $MOH$, is given by the expression
$ K_b=\frac{[M^{+}][OH^{-}]}{[MOH]} $
Values of ionisation constant of some weak bases at a particular temperature are given below:
| Base | Dimethylamine | Urea | Pyridine | Ammonia |
|---|---|---|---|---|
| $K_b$ | $5.4 \times 10^{-4}$ | $1.3 \times 10^{-14}$ | $1.77 \times 10^{-9}$ | $1.77 \times 10^{-5}$ |
Arrange the bases in decreasing order of the extent of their ionisation at equilibrium. Which of the above base is the strongest?
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Answer: * Order of extent of ionisation at equilibrium is as follows :
Dimethylamine $>$ Ammonia $>$ Pyridine $>$ Urea
- Since dimethylamine will ionise to the maximum extent it is the strongest base out of the four given bases.
26. Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases?
$OH^{-}, RO^{-}, CH_3 COO^{-}, Cl^{-}$
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Answer: $RO^{-}>OH^{-}>CH_3 COO^{-}>Cl^{-}$27. Arrange the following in increasing order of $pH$.
$ KNO_3(a q), CH_3 COONa(a q), NH_4 Cl(a q), C_6 H_5 COONH_4(a q) $
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Answer: $NH_4 Cl<C_6 H_5 COONH_4<KNO_3<CH_3 COONa$28. The value of $K_c$ for the reaction $2 HI(g) \rightarrow H_2(g)+I_2(g)$ is $1 \times 10^{-4}$
At a given time, the composition of reaction mixture is
$[HI]=2 \times 10^{-5} mol, \quad[H_2]=1 \times 10^{-5} mol$ and $[I_2]=1 \times 10^{-5} mol$
In which direction will the reaction proceed?
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Answer: At a given time the reaction quotient $Q$ for the reaction will be given by the expression.
$ \begin{aligned} Q & =\frac{[H_2][I_2]}{[H I]^{2}} \\ & =\frac{1 \times 10^{-5} \times 1 \times 10^{-5}}{(2 \times 10^{-5})^{2}}=\frac{1}{4} \\ & =0.25=2.5 \times 10^{-1} \end{aligned} $
As the value of reaction quotient is greater than the value of $K_c$ i.e. $1 \times 10^{-4}$ the reaction will proceed in the reverse direction.
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.
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Answer: Concentration of $10^{-8} mol dm^{-3}$ indicates that the solution is very dilute. Hence, the contribution of $H_3 O^{+}$concentration from water is significant and should also be included for the calculation of $pH$.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 a 100 times?
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Answer: (i) $pH=5$
$[H^{+}]=10^{-5} mol L^{-1}$
On 100 times dilution
$[H^{+}]=10^{-7} mol L^{-1}$
On calculating the $pH$ using the equation $pH=-\log [H^{+}]$, value of $pH$ comes out to be 7. It is not possible. This indicates that solution is very dilute. Hence,
Total hydrogen ion concentration
$=[H^{+}]$
$ \begin{aligned} & =\begin{matrix} \text{ Contribution of } \\ H_3 O^{+} \text{ion } \\ \text{ concentration } \\ \text{ of acid } \end{matrix} \\ & =10^{-7}+10^{-7} . \end{aligned}+\begin{aligned} & \text{ Contribution of } \\ & H_3 O^{+} \text{ion } \\ & \text{ concentration } \\ & \text{ of water } \end{aligned} $
$pH=2 \times 10^{-7}=7-\log 2=7-0.3010=6.6990$
31. A sparingly soluble salt gets precipitated only when the product of 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 $H_2 SO_4$.
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Answer: At $t=0$
At equilibrium in water
At equilibrium in the presence of sulphuric acid $\begin{matrix} \mathbf{B a S O}_4(\mathbf{s}) & \rightarrow Ba^{2+}(a q) & +\underset{4}{\mathbf{S O}_4^{2-}}(a q) \\ 1 & 0 & 0\end{matrix} $
$\begin{matrix} 1-S & S & S\end{matrix} $
$S \quad(S+0.01)$
$K _{sp}$ for $BaSO_4$ in water $=[Ba^{2+}][SO_4^{2-}]=(S)(S)=S^{2}$
But $S=8 \times 10^{-4} mol dm^{-3}$
$\therefore K _{sp}=(8 \times 10^{-4})^{2}=64 \times 10^{-8}$ ……(1)
The expression for $K _{sp}$ in the presence of sulphuric acid will be as follows : $K _{\text{sp }}=(S)(S+0.01)$ ……. (2)
Since value of $K _{sp}$ will not change in the presence of sulphuric acid, therefore from (1) and (2)
(S) $(S+0.01)=64 \times 10^{-8}$
$S^{2}+0.01 S=64 \times 10^{-8}$
$S^{2}+0.01 S-64 \times 10^{-8}=0$
$ \begin{aligned} & S=\frac{-0.01 \pm \sqrt{(0.01)^{2}+(4 \times 64 \times 10^{-8})}}{2} \\ &=\frac{-0.01 \pm \sqrt{10^{-4}+(256 \times 10^{-8})}}{2} \\ &=\frac{-0.01 \pm \sqrt{10^{-4}(1+256 \times 10^{-2})}}{2} \\ &=\frac{-0.01 \pm 10^{-2} \sqrt{1+0.256}}{2} \\ &=\frac{-0.01 \pm 10^{-2} \sqrt{1.256}}{2} \\ &=\frac{-10^{-2}+(1.12 \times 10^{-2})}{2} \\ &=\frac{(-1+1.12) \times 10^{-2}}{2}=\frac{0.12}{2} \times 10^{-2} \\ &=6 \times 10^{-4} mol dm^{-3} \end{aligned} $
32. $pH$ of $0.08 mol dm^{-3} HOCl$ solution is 2.85. Calculate its ionisation constant.
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Answer: $pH$ of $HOCl=2.85$
But, $-pH=\log [H^{+}]$
$\therefore-2.85=\log [H^{+}]$
$\overline{3} .15=\log [H^{+}]$
$[H^{+}]=1.413 \times 10^{-3}$
For weak mono basic acid $[H^{+}]=\sqrt{K_a \times C}$
$ \begin{aligned} K_a & =\frac{[H^{+}]^{2}}{C}=\frac{(1.413 \times 10^{-3})^{2}}{0.08} \\ & =24.957 \times 10^{-6}=2.4957 \times 10^{-5} \end{aligned} $
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.
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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 concentration of solution $B=10^{-4} mol L^{-1}$
On mixing one litre of each solution, total volume $=1 L+1 L=2 L$
Amount of $H^{+}$ions in $1 L$ of Solution $A=$ Concentration $\times$ volume $V$
$ =10^{-6} mol \times 1 L $
Amount of $H^{+}$ions in $1 L$ of solution $B=10^{-4} mol \times 1 L$
$\therefore$ Total amount of $H^{+}$ions in the solution formed by mixing solutions $A$ and $B$ is $(10^{-6} mol+10^{-4} mol)$
This amount is present in 2L solution.
$ \begin{aligned} & \therefore \text{ Total }[H^{+}]=\frac{10^{-4}(1+0.01)}{2}=\frac{1.01 \times 10^{-4}}{2} mol L^{-1}=\frac{1.01 \times 10^{-4}}{2} mol L^{-1} \\ & =0.5 \times 10^{-4} mol L^{-1} \\ & =5 \times 10^{-5} mol L^{-1} \\ & pH=-\log [H^{+}]=-\log (5 \times 10^{-5}) \\ & =-[\log 5+(-5 \log 10)] \\ & =-\log 5+5 \\ & =5-\log 5 \\ & =5-0.6990 \\ & =4.3010=4.3 \end{aligned} $
34. The solubility product of $Al(OH)_3$ is $2.7 \times 10^{-11}$. Calculate its solubility in $gL^{-1}$ and also find out $pH$ of this solution. (Atomic mass of $Al=27 u$ ).
Show Answer
Answer: Let $S$ be the solubility of $Al(OH)_3$.
$ Al(OH)_3 \leftrightharpoons Al^{3+}(aq)+30 H^{-}(aq) $
| Concentration of species at $t=0$ | 1 | 0 | 0 |
|---|---|---|---|
| Concentration of various species at equilibrium | 1-S | S | 3S |
$ \begin{aligned} & K _{s p}=[Al^{3+}][OH^{-}]^{3}=(S)(3 S)^{3}=27 S^{4} \\ & S^{4}=\frac{K _{s p}}{27}=\frac{27 \times 10^{-11}}{27 \times 10}=1 \times 10^{-12} \\ & S=1 \times 10^{-3} mol L^{-1} \end{aligned} $
(i) Solubility of $Al(OH)_3$
Molar mass of $Al(OH)_3$ is $78 g$. Therefore,
Solubility of $Al(OH)_3$ in $g L^{-1}=1 \times 10^{-3} \times 78 g L^{-1}=78 \times 10^{-3} g L^{-1}$
$ =7.8 \times 10^{-2} g L^{-1} $
(ii) $\mathbf{p H}$ of the solution
$ \begin{aligned} & S=1 \times 10^{-3} mol L^{-1} \\ & {[OH^{-}]=3 S=3 \times 1 \times 10^{-3}=3 \times 10^{-3}} \\ & \text{ pOH }=3-\log 3 \\ & \text{ pH }=14-\text{ pOH }=11+\log 3=11.4771 \end{aligned} $
35. Calculate the volume of water required to dissolve $0.1 g$ lead (II) chloride to get a saturated solution. $(K _{sp}.$ of $PbCl_2=3.2 \times 10^{-8}$, atomic mass of $.Pb=207 u)$.
Show Answer
Answer: $K _{\text{sp }}$ of $PbCl_2=3.2 \times 10^{-8}$
Let $S$ be the solubility of $PbCl_2$.
$ PbCl_2(s) \leftrightharpoons Pb^{2+}(aq)+2 Cl^{-}(aq) $
| Concentration of species at $t=0$ | 1 | 0 | 0 |
|---|---|---|---|
| Concentration of various species at equilibrium | 1-S | S | 2S |
$K _{\text{sp }}=[Pb^{2+}][Cl^{-}]^{2}=(S)(2 S)^{2}=4 S^{3}$
$K _{sp}=4 S^{3}$
$S^{3}=\frac{K _{sp}}{4}=\frac{3.2 \times 10^{-8}}{4} mol L^{-1}=8 \times 10^{-9} mol L^{-1}$
$S=\sqrt[3]{8 \times 10^{-9}}=2 \times 10^{-3} mol L^{-1} \quad \therefore S=2 \times 10^{-3} mol L^{-1}$
Molar mass of $PbCl_2=278$
$\therefore$ Solubility of $PbCl_2$ in $g L^{-1}=2 \times 10^{-3} \times 278 g L^{-1}$
$ \begin{aligned} & =556 \times 10^{-3} g L^{-1} \\ & =0.556 g L^{-1} \end{aligned} $
To get saturated solution, $0.556 g$ of $PbCl_2$ is dissolved in $1 L$ water.
$0.1 g PbCl_2$ is dissolved in $\frac{0.1}{0.556} L=0.1798 L$ water.
To make a saturated solution, dissolution of $0.1 g PbCl_2$ in $0.1798 L \approx 0.2 L$ of water will be required.
36. A reaction between ammonia and boron trifluoride is given below:
$: NH_3+BF_3 \longrightarrow H_3 N: BF_3$
Identify the acid and base in this reaction. Which theory explains it? What is the hybridisation of $B$ and $N$ in the reactants?
37. Following data is given for the reaction: $CaCO_3(s) \to CaO(s)+CO_2(g)$
$ \begin{aligned} & \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} \end{aligned} $
Predict the effect of temperature on the equilibrium constant of the above reaction.
IV. Matching Type
Show Answer
Answer: $\Delta_r H^{\ominus}=\Delta_f H^{\ominus}[CaO(s)]+\Delta_f H^{\ominus}[CO_2(g)]-\Delta_f H^{\ominus}[CaCO_3(s)]$
$\therefore \Delta_r H^{\ominus}=178.3 kJ mol^{-1}$
The reaction is endothermic. Hence, according to Le-Chatelier’s principle, reaction will proceed in forward direction on increasing temperature.
IV. Matching Type
38. Match the following equilibria with the corresponding condition
| (i) Liquid $\rightarrow$ Vapour | (a) Saturated solution |
|---|---|
| (ii) Solid $\leftrightharpoons$ Liquid | (b) Boiling point |
| (iii) Solid $\leftrightharpoons$ Vapour | (c) Sublimation point |
| (iv) Solute $(s) \leftrightharpoons$ Solute (solution) | (d) Melting point |
| (e) Unsaturated solution |
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Answer: (i) $\to$ (b) $\quad$ (ii) $\to$ (d) $\quad$ (iii) $\to$ (c) $\quad$ (iv) $\to$ (a)39. For the reaction: $N_2(g)+3 H_2(g) \leftrightharpoons 2 NH_3(g)$
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 $K_c$ are written in Column II. Match the following reactions with the corresponding equilibrium constant
| Column I (Reaction) | Column II (Equilibrium constant) |
|---|---|
| (i) $2 N_2(g)+6 H_2(g) \rightarrow 4 NH_3(g)$ | (a) $2 K_c$ |
| (ii) $\quad 2 NH_3(g) \rightarrow N_2(g)+3 H_2(g)$ | (b) $\quad K_c^{\frac{1}{2}}$ |
| (iii) $\quad \frac{1}{2} N_2(g)+\frac{3}{2} H_2(g) \rightarrow NH_3(g)$ | (c) $\frac{1}{K_c}$ |
| (d) $K_c^{2}$ |
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Answer: (i) $\to$ (d) $\quad$ (ii) $\to$ (c) $\quad$ (iii) $\to$ (b)40. Match standard free energy of the reaction with the corresponding equilibrium constant
| (i) $\Delta G^{\ominus}>0$ | (a) $K>1$ |
|---|---|
| (ii) $\Delta G^{\ominus}<0$ | (b) $K=1$ |
| (iii) $\Delta G^{\ominus}=0$ | (c) $K=0$ |
| (d) $K<1$ |
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Answer: (i) $\to$ (d) $\quad$ (ii) $\to$ (a) $\quad$ (iii) $\to$ (b)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) $H_2 CO_3$ |
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Answer: (i) $\to$ (b) $\quad$ (ii) $\to$ (e) $\quad$ (iii) $\to$ (c) $\quad$ (iv) $\to$ (d)42. Match the following graphical variation with their description
A
(i)
(ii)
(iii)
(a) Variation in product concentration with time
(b) Reaction at equilibrium
(c) Variation in reactant concentration with time
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Answer: (i) $\to$ (c) $\quad$ (ii) $\to$ (a) $\quad$ (iii) $\to$ (b)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) $\Delta G<0, K>1$ |
V. Assertion and Reason Type
In the following questions 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.
Show Answer
Answer: (i) $\to$ (b) and (c) $\quad$ (ii) $\to$ (d) $\quad$ (iii) $\to$ (a)
V. Assertion and Reason Type
44. 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 periodic table, $H-A$ bond strength is a more important factor in determining acidity of an 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.
45. 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 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.
46. 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 correct explanation of $A$.
(ii) Both $A$ and $R$ are true but $R$ is not correct explanation of $A$.
47. 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 correct explanation of $A$.
(ii) Both $A$ and $R$ are true but $R$ is not correct explanation of $A$.
(iii) $A$ is true but $R$ is false.
(iv) Both $A$ and $R$ are false.
48. Assertion (A) : Aqueous solution of ammonium carbonate is basic.
Reason (R) : Acidic/basic nature of a salt solution of a salt of 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 correct explanation of $A$.
(ii) Both $A$ and $R$ are true but $R$ is not correct explanation of $A$.
(iii) $A$ is true but $R$ is false.
(iv) Both $A$ and $R$ are false.
49. 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 correct explanation of $A$.
(ii) Both $A$ and $R$ are true but $R$ is not correct explanation of $A$.
(iii) $A$ is false but $R$ is true.
(iv) Both $A$ and $R$ are false.
50. Assertion (A): In the dissociation of $PCl_5$ at constant pressure and temperature addition of helium at equilibrium increases the dissociation of $PCl_5$.
Reason (R) : Helium removes $Cl_2$ from the field of action.
(i) Both $A$ and $R$ are true and $R$ is correct explanation of $A$.
(ii) Both $A$ and $R$ are true but $R$ is not correct explanation of $A$.
(iii) $A$ is true but $R$ is false.
(iv) Both $A$ and $R$ are false.
VI. Long Answer Type
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.
Show Answer
Answer: (i) G $ _{c}<K_c$
(ii) $Q_c>K_c$
(iii) $\beta_c=K_c$
where, $Q_c$ is reaction quotient in terms of concentration and $K_c$ is equilibrium constant.
52. On the basis of Le Chatelier principle explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction.
$N_2(g)+3 H_2(g) \rightarrow 2 NH_3(g) \quad \Delta H=-92.38 kJ mol^{-1}$
What will be the effect of addition of argon to the above reaction mixture at constant volume?
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.
Show Answer
Answer: $[.$ Hint : $A_x^{p+} B_y^{q-} \rightarrow x A^{p+}(aq)+y B^{q-}(aq)$
$S$ moles of $A_x B_y$ dissolve to give $x S$ moles of $A^{p+}$ and $y S$ moles of $B^{q}$.]
54. Write a relation between $\Delta G$ 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 increase in pressure in terms of reaction quotient $Q$. for the reaction: $CO(g)+3 H_2(g) \rightarrow CH_4(g)+H_2 O(g)$
Show Answer
Answer: $\Delta G=\Delta G^{\ominus}+R T \ln Q$
$\Delta G^{\ominus}=$ Change in free energy as the reaction proceeds
$\Delta G=$ Standard free energy change
$Q=$ Reaction quotient
$R=$ Gas constant
$T=$ Absolute temperature
Since $\Delta G^{\ominus}=-R T \ln K$
$\therefore \quad \Delta G=-R T \ln K+R T \ln Q=RT \ln \frac{Q}{K}$
If $Q<K, \Delta G$ will be negative. Reaction proceeds in the forward direction.
If $Q=K, \Delta G=0$, no net reaction.
[Hint: Next relate $Q$ with concentration of $CO, H_2, CH_4$ and $H_2 O$ in view of reduced volume (increased pressure). Show that $Q<K$ and hence the reaction proceeds in forward direction.]
