Chapter 7: Equilibrium

Q1Written answer

We know that the relationship between ${{\mathbf{K}}_{\mathbf{c}}}$ and $\mathbf{Kp}$ is $\mathbf{Kp}={{\mathbf{K}}_{\mathbf{c}}}{{\left( \mathbf{RT} \right)}^{\Delta \mathbf{n}}}$ What would be the value of ∆n for the reaction-

$\mathbf{N}{{\mathbf{H}}_{\mathbf{4}}}\mathbf{Cl}\left( \mathbf{s} \right)\to \mathbf{N}{{\mathbf{H}}_{\mathbf{3}}}\left(\mathbf{g}\right)+\mathbf{HCl}\text{ }\left( \mathbf{g} \right)$

(i) 1

(ii) 0.5

(iii) 1.5

(iv) 2

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Worked solution

Ans: $\text{ }\!\!\Delta\!\!\text{ n}$= number of gaseous moles of products -number of gaseous moles of reactants

= 2-0= 2

Hence, option(iv) is the correct answer.

Q2Written answer

For the reaction ${{\mathbf{H}}_{\mathbf{2}}}\left( \mathbf{g} \right)+{{\mathbf{I}}_{\mathbf{2}}}\left( \mathbf{g} \right)\to \mathbf{2HI}\left( \mathbf{g} \right)$ the standard free energy is $\mathbf{\Delta }{{\mathbf{G}}^{\odot }}$> 0. The equilibrium constant (K) would be __________.

(i) K = 0

(ii) K > 1

(iii) K = 1

(iv) K < 1

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Worked solution

Ans: The given chemical reaction is:

${{\text{H}}_{\text{2}}}\left( \text{g} \right)\text{+}{{\text{I}}_{\text{2}}}\left( \text{g} \right)\to \text{2HI}\left( \text{g} \right)$

By, thermodynamical equation of equilibrium,

$\text{ }\!\!\Delta\!\!\text{ }{{\text{G}}^{\odot }}$=−RTlnK>0

⇒−RTlnK>0

Now, R>0&T>0

⇒−lnK>0

⇒K<1

Hence, option(iv) is the correct answer.

Q3Written answer

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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Ans: A physical equilibrium is having dynamic nature, both forward and reverse process occur at equal rates. Some kind of motion is always present. The individual molecules continuously move from one phase to another. Hence, option (iii) is the correct answer.

Q4Written answer

$\mathbf{PC}{{\mathbf{l}}_{\mathbf{5}}},\text{}\mathbf{PC}{{\mathbf{l}}_{\mathbf{3}}}\text{}\mathbf{and}\text{ }\mathbf{C}{{\mathbf{l}}_{\mathbf{2}}}$ are at equilibrium at 500K in a closed container and their concentrations are $\mathbf{0}.\mathbf{8}\text{}\times\text{}\mathbf{1}{{\mathbf{0}}^{\mathbf{3}\text{}}}\mathbf{mol}\text{}{{\mathbf{L}}^{\mathbf{1}}},\text{}\mathbf{1}.\mathbf{2}\text{ }\times \text{ }\mathbf{1}{{\mathbf{0}}^{\mathbf{3}}}\text{ }\mathbf{mol}\text{ }{{\mathbf{L}}^{\mathbf{1}}}$ $\text{ }\mathbf{and}\text{ }\mathbf{1}.\mathbf{2}\text{ }\times \text{ }\mathbf{1}{{\mathbf{0}}^{\mathbf{3}}}\text{ }\mathbf{mol}\text{ }{{\mathbf{L}}^{\mathbf{1}}}$ respectively. The value of ${{\mathbf{K}}_{\mathbf{c}}}$ for the reaction-

$\mathbf{PC}{{\mathbf{l}}_{\mathbf{5}\text{ }}}\left( \mathbf{g} \right)\to \mathbf{PC}{{\mathbf{l}}_{\mathbf{3}}}\text{ }\left( \mathbf{g} \right)\text{ }+\text{ }\mathbf{C}{{\mathbf{l}}_{\mathbf{2}}}\text{ }\left( \mathbf{g} \right)$will be

(i) 1.8 × $\mathbf{1}{{\mathbf{0}}^{\mathbf{3}\text{ }}}\mathbf{mol}\text{ }{{\mathbf{L}}^{\mathbf{1}}}$

(ii) 1.8 × $\mathbf{1}{{\mathbf{0}}^{-\mathbf{3}\text{ }}}$

(iii) 1.8 × $\mathbf{1}{{\mathbf{0}}^{\mathbf{3}}}\mathbf{L}\text{ }\mathbf{mo}{{\mathbf{l}}^{-1}}$

(iv) 0.55 × $\mathbf{1}{{\mathbf{0}}^{\mathbf{4}}}$

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Worked solution

Ans: The chemical equation is given as-

${{K}_{c}}=\frac{\left[ PC{{l}_{5}} \right]\left[ PC{{l}_{3}} \right]}{\left[ C{{l}_{2}} \right]}$

=$\text{0}\text{.8 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{--3 }}}\text{mol }{{\text{L}}^{\text{--1}}}\times \text{ 1}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{--3}}}\text{ mol }{{\text{L}}^{\text{--1}}}$/$\text{1}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{--3}}}\text{ mol }{{\text{L}}^{\text{--1}}}$

=$\text{1}\text{.8}\!\!\times\!\!\text{1}{{\text{0}}^{\text{-3}}}\text{mol}{{\text{L}}^{\text{-1}}}$.

Hence, option(iii) is the correct answer.

Q5Written answer

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 ∆H value decreases as the temperature increases.

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Ans: Oxalic acid, reacts with ferric ions to form the stable complex ion$\left[ \text{Fe}{{\left( {{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ } \right)}_{\text{3}}} \right]{{\text{ }}^{\text{3 -}}}$, thus decreasing the concentration of free ferric ions. Because the concentration of ${{\left[ \text{Fe}\left( \text{SCN} \right) \right]}^{\text{2+}}}$decreases, as a result the intensity of red colour decreases. Hence, option (ii) is the correct answer.

Q6Written answer

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.

\[[Co (H_{2}O)_{6} ]_{3} +(aq) +4Cl^{-}(aq) \rightarrow [CoCl_{4}]^{2-}(aq) + 6H_{2}O (I) \]

(i) ∆H > 0 for the reaction

(ii) ∆H < 0 for the reaction

(iii) ∆H = 0 for the reaction

(iv) The sign of ∆H cannot be predicted on the basis of this information.

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Worked solution

Ans: On cooling the mixture reaction moves towards backward direction it means it is an endothermic reaction i.e., $\Delta \text{H}$> 0. Hence, option (i) is the correct answer.

Q7Written answer

The pH of neutral water at 25°C is 7.0. As the temperature increases, ionisation of water increases, however, the concentration of ${{\mathbf{H}}^{+}}$ions and $\mathbf{O}{{\mathbf{H}}^{-}}$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

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Worked solution

Ans: As Kw increases Ans: As Kw increases $\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]>{{10}^{-14}}.$ As $\left[ {{H}^{+}} \right]=\left[ O{{H}^{-}} \right]$ or $\left[ {{H}^{+}} \right]=\left[ O{{H}^{-}} \right]$ $\left[ {{H}^{+}} \right]2={{110}^{-14}}$ or $\left[ {{H}^{+}} \right]>{{10}^{-7}}$

M pH < 7. Hence, option(iii) is the correct answer.

Q8Written answer

The ionisation constant of an acid, ${{\mathbf{K}}_{\mathbf{a}}}$, is the measure of strength of an acid. The ${{\mathbf{K}}_{\mathbf{a}}}$ values of acetic acid, hypochlorous acid and formic acid are respectively. Which of the following orders of pH of 0.1 mol $\mathbf{d}{{\mathbf{m}}^{\mathbf{3}}}$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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Ans: Higher the value of ${{\text{K}}_{\text{a}}}$ higher will be the acidic strength ${{\text{K}}_{\text{a}}}$ value for formic acid is highest and that will be of hypochlorous acid is lowest. Hence, option(iv) is the correct answer.

Q9Written answer

${{\mathbf{K}}_{\mathbf{a}}}_{_{1\text{ }}},{{\mathbf{K}}_{\mathbf{a}}}_{_{\mathbf{2 }}}\mathbf{and}{{\mathbf{K}}_{\mathbf{a}}}_{_{\mathbf{3}}}$ are the respective ionisation constants for the following reactions.

${{\mathbf{H}}_{\mathbf{2}}}\mathbf{S }\to {{\mathbf{H}}^{\mathbf{+}}}\mathbf{ + H}{{\mathbf{S}}^{\mathbf{}}}$

$\mathbf{H}{{\mathbf{S}}^{\mathbf{}}}\mathbf{ }\to \mathbf{ }{{\mathbf{H}}^{\mathbf{+}}}\mathbf{ + }{{\mathbf{S}}^{\mathbf{2}}}\mathbf{ }$

${{\mathbf{H}}_{\mathbf{2}}}\mathbf{S }\to \mathbf{2}{{\mathbf{H}}^{\mathbf{+}}}\mathbf{ + }{{\mathbf{S}}^{\mathbf{2}}}$

The correct relationship between ${{\mathbf{K}}_{\mathbf{a}}}_{_{1\text{ }}}$ $,{{\mathbf{K}}_{\mathbf{a}}}_{_{\mathbf{2 }}}$and ${{\mathbf{K}}_{\mathbf{a}}}_{_{\mathbf{3}}}$

$\left( \mathbf{i} \right)\text{ }{{\mathbf{K}}_{\mathbf{a}}}_{_{\mathbf{3}}}\text{ }=\text{ }{{\mathbf{K}}_{\mathbf{a}}}_{_{1\text{ }}}\text{ }\times \text{ }{{\mathbf{K}}_{\mathbf{a}}}_{_{\mathbf{2 }}}$

$\left( \mathbf{ii} \right)\text{ }{{\mathbf{K}}_{\mathbf{a}}}_{_{\mathbf{3}}}=\text{ }{{\mathbf{K}}_{\mathbf{a}}}_{_{1\text{ }}}\text{ }+{{\mathbf{K}}_{\mathbf{a}}}_{_{\mathbf{2 }}}$

$\left( \mathbf{iii} \right){{\mathbf{K}}_{\mathbf{a}}}_{_{\mathbf{3}}}=\text{ }{{\mathbf{K}}_{\mathbf{a}}}_{_{1\text{ }}}{{\mathbf{K}}_{\mathbf{a}}}_{_{\mathbf{2 }}}$

$\left( \mathbf{iv} \right)\text{ }{{\mathbf{K}}_{\mathbf{a}}}_{_{\mathbf{3}}}\text{ }=\text{ }{{\mathbf{K}}_{\mathbf{a}}}_{_{1\text{ }}}/\text{ }{{\mathbf{K}}_{\mathbf{a}}}_{_{\mathbf{2 }}}$

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Ans:

${{\text{H}}_{\text{2}}}\text{S }\to {{\text{H}}^{\text{+}}}\text{ + H}{{\text{S}}^{\text{--}}}$ ${{\text{K}}_{\text{a}}}_{_{\text{1 }}}$

$\text{H}{{\text{S}}^{\text{--}}}\to \text{ }{{\text{H}}^{\text{+}}}\text{ + }{{\text{S}}^{\text{2--}}}\text{ }$ ${{\text{K}}_{\text{a}}}_{_{\text{2 }}}$

${{\text{H}}_{\text{2}}}\text{S}\to \text{2}{{\text{H}}^{\text{+}}}\text{ + }{{\text{S}}^{\text{2--}}}$ ${{\text{K}}_{\text{a}}}_{_{\text{3}}}$

${{\text{K}}_{{{\text{a}}_{1}}}}=\frac{\left[ \text{H}{{\text{S}}^{-}} \right]\left[ {{\text{H}}^{+}} \right]}{\left[ {{\text{H}}_{2}}\text{S} \right]}$.

${{\text{K}}_{{{\text{a}}_{2}}}}=\frac{\left[ {{\text{H}}^{+}} \right]\left[ {{\text{S}}^{2-}} \right]}{\left[ \text{H}{{\text{S}}^{-}} \right]}$

${{\text{K}}_{\text{a}}}_{_{\text{3}}}$= ${{\text{K}}_{\text{a}}}_{_{\text{2 }}}$.${{\text{K}}_{\text{a}}}_{_{\text{1 }}}$

Hence, option(i) is the correct answer.

Q10Written answer

Acidity of $\mathbf{B}{{\mathbf{F}}_{\mathbf{3}}}$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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Worked solution

Ans: $\text{B}{{\text{F}}_{\text{3}}}$ is an electron deficient species that is why it is a Lewis acid. (all Lewis acid can accept a pair of electrons). Hence, option (iii) is the correct answer.

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Multiple Choice Questions (Type-I)