Answer

(d) The relationship between $K_{p}$ and $K_{c}$ is

$$ K_{p}=K_{c}(R T)^{\Delta n} $$

where, $\Delta n=$ (number of moles of gaseous products) - (number of moles of gaseous reactants)

For the reaction,

$$ \begin{gathered} NH_{4} Cl(s) \rightleftharpoons NH_{3}(g)+HCl(g) \\ \Delta n=2-0=2 \end{gathered} $$

• Option (a) is incorrect because $\Delta n$ is not equal to 1. For the reaction $NH_{4}Cl(s) \rightleftharpoons NH_{3}(g) + HCl(g)$, the number of moles of gaseous products is 2 (1 mole of $NH_{3}$ and 1 mole of $HCl$), and the number of moles of gaseous reactants is 0. Therefore, $\Delta n = 2 - 0 = 2$.

• Option (b) is incorrect because $\Delta n$ is not equal to 0.5. For the reaction $NH_{4}Cl(s) \rightleftharpoons NH_{3}(g) + HCl(g)$, the number of moles of gaseous products is 2 (1 mole of $NH_{3}$ and 1 mole of $HCl$), and the number of moles of gaseous reactants is 0. Therefore, $\Delta n = 2 - 0 = 2$.

• Option (c) is incorrect because $\Delta n$ is not equal to 1.5. For the reaction $NH_{4}Cl(s) \rightleftharpoons NH_{3}(g) + HCl(g)$, the number of moles of gaseous products is 2 (1 mole of $NH_{3}$ and 1 mole of $HCl$), and the number of moles of gaseous reactants is 0. Therefore, $\Delta n = 2 - 0 = 2$.

Answer

(d) $\Delta G^{\ominus}$ and $K$ are related as

$$ \Delta G^{\ominus}=-R T \ln K_{C} $$

when $G^{\ominus}>0$ means $\Delta G^{\circ}$ is positive. This can be so only if $\ln K_{c}$ is negative i.e., $K_{c}<1$.

• Option (a) $K=0$: This is incorrect because if $K=0$, it would imply that the reaction does not proceed at all, meaning no products are formed. However, a positive $\Delta G^{\ominus}$ indicates that the reaction is not spontaneous, but it does not mean that the reaction does not occur at all.

• Option (b) $K>1$: This is incorrect because a $K>1$ would imply that the reaction favors the formation of products at equilibrium, which would correspond to a negative $\Delta G^{\ominus}$. Since $\Delta G^{\ominus}>0$ in this case, the reaction does not favor the formation of products.

• Option (c) $K=1$: This is incorrect because if $K=1$, it would imply that the reaction is at equilibrium with no net change in the concentrations of reactants and products, corresponding to $\Delta G^{\ominus}=0$. However, the given condition is $\Delta G^{\ominus}>0$, indicating that the reaction is not at equilibrium and does not favor the formation of products.

Answer

(c) At the stage of equilibria involving physical processes like melting of ice and freezing of water etc., process does not stop but the opposite processes i.e., forward and reverse process occur with the same rate.

• (a) Equilibrium is possible only in a closed system at a given temperature: This statement is actually correct. For equilibrium to be maintained, the system must be closed so that no matter is exchanged with the surroundings, and the temperature must be constant to ensure that the rates of the forward and reverse processes remain equal.

• (b) All measurable properties of the system remain constant: This statement is also correct. At equilibrium, properties such as pressure, temperature, and concentration remain constant over time because the rates of the forward and reverse processes are equal.

• (d) The opposing processes occur at the same rate and there is dynamic but stable condition: This statement is correct as well. At equilibrium, the forward and reverse processes occur at the same rate, leading to a dynamic but stable condition where the macroscopic properties of the system do not change.

Answer

(b) For the reaction,

$$ PCl_{5} \rightleftharpoons PCl_{3}+Cl_{2} $$

At $500 \mathrm{~K}$ in a closed container, $\left[\mathrm{PCl}_{5}\right]=0.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$

$$ \begin{aligned} {\left[PCl_{3}\right] } & =1.2 \times 10^{-3} mol L^{-1} \\ {\left[Cl_{2}\right] } & =1.2 \times 10^{-3} mol L^{-1} \\ K_C & =\frac{\left[PCl_3 \right]\left[Cl_2\right]}{\left[PCl_5 \right]}=\frac{\left(1.2 \times 10^{-3}\right) \times\left(1.2 \times 10^{-3}\right)}{\left(0.8 \times 10^{-3}\right)} \\ & =1.8 \times 10^{-3} \end{aligned} $$

• Option (a): $1.8 \times 10^{3} \mathrm{~mol} \mathrm{~L}^{-1}$ is incorrect because the calculated equilibrium constant $K_c$ is $1.8 \times 10^{-3}$, not $1.8 \times 10^{3}$. The value is off by a factor of $10^6$.

• Option (c): $1.8 \times 10^{-3} \mathrm{~mol}^{1} \mathrm{~L}$ is incorrect because the units are not appropriate for the equilibrium constant $K_c$ in this context. The equilibrium constant $K_c$ is dimensionless for this reaction, as it is a ratio of concentrations.

• Option (d): $0.55 \times 10^{4}$ is incorrect because the calculated equilibrium constant $K_c$ is $1.8 \times 10^{-3}$, not $0.55 \times 10^{4}$. The value is significantly different from the correct value.

Answer

(b) In the reaction, $\mathrm{Fe}^{3+}+\mathrm{SCN}^{-} \rightleftharpoons \underset{(\mathrm{Red})}{\mathrm{FeSCN}^{2+}}$

When oxalic acid is added it combines with $\mathrm{Fe}^{3+}$ ions, then, equilibrium shifts towards backward direction and intensity of red colour decreases.

• (a) This statement is correct because in a perfectly insulated flask, the system is in thermal equilibrium, and no heat exchange occurs with the surroundings. Therefore, the mass of ice and water remains constant over time.

• (c) This statement is correct because a catalyst only speeds up the rate at which equilibrium is reached; it does not affect the equilibrium constant itself.

• (d) This statement is correct because for an exothermic reaction (negative $\Delta H$), increasing the temperature shifts the equilibrium position to favor the reactants, thereby decreasing the equilibrium constant.

Answer

(a) In the reaction,

$ [Co \underset{\text(Pink)}{(H_2O)_6}]^{+3} _{(aq)} + 4Cl^- _{(aq)} \leftrightharpoons [Co \underset{\text{Blue}}{Cl_4}]^{2-} _{(aq)} + 6H _2O _{(l)} $

On cooling, the equilibrium shifts backward direction or on heating, the equilibrium shifts forward direction. Hence, reaction is endothermic. i.e., $\Delta H>0$.

• (b) $\Delta H<0$ for the reaction: This option is incorrect because the reaction is endothermic, as indicated by the fact that the equilibrium shifts in the forward direction upon heating. An exothermic reaction would shift in the reverse direction upon heating.

• (c) $\Delta H=0$ for the reaction: This option is incorrect because the reaction is temperature-dependent, as evidenced by the color change with temperature. If $\Delta H$ were zero, the equilibrium position would not change with temperature.

• (d) The sign of $\Delta H$ cannot be predicted on the basis of this information: This option is incorrect because the information provided (color change with temperature) is sufficient to determine that the reaction is endothermic, indicating that $\Delta H>0$.

Answer

(c) The $\mathrm{pH}$ of neutral water at $25^{\circ} \mathrm{C}$ is 7.0 .

With rise in temperature, $\mathrm{pH}$ of pure water decreases and it become less than 7 at $60^{\circ} \mathrm{C}$.

• Option (a) Equal to 7.0: This is incorrect because the ionization of water increases with temperature, leading to an increase in the concentration of $\mathrm{H}^{+}$ ions. Since $\mathrm{pH}$ is the negative logarithm of the $\mathrm{H}^{+}$ ion concentration, an increase in $\mathrm{H}^{+}$ concentration results in a decrease in $\mathrm{pH}$. Therefore, the $\mathrm{pH}$ of pure water at $60^{\circ} \mathrm{C}$ will be less than 7.0, not equal to 7.0.

• Option (b) Greater than 7.0: This is incorrect because an increase in temperature increases the ionization of water, which increases the concentration of $\mathrm{H}^{+}$ ions. A higher concentration of $\mathrm{H}^{+}$ ions means a lower $\mathrm{pH}$ value. Therefore, the $\mathrm{pH}$ of pure water at $60^{\circ} \mathrm{C}$ will be less than 7.0, not greater than 7.0.

• Option (d) Equal to zero: This is incorrect because a $\mathrm{pH}$ of zero would imply an extremely high concentration of $\mathrm{H}^{+}$ ions, which is not the case for pure water even at elevated temperatures. The $\mathrm{pH}$ of pure water at $60^{\circ} \mathrm{C}$ will be less than 7.0 but not as low as zero.

Thinking Process

This problem is based upon the relationship between ionisation constant $\left(K_{a}\right)$ and $p H$ i.e, $K_{a} \propto \frac{1}{p H}$. Greater the $K_{a}$ lesser the value of $p H$ and vice-versa.

Answer

(d) As the acidity or $K_{a}$ value increases, $\mathrm{pH}$ decreases, thus, the order of $\mathrm{pH}$ value of the acids is

$$ \begin{gathered} \text { Hypochlorous acid< Acetic acid < Formic acid } \\ \left(3.8 \times 10^{-8}\right) \quad\left(1.74 \times 10^{-5}\right) \quad\left(18 \times 10^{-4}\right) \end{gathered} $$

• Option (a) Acetic acid > hypochlorous acid > formic acid: This option is incorrect because it suggests that acetic acid has a higher pH than hypochlorous acid and formic acid. However, acetic acid has a higher ionization constant ($K_a = 1.74 \times 10^{-5}$) than hypochlorous acid ($K_a = 3.0 \times 10^{-8}$), meaning acetic acid is stronger and should have a lower pH than hypochlorous acid. Additionally, formic acid has the highest $K_a$ value ($1.8 \times 10^{-4}$), indicating it is the strongest acid among the three and should have the lowest pH.

• Option (b) Hypochlorous acid > acetic acid > formic acid: This option is incorrect because it suggests that hypochlorous acid has a higher pH than acetic acid and formic acid. Hypochlorous acid has the lowest ionization constant ($K_a = 3.0 \times 10^{-8}$), making it the weakest acid among the three and should have the highest pH. Acetic acid, with a $K_a$ of $1.74 \times 10^{-5}$, is stronger than hypochlorous acid but weaker than formic acid, so its pH should be between that of hypochlorous acid and formic acid. Formic acid, with the highest $K_a$ value ($1.8 \times 10^{-4}$), should have the lowest pH.

• Option (c) Formic acid > hypochlorous acid > acetic acid: This option is incorrect because it suggests that formic acid has a higher pH than hypochlorous acid and acetic acid. Formic acid has the highest ionization constant ($K_a = 1.8 \times 10^{-4}$), making it the strongest acid among the three and should have the lowest pH. Hypochlorous acid, with the lowest $K_a$ value ($3.0 \times 10^{-8}$), is the weakest acid and should have the highest pH. Acetic acid, with a $K_a$ of $1.74 \times 10^{-5}$, should have a pH between that of hypochlorous acid and formic acid.

Thinking Process

To find out the correct relationship between three ionisation constants $\left(K_{a_{1}}, K_{a_{2}}\right.$ and $\left.K_{a_{3}}\right)$ this must be keep in mind that when two reactions are added, their equilib:

Answer

(a) For the reaction,

$$ \begin{aligned} H_{2} ~S & \rightleftharpoons H^{+} + HS^{-} \\ K_{a_{1}} & =\frac{\left[H^{+}\right]\left[HS^{-}\right]}{\left[H_{2} ~S \right]} \end{aligned} $$

For the reaction,

$$ \mathrm{HS}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{S}^{2-} $$

$$ K_{\mathrm{a}_{2}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{S}^{2-}\right]}{\left[\mathrm{HS}^{-}\right]} $$

When, the above two reactions are added, their equilibrium constants are multiplied, thus

Hence,

$$ K_{a_{3}}=\frac{\left[H^{+}\right]^{2}\left[~S^{2-}\right]}{\left[H_{2} ~S \right]}=K_{a_{1}} \times K_{a_{2}} $$

$$K_{a_{3}}=K_{a_{1}} \times K_{a_{2}}$$

• Option (b) $K_{a_{3}}=K_{a_{1}}+K_{a_{2}}$: This is incorrect because the equilibrium constants of reactions are not additive. When two reactions are added, their equilibrium constants are multiplied, not added.

• Option (c) $K_{a_{3}}=K_{a_{1}}-K_{a_{2}}$: This is incorrect because the equilibrium constants of reactions are not subtracted. The relationship between equilibrium constants when reactions are combined involves multiplication, not subtraction.

• Option (d) $K_{a_{3}}=K_{a_{1}} / K_{a_{2}}$: This is incorrect because the equilibrium constants of reactions are not divided when reactions are combined. The correct relationship involves the multiplication of the equilibrium constants, not division.

Answer

(c) GN Lewis in 1923 defined an acid as a species which accepts an electron pair and base which donates an electron pair.As $\mathrm{BF}_{3}$ is an electron deficient compound, hence, it is a Lewis acid.

• (a) Arrhenius concept: The Arrhenius concept defines acids as substances that increase the concentration of hydrogen ions ($\mathrm{H}^+$) in aqueous solution. $\mathrm{BF}_3$ does not release $\mathrm{H}^+$ ions in solution, so it does not fit this definition.

• (b) Bronsted Lowry concept: The Bronsted-Lowry concept defines acids as proton donors and bases as proton acceptors. $\mathrm{BF}_3$ does not donate a proton ($\mathrm{H}^+$), so it does not fit this definition.

• (d) Bronsted Lowry as well as Lewis concept: While $\mathrm{BF}_3$ fits the Lewis concept as an electron pair acceptor, it does not fit the Bronsted-Lowry concept as it does not donate a proton. Therefore, it cannot be explained by both concepts simultaneously.

Answer

(c) When the concentration of $\mathrm{NH}_{4} \mathrm{OH}$ (weak base) is higher than the strong acid $(\mathrm{HCl}), a$ mixture of weak base and its conjugate acid is obtained, which acts as basic buffer.

$ \quad\quad\quad NH_4OH + HCL \longrightarrow NH_4Cl + H_2O $

Initial $\quad\quad$ 0.1 M $\quad$ 0.05 M $\quad\quad\quad$ 0

After
reaction $\quad$ 0.05 M $\quad$ 0 $\quad\quad\quad$ 0.05 M

• (a) The mixture of $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NH}_{4} \mathrm{OH}$ and $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}$ will not produce a buffer solution because the strong acid (HCl) will completely neutralize the weak base (NH₄OH), resulting in a solution of the salt (NH₄Cl) and water, without any remaining weak base or its conjugate acid.

• (b) The mixture of $0.05 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NH}_{4} \mathrm{OH}$ and $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}$ will not produce a buffer solution because the amount of strong acid (HCl) is in excess compared to the weak base (NH₄OH). This will result in complete neutralization of the weak base and an excess of strong acid, leading to an acidic solution rather than a buffer.

• (d) The mixture of $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{CH}_{4} \mathrm{COONa}$ and $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NaOH}$ will not produce a buffer solution because both components are salts of strong bases (NaOH) and weak acids (CH₄COOH). The strong base (NaOH) will react with the weak acid salt (CH₄COONa) to form a solution that is not a buffer.

Answer

(d) Among the given solvent, $AgCl$ is most soluble in aqueous ammonia solution. $AgCl$ react with aqueous ammonia to form a complex, $\left[Ag\left(NH_{3}\right)_{2}\right]^{+} Cl^{-}$.

• (a) $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{AgNO}_{3}$ solution: Silver nitrate ($\mathrm{AgNO}_{3}$) provides a common ion effect due to the presence of $\mathrm{Ag}^{+}$ ions, which decreases the solubility of $\mathrm{AgCl}$ by shifting the equilibrium towards the solid $\mathrm{AgCl}$.

• (b) $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}$ solution: Hydrochloric acid ($\mathrm{HCl}$) provides a common ion effect due to the presence of $\mathrm{Cl}^{-}$ ions, which decreases the solubility of $\mathrm{AgCl}$ by shifting the equilibrium towards the solid $\mathrm{AgCl}$.

• (c) $\mathrm{H}_{2} \mathrm{O}$: While $\mathrm{AgCl}$ is somewhat soluble in water, it does not form any complex ions in pure water, resulting in lower solubility compared to when it forms complexes in other solvents like aqueous ammonia.

Answer

(a) Given that,

$$ K_{a}=1.74 \times 10^{-5} $$

Concentration of $\mathrm{CH}_{3} \mathrm{COOH}=0.01 \mathrm{~mol} \mathrm{dm}^{-3}$

$$ \begin{aligned} {\left[\mathrm{H}^{+}\right] } & =\sqrt{K_{a} \cdot \mathrm{C}} \\ & =\sqrt{1.74 \times 10^{-5} \times 0.01}=4.17 \times 10^{-4} \\ \mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right] \\ & =-\log \left(4.17 \times 10^{-4}\right)=3.4 \end{aligned} $$

• Option (b) 3.6: This value is incorrect because the calculated concentration of hydrogen ions $([H^+])$ for $(0.01 , \text{mol dm}^{-3} , \text{CH}_3\text{COOH})$ with $(K_a = 1.74 \times 10^{-5})$ results in a pH of 3.4, not 3.6. The value 3.6 would correspond to a lower concentration of hydrogen ions than what is actually present.

• Option (c) 3.9: This value is incorrect because it suggests an even lower concentration of hydrogen ions than option (b). Given the $(K_a)$ and the concentration of acetic acid, the pH of 3.9 would imply a much weaker acidic solution than calculated.

• Option (d) 3.0: This value is incorrect because it suggests a higher concentration of hydrogen ions than what is calculated. A pH of 3.0 would correspond to a higher $([H^+])$ concentration, which is not supported by the given $(K_a)$ and acetic acid concentration.

Answer

(c) Given that,

$$ K_{a} \text { for } \mathrm{CH}_{3} \mathrm{COOH}=1.8 \times 10^{-5} $$

$$ K_{b} \text { for } \mathrm{NH}_{4} \mathrm{OH}=1.8 \times 10^{-5} $$

Ammonium acetate is a salt of weak acid and weak base. For such salts

$$ \begin{aligned} pH & =7+\frac{pK _{a} - p K _b}{2} \\ & =7+\frac{\left[-\log 1.8 \times 10^{-5}\right]-\left[-\log 1.8 \times 10^{-5}\right]}{2} \\ & =7+\frac{4.74-4.74}{2}=7.00 \end{aligned} $$

• Option (a) 7.005: This option is incorrect because the calculation for the pH of ammonium acetate, which is a salt of a weak acid and a weak base, shows that the pH is exactly 7. The given values of $( K_a )$ and $( K_b )$ are equal, leading to a neutral pH of 7. There is no basis for the slight deviation to 7.005.

• Option (b) 4.75: This option is incorrect because it represents the pH of a solution where the $( K_a )$ of the weak acid (acetic acid) is considered alone, without taking into account the $( K_b )$ of the weak base (ammonium hydroxide). For ammonium acetate, both $( K_a )$ and $( K_b )$ are equal, resulting in a neutral pH of 7, not 4.75.

• Option (d) Between 6 and 7: This option is incorrect because the pH calculation for ammonium acetate, given that $( K_a )$ and $( K_b )$ are equal, results in a pH of exactly 7. There is no reason for the pH to fall between 6 and 7 when the values of $( K_a )$ and $( K_b )$ are the same.

Answer

(a) As we know that

$$ \Delta G^{\ominus}=-R T \ln K $$

At the stage of half completion of the reaction,

$$ \begin{aligned} A \rightleftharpoons B,[A] & =[B] \\ \text{Therefore, }K & =1 . \\ \text{Thus, } \Delta G^{\ominus} & =0 \end{aligned} $$

• Option (b) $\Delta G^{\ominus}>0$: This option is incorrect because at the stage of half completion of the reaction, the equilibrium constant $( K )$ is equal to 1. Since $(\Delta G^{\ominus} = -RT \ln K)$, and $(\ln 1 = 0)$, it follows that $(\Delta G^{\ominus} = 0)$, not greater than 0.

• Option (c) $\Delta \mathrm{G}^{\ominus}<0$: This option is incorrect for the same reason as option (b). At the stage of half completion, $( K = 1 )$ and $(\ln 1 = 0)$, so $(\Delta G^{\ominus} = 0)$, not less than 0.

• Option (d) $\Delta G^{\ominus}=-R T \ln K$: While this equation is correct in general, it does not specifically address the condition of half completion of the reaction. At half completion, $( K = 1 )$, and thus $(\Delta G^{\ominus} = 0)$. Therefore, this option does not directly answer the question about the stage of half completion.

Answer

(a) In the reaction, $\quad N_{2}(g)+3 H_{2}(g) \rightleftharpoons 2 NH_{3}(g)$

If the total pressure at which the equilibrium is established, is increased without changing the temperature, $K$ will remain same. $K$ changes only with change in temperature.

• (b) K will decrease: This is incorrect because the equilibrium constant ( K ) for a given reaction is only affected by changes in temperature, not by changes in pressure. Increasing the pressure does not change the value of ( K ).

• (c) K will increase: This is incorrect because, similar to the previous point, the equilibrium constant ( K ) is not influenced by changes in pressure. It remains constant as long as the temperature is unchanged.

• (d) K will increase initially and decrease when pressure is very high: This is incorrect because the equilibrium constant ( K ) does not depend on the pressure at all. It is solely a function of temperature. Therefore, ( K ) will neither increase nor decrease with changes in pressure.

Answer

(b) The given compounds are

$$ \underset{\text{(Maximum b.p)}}{Water} < \quad acetone < \underset{\text{(Maximum b.p)}}{ether} $$

Greater the boiling point, lower is the vapour pressure of the solvent. Hence, the correct order of vapour pressure will be

Water < acetone <ether.

• Option (a) Water < ether < acetone: This option is incorrect because it suggests that ether has a lower vapor pressure than acetone. However, since ether has the lowest boiling point among the three, it should have the highest vapor pressure. Therefore, ether should not be placed between water and acetone in terms of vapor pressure.

• Option (c) Ether < acetone < water: This option is incorrect because it suggests that ether has the lowest vapor pressure and water has the highest vapor pressure. Given that water has the highest boiling point, it should have the lowest vapor pressure. Conversely, ether, with the lowest boiling point, should have the highest vapor pressure. This order is completely reversed.

• Option (d) Acetone < ether < water: This option is incorrect because it suggests that acetone has a lower vapor pressure than ether. Since acetone has a higher boiling point than ether, it should have a lower vapor pressure. Therefore, acetone should not be placed before ether in terms of vapor pressure.

Answer

(a) For the reaction, $\frac{1}{2} H_{2}(g)+\frac{1}{2} I_{2}(g) \rightleftharpoons HI(g)$

$$ K_{c}=\frac{[HI]}{\left[H_{2}\right]^{1 / 2}\left[I_{2}\right]^{1 / 2}}=5 $$

Thus, for the reaction,

$$ \begin{aligned} 2 HI(g) & \rightleftharpoons H_{2}(g)+ I_{2}(g) \\ K_{c_{1}} & =\frac{\left[H_{2}\right]\left[I_{2}\right]}{[HI]^{2}}=\left(\frac{1}{K_c}\right)^{2}=\left(\frac{1}{5}\right)^{2}=\frac{1}{25}=0.04 \end{aligned} $$

• Option (b) 0.4: This value is incorrect because it does not correctly represent the inverse square of the given equilibrium constant ( K_c ). The correct calculation should be (\left(\frac{1}{5}\right)^2 = 0.04), not 0.4.

• Option (c) 25: This value is incorrect because it represents the square of the given equilibrium constant ( K_c ) rather than the inverse square. The correct calculation for the reverse reaction should be (\left(\frac{1}{5}\right)^2 = 0.04), not (5^2 = 25).

• Option (d) 2.5: This value is incorrect because it does not follow the correct mathematical relationship for the equilibrium constant of the reverse reaction. The correct calculation should be (\left(\frac{1}{5}\right)^2 = 0.04), not 2.5.

Thinking Process

At constant volume, the equilibrium remain unaffected on addition of small amount of inert gas like argon, nean, Kruspton, etc.

Answer

(d) In these reactions, at constant volume

$$ \begin{aligned} H_{2}(g)+ I_{2}(g) & \rightleftharpoons 2 HI(g) \\ PCl_{5}(g) & \rightleftharpoons PCl_{3}(g)+ Cl_{2}(g) \\ N_{2}(g)+3 H_{2}(g) & \rightleftharpoons 2 NH_{3}(g) \end{aligned} $$

The equilibrium constant $(K)$ remains unaffected on addition of inert gas in all the three cases.

• Option (a) is incorrect because the equilibrium constant ( K ) remains unaffected by the addition of an inert gas at constant volume. The partial pressures of the reacting gases do not change, so the position of equilibrium does not shift.

• Option (b) is incorrect because the equilibrium constant ( K ) remains unaffected by the addition of an inert gas at constant volume. The partial pressures of the reacting gases do not change, so the position of equilibrium does not shift.

• Option (c) is incorrect because the equilibrium constant ( K ) remains unaffected by the addition of an inert gas at constant volume. The partial pressures of the reacting gases do not change, so the position of equilibrium does not shift.

Answer

$(a, c, d)$

For the reaction, $\quad N_{2} O_{4}(g) \rightleftharpoons 2 NO_{2}(g)$

At $\quad 400 \mathrm{~K}, K=50$

At $\quad 500 \mathrm{~K}, \mathrm{~K}=1700$

(a) As the value of $K$ increase with increase of temperature and $K=\frac{K_{f}}{K_{b}}$, this means that $K_{f}$ increases, i.e., forward reaction is favoured. Hence, reaction is endothermic.

(c) Since, number of moles of gaseous products are greater than the number of moles of gaseous reactants. Thus, higher pressure favours the backward reaction, i.e., more $N_{2} O_{4}(g)$ will be obtained, if $P_{\text {product }}>P_{\text {reactant }}$.

(d) As reaction is accompanied by increase in the number of moles, entropy increases.

• (b) The reaction is exothermic: This option is incorrect because the value of the equilibrium constant ( K ) increases with an increase in temperature, indicating that the forward reaction is favored at higher temperatures. According to Le Chatelier’s principle, this behavior is characteristic of an endothermic reaction, not an exothermic one. In an exothermic reaction, the equilibrium constant would decrease with an increase in temperature.

Answer

$(a, d)$

At a particular temperature and atmospheric pressure, the solid and liquid phases of a pure substance can exist as Solid $\rightleftharpoons$ liquid.

They exists at normal melting point or normal freezing point.

• Equilibrium temperature: This term is too general and does not specifically refer to the temperature at which the solid and liquid phases of a pure substance coexist. It could refer to any temperature at which two phases are in equilibrium, not necessarily solid and liquid.

• Boiling point: This term refers to the temperature at which a liquid turns into a gas (vaporizes) at a given pressure, not the temperature at which a solid and liquid coexist.

Answer

Note If Bronsted acid is a strong acid then its conjugate base is a weak base and vice-versa. Generally, the conjugate acid has one extra proton and each conjugate base has one less proton.

Answer

Explanation for the given statement on the basis of ionisation and effect upon the concentration of sodium chloride is given below

(i) Sugar being a non-electrolyte does not ionise in water whereas $\mathrm{NaCl}$ ionises completely in water and produces $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$ion which help in the conduction of electricity.

(ii) When concentration of $\mathrm{NaCl}$ is increased, more $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$ions will be produced. Hence, conductance or conductivity of the solution increases.

Answer

$BF_{3}$ is an electron deficient compound and hence acts as Lewis acid. $NH_{3}$ has one lone pair which it can donate to $BF_{3}$ and form a coordinate bond. Hence, $NH_{3}$ acts as a Lewis base.

$$ H_{3} ~N: \longrightarrow BF_{3} $$

Answer

Given that, ionisation constant of a weak base $\mathrm{MOH}$

$$ K_{b}=\left[M^{+}\right]\left[\mathrm{OH}^{-}\right][\mathrm{MOH}] . $$

Larger the ionisation constant $\left(K_{b}\right)$ of a base, greater is its ionisation and stronger the base. Hence, dimethyl amine is the strongest base.

$$ K_{b} \text { Dimethyl amine }>\underset{5.4 \times 10^{-4}}{\text { ammonia }}>\underset{1.77 \times 10^{-5}}{1.77 \times 10^{-9}}>\underset{1.3 \times 10^{-14}}{\text { urea }} $$

Answer

Conjugate acid of the given bases are $H_{2} O, ROH, CH_{3} COOH$ and $HCl$. Order of their acidic strength is

$$ HCl>CH_{3} COOH>H_{2} O>ROH $$

Hence, order of basic strength of their conjugate bases is

$$ \mathrm{Cl}^{-}<\mathrm{CH}_{3} \mathrm{COO}^{-}<\mathrm{OH}^{-}<\mathrm{RO}^{-} $$

Answer

(i) $KNO_{3}$ is a salt of strong acid $\left(HNO_{3}\right)$ strong base $(KOH)$, hence its aqueous solution is neutral; $pH=7$.

(ii) $CH_{3} COONa$ is a salt of weak acid $\left(CH_{3} COOH \right)$ and strong base $(NaOH)$, hence, its aqueous solution is basic; $pH>7$.

(iii) $NH_{4} Cl$ is a salt of strong acid $(HCl)$ and weak base $\left(NH_{4} OH\right)$ hence, its aqueous solution is acidic; $pH<7$.

(iv) $C_{6} H_{5} COONH_{4}$ is a salt of weak acid, $C_{6} H_{5} COOH$ and weak base, $NH_{4} OH$. But $NH_{4} OH$ is slightly stronger than $C_{6} H_{5} COOH$. Hence, $pH$ is slightly $>7$.

Therefore, increasing order of $\mathrm{pH}$ of the given salts is,

$$ NH_{4} Cl<C_{6} H_{5} COONH_{4}>KNO_{3}<CH_{3} COONa $$

Answer

Given that,

$$ \begin{aligned} {[HI] } & =2 \times 10^{-5} ~mol \\ {\left[H_{2}\right] } & =1 \times 10^{-5} ~mol \\ {\left[I_{2}\right] } & =1 \times 10^{-5} ~mol \end{aligned} $$

At a given time, the reaction quotient $Q$ for the reaction will be given by the expression

$$ \begin{aligned} Q & =\frac{\left[H_{2}\right]\left[I_{2}\right]}{[HI]^{2}} \\ & =\frac{1 \times 10^{-5} \times 1 \times 10^{-5}}{\left(2 \times 10^{-5}\right)^{2}}=\frac{1}{4} \\ & =0.25=2.5 \times 10^{-1} \end{aligned} $$

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 reaction.

Answer

Concentration $10^{-8} ~mol dm^{-3}$ indicates that the solution is very dilute. So, we cannot neglect the contribution of $H_{3} O^{+}$ions produced from $H_{2} O$ in the solution. Total $\left[H_{3} O^{+}\right]=10^{-8}+10^{-7} M$. From this we get the value of $pH$ close to 7 but less than 7 because the solution is acidic.

From calculation, it is found that $pH$ of $10^{-8} ~mol dm^{-3}$ solution of $HCl$ is equal to 6.96 .

Answer

Given that,

$$ \begin{aligned} \mathrm{pH} & =5 \\ {\left[\mathrm{H}^{+}\right] } & =10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \end{aligned} $$

On diluting the solution 100 times $\left[\mathrm{H}^{+}\right]=\frac{10^{-5}}{100}=10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}$

On calculating the $\mathrm{pH}$ using the equation $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$, value of $\mathrm{pH}$ comes out to be 7. It is not possible. This indicates that solution is very dilute.

Hence, $\quad$ Total $\mathrm{H}^{+}$ion concentration $=\mathrm{H}^{+}$ions from acid $+\mathrm{H}^{+}$ion from water

$$ \begin{aligned} {\left[\mathrm{H}^{+}\right] } & =10^{-7}+10^{-7}=2 \times 10^{-7} \mathrm{M} \\ \mathrm{pH} & =-\log \left[2 \times 10^{-7}\right] \\ \mathrm{pH} & =7-0.3010=6.699 \end{aligned} $$

Answer

$$ \begin{aligned} BaSO_{4}(~s) & \rightleftharpoons Ba^{2+}(aq)+SO_{4}^{2-}(aq) \\ K_{sp} \text { for } BaSO_{4} & =\left[Ba^{2+}\right]\left[SO_{4}^{2-}\right]=s \times s=s^{2} \\ but \quad s & =8 \times 10^{-4} ~mol dm ^{-3} \\ \therefore \quad K_{sp} & =\left(8 \times 10^{-4}\right)^{2}=64 \times 10^{-8} \end{aligned} $$

In the presence of $0.01 MH_{2} SO_{4}$, the expression for $K_{sp}$ will be

$$ \begin{aligned} & K_{sp}=\left[Ba^{2+}\right]\left[SO_{4}^{2-}\right] \\ & K_{sp}=(s) \cdot(s+0.01) \quad\left(0.01 M SO_{4}^{2-} \text { ions from } 0.01 M H_{2} SO_{4}\right) \end{aligned} $$

$$ \begin{aligned} 64 \times 10^{-8} & =s \cdot(s+0.01) \\ s^{2}+0.01 s-64 \times 10^{-8} & =0 \end{aligned} $$

$$ \begin{aligned} S & =\frac{-0.01 \pm \sqrt{(0.01)^{2}+\left(4 \times 64 \times 10^{-8}\right)}}{2} \\ & =\frac{-0.01 \pm \sqrt{10^{-4}+\left(256 \times 10^{-8}\right)}}{2} \\ & =\frac{-0.01 \pm \sqrt{10^{-4}\left(1+256 \times 10^{-4}\right)}}{2} \\ & =\frac{-0.01 \pm 10^{-2} \sqrt{1+0.0256}}{2}=\frac{10^{-2}(-1 \pm 1.012719)}{2} \\ & =5 \times 10^{-3}(-1+1.012719)=6.4 \times 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3} \end{aligned} $$

Note $s«<0.01, s 0, s+0.01 \approx 0.01$ and $64 \times 10^{-8}=s \times 0.01$

$$ s=\frac{64 \times 10^{-8}}{0.01}=6.4 \times 10^{-5} $$

Thinking Process

To solve this problem, we use two steps

Step I Find out the concentration of hydrogen ion $\left[\mathrm{H}^{+}\right]$through the formula $-\mathrm{pH}=\log \left[\mathrm{H}^{+}\right]$

Step II Afterward, calculate the $K_{a}$ of $HOCl$ which is weak monobasic acid by using the formula $K_{a}=\frac{\left[H^{+}\right]^{2}}{C}$. where, $C$ is concentration of the solution

Answer

$\mathrm{pH}$ of $\mathrm{HOCl}=2.85$

$$ \begin{aligned} & \text { But, } \quad-\mathrm{pH}=\log \left[\mathrm{H}^{+}\right] \\ & \therefore \quad-2.85=\log \left[\mathrm{H}^{+}\right] \\ & \Rightarrow \quad \overline{3} .15=\log \left[\mathrm{H}^{+}\right] \\ & \Rightarrow \quad\left[\mathrm{H}^{+}\right]=1.413 \times 10^{-3} \\ & \text{for weak monobasic acid } [H^+] = \sqrt{K_a \times C}\\ & \Rightarrow \quad K_{a}=\frac{\left[H^{+}\right]^{2}}{C}=\frac{\left(1.413 \times 10^{-3}\right)^{2}}{0.08} \\ & =24.957 \times 10^{-6}=2.4957 \times 10^{-5} \end{aligned} $$

Answer

$\mathrm{pH}$ of solution $A=6$. Hence, $\left[\mathrm{H}^{+}\right]=10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$

$\mathrm{pH}$ of solution $B=4$. Hence, $\left[\mathrm{H}^{+}\right]=10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$

On mixing $1 \mathrm{~L}$ of each solution, molar concentration of total $\mathrm{H}^{+}$is halved.

Total,

$$ \begin{aligned} & {\left[\mathrm{H}^{+}\right]=\frac{10^{-6}+10^{-4}}{2} \mathrm{~mol} \mathrm{~L}^{-1}} \\ & {\left[\mathrm{H}^{+}\right]=\frac{1.01 \times 10^{-4}}{2}=5.05 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}} \\ & {\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}} \\ & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \quad \Rightarrow \mathrm{pH}=-\log \left(5.0 \times 10^{-5}\right) \\ & \mathrm{pH}=-[\log 5+(-5 \log 10)] \Rightarrow \mathrm{pH}=-\log 5+5 \\ & \mathrm{pH}=5-\log 5=5-0.6990 \Rightarrow \mathrm{pH}=4.3010 \approx 4.3 \end{aligned} $$

Thus, the $\mathrm{pH}$ of resulting solution is 4.3 .

Answer

Let $S$ be the solubility of $\mathrm{Al}(\mathrm{OH})_{3}$.

Concentration of species at $t=0$

Concentration of various species at equilibrium

$$ \begin{aligned} K_{\mathrm{sp}} & =\left[\mathrm{Al}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3}=(\mathrm{S})(3 \mathrm{~S})^{3}=27 \mathrm{~S}^{4} \\ \mathrm{~S}^{4} & =\frac{K_{\mathrm{sp}}}{27}=\frac{2.7 \times 10^{-11}}{27}=1 \times 10^{-12} \\ S & =1 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \end{aligned} $$

(i) Solubility of $\mathrm{Al}(\mathrm{OH})_{3}$

Molar mass of $\mathrm{Al}(\mathrm{OH})_{3}$ is $78 \mathrm{~g}$. Therefore,

Solubility of $\mathrm{Al}(\mathrm{OH})_{3}$ in $\mathrm{g}^{-1}=1 \times 10^{-3} \times 78 \mathrm{~g} \mathrm{~L}^{-1}=78 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}$

$$ =7.8 \times 10^{-2} \mathrm{~g} \mathrm{~L}^{-1} $$

(ii) $\mathrm{pH}$ of the solution $\quad S=1 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$

$$ \begin{aligned} {\left[\mathrm{OH}^{-}\right] } & =3 \mathrm{~S}=3 \times 1 \times 10^{-3}=3 \times 10^{-3} \\ \mathrm{pOH} & =3-\log 3 \end{aligned} $$

$$ pH = 14 - pOH = 11 + log 3 = 11 4771 $$

Answer

Suppose, solubility of $\mathrm{PbCl}_{2}$ in water is $s \mathrm{~mol} \mathrm{~L}^{-1}$

$$ \begin{aligned} \mathrm{PbCl} _2(\mathrm{~s}) & \rightleftharpoons \mathrm{Pb} ^{2+}(\mathrm{aq})+2 \mathrm{Cl} ^{-}(\mathrm{aq}) \\ \left (1-\mathrm{s})\mathrm{K} _{\mathrm{sp}}\right. & =\left[\mathrm{Pb} ^{2 \mathrm{~s}}\right] \cdot\left[\mathrm{Cl} ^{-}\right] ^2 \\ \mathrm{~K} _{\mathrm{sp}} & =[\mathrm{s}][2 \mathrm{~s}]^2=4 \mathrm{~s} ^3 \\ 3.2 \times 10 ^{-8} & =4 \mathrm{~s}^3 \\ \mathrm{~s}^3 & =\frac{3.2 \times 10^{-8}}{4}=0.8 \times 10^{-8} \\ s^3 & =8.0 \times 10^{-9} \end{aligned} $$

Solubility of $$ \begin{aligned} & \mathrm{PbCl} _2, \mathrm{~S}=2 \times 10 ^{-3} \mathrm{~mol} \mathrm{~L} ^{-1} \\ \end{aligned} $$

Solubility of $$ \mathrm{PbCl} _2 \text { in } \mathrm{gL}^{-1}=278 \times 2 \times 10^{-3}=0.556 \mathrm{~g} \mathrm{~L}^{-1} $$ $\left(\because\right.$ Molar mass of $\left.\mathrm{PbCl}_2=207+(2 \times 35.5)=278\right)$ $0.556 \mathrm{~g}$ of $\mathrm{PbCl} _2$ dissolve in $1 \mathrm{~L}$ of water.

$\therefore \quad 0.1 \mathrm{~g}$ of $\mathrm{PbCl} _2$ will dissolve in $=\frac{1 \times 0.1}{0.556} \mathrm{~L}$ of water $$ =0.1798 \mathrm{~L} $$

To make a saturated solution, dissolution of $0.1 \mathrm{~g} \mathrm{PbCl}_{2}$ in $0.1798 \mathrm{~L} \approx 0.2 \mathrm{~L}$ of water will be required.

Answer

Although $BF_{3}$ does not have a proton but acts as Lewis acid as it is an electron deficient compound. It reacts with $NH_{3}$ by accepting the lone pair of electrons from $NH_{3}$ and complete its octet. The reaction can be represented by

$$ BF_{3}+: NH_{3} \longrightarrow BF_{3} \leftarrow NH_{3} $$

Lewis electronic theory of acids and bases can explain it. Boron in $BF_{3}$ is $s p^{2}$ hybridised where $N$ in $NH_{3}$ is $s p^{3}$ hybridised.

Answer

Given that,

$\Delta_{f} H^{\ominus}[\mathrm{CaO}(\mathrm{s})]=-635.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta_{f} H^{\ominus}\left[\mathrm{CO}_{2}(g)\right]=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta_{f} H^{\ominus}\left[\mathrm{CaCO}_{3}(\mathrm{~s})\right]=-1206.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$

In the reaction,

$$ \begin{gathered} CaCO_{3}(s) \rightleftharpoons CaO(s)+CO_{2}(g) \\ \quad \Delta_{f} H^{\ominus}=\Delta_{f} H^{\ominus}[CaO(s)]+\Delta_{f} H^{\ominus}\left[CO_{2}(g)\right]-\Delta_{f} H^{\ominus}\left[CaCO_{3}(s)\right] \\ \therefore \quad \Delta_{f} H^{\ominus}=-635.1+(-393.5)-(-1206.9)=178.3 kJmol^{-1} \end{gathered} $$

Because $\Delta H$ value is positive, so the reaction is endothermic. Hence, according to Le-Chatelier’s principle, reaction will proceed in forward direction on increasing temperature. Thus, the value of equilibrium constant for the reaction increases.

Answer

A. $\rightarrow(2)$

B. $\rightarrow(4)$

C. $\rightarrow$ (3)

D. $\rightarrow(1)$

A. Liquid $\rightleftharpoons$ Vapour equilibrium exists at the boiling point.

B. Solid $\rightleftharpoons$ Liquid equilibrium exists at the melting point.

C. Solid $\rightleftharpoons$ Vapour equilibrium exists at the sublimation point.

D. Solute $(s) \rightleftharpoons$ Solute (solution) equilibrium exists at saturated solution.

Answer

A. $\rightarrow(4)$

B. $\rightarrow(3)$

C. $\rightarrow(2)$

For the reaction,

$$ N_{2}(g)+3 H_{2}(g) \rightleftharpoons 2 NH_{3}(g) $$

Equilibrium constant $K_C=\frac{\left[NH_3 \right]^2}{\left[N_2 \right]\left[H_2 \right]^3}$

A. The given reaction $\left[2 N_{2}(g)+6 H_{2}(g) \rightleftharpoons 4 NH_{3}(g)\right]$ is twice the above reaction. Hence, $K=K_{c}^{2}$

B. The reaction $\left[2 NH_{3}(g) \rightleftharpoons N_{2}(g)+3 H_{2}(g)\right]$ is reverse of the above reaction. Hence, $K=\frac{1}{K_{c}}$

C. The reaction $\left[\frac{1}{2} ~N_{2}(g)+\frac{3}{2} H_{2}(g) \rightleftharpoons NH_{3}(g)\right]$ is half of the above reaction. Hence, $K=\sqrt{K_{c}}=K_{c}^{\frac{1}{2}}$.

Answer

A. $\rightarrow(4)$

B. $\rightarrow(1)$

C. $\rightarrow(2)$

As we know that, $\Delta G^{\ominus}=-R T \ln K$

A. If $\Delta G^{\ominus}>0$, i.e., $\Delta G^{\circ}$ is positive, then $\ln K$ is negative i.e., $K<1$.

B. If $\Delta G^{\ominus}<0$, i.e., $\Delta G^{\circ}$ is negative then $\ln K$ is positive i.e., $K>1$.

C. If $\Delta G^{\ominus}=0, \ln K=0$, i.e., $K=1$.

Answer

A. $\rightarrow(2)$

B. $\rightarrow(5)$

C. $\rightarrow(3)$

D. $\rightarrow$ (4)

As conjugate acid $\rightarrow$ Base $+H ^{+}$

A. $NH_{3} + H^{+} \longrightarrow NH_{4}^{+}$

B. $HCO_{3}^{-}+H^{+} \longrightarrow H_{2} CO_{3}$

C. $H_{2} O+H^{+} \longrightarrow H_{3} O^{+}$

D. $HSO_{4} ^{-}+H^{+} \longrightarrow H_{2} SO_{4}$

Answer

A. $\rightarrow(3)$

B. $\rightarrow(1)$

C. $\rightarrow(2)$

A. Graph (A) represents variation of reactant concentration with time.

B. Graph (B) represents variation of product concentration with time.

C. Graph $(\mathrm{C})$ represents reaction at equilibrium.

Answer

A. $\rightarrow(2,3)$

B. $\rightarrow$ (4)

C. $\rightarrow(1)$

A. $\Delta G\left(\Delta G^{\ominus}\right)$ is 0 , reaction has achieved equilibrium: at this point, there is no longer any free energy left to drive the reaction.

B. If $\Delta G<0$, then $K>1$ which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present predominantly.

C. If $\Delta G>0$, then $K<1$, which implies a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.

Assertion and Reason

In the following questions a statement of Assertion (A) followed by a statement of Reason $(\mathrm{R})$ is given. Choose the correct option out of the choices given below in each question.

Answer

(a) Both assertion and reason are true and reason is the correct explanation of assertion.

In the hydrogen halides, the HI is strongest acid while HF is the weak acid. It is because while comparing acids formed by the elements belonging to the same group of periodic table, $\mathrm{H}$ - A bond strength is a more important factor in determining acidity of an acid than the polar nature of the bond.

Answer

(a) Both assertion and reason are true and reason is correct explanation of assertion. A solution containing a mixture of acetic acid and the sodium acetate acts as a buffer solution as it maintains a constant value of $\mathrm{pH}(=4.75)$ and its $\mathrm{pH}$ is not affected on addition of small amounts of acid or alkali.

Answer

(b) Both assertion and reason are true but reason is not correct explanation of assertion. $\mathrm{HCl}$ gives the common $\mathrm{H}^{+}$ions and hence ionisation equilibrium $\mathrm{H}_{2} \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}$ is suppressed.

Answer

(c) Assertion is true but reason is false.

Equilibrium constant of a reaction depends upon temperature.

Answer

(a) Both assertion and reason are true and reason is the correct explanation of assertion.

If $K_{b}$ of $NH_{4} OH>K_{a}$ of $H_{2} CO_{3}$

The solution is basic.

or, if $K_{a}$ of $H_{2} CO_{3}>K_{b}$ of $NH_{4} OH$; the solution is acidic.

Answer

(b) Both assertion and reason are true but reason is not correct explanation of assertion. Ammonium acetate is a salt of weak acid $\left(CH_{3} COOH \right)$ and weak base $\left(NH_{4} OH \right)$.

Answer

(c) Assertion is true but reason is false.

$$ PCl_{5} \longrightarrow PCl_{3}+Cl_{2} $$

At constant pressure, when helium is added to the equilibrium, volume increases. Thus, in order to maintain the $K$ constant, degree of dissociation of $\mathrm{PCl}_{5}$ increases. Helium is unreactive towards chlorine gas.

Answer

Prediction of the following stages of a reaction by comparing the value of $K_{c}$ and $Q_{c}$ are

(i) If $Q_{C}<K_{C}$, the reaction will proceed in the direction of the products (forward reaction).

(ii) If $Q_{c}>K_{c}$, the reaction will proceed in the direction of reactants (reverse reaction).

(iii) If $Q_{C}=K_{c}$, the reaction mixture is already at equilibrium.

Answer

$$ N_{2}(g)+3 H_{2}(g) \rightleftharpoons 2 NH_{3}(g) ; \Delta H=-92.38 ~kJ ~mol^{-1} $$

It is an exothermic process as $\Delta H$ is negative.

Effect of temperature According to Le-Chatelier’s principle, low temperature is favourable for high yield of ammonia, but practically very low temperatures slow down the reaction. So, optimum temperature $700 \mathrm{~K}$ is favourable in attainment of equilibrium.

Effect of pressure Similarly, high pressure about 200 atm is favourable for high yield of ammonia. On increasing pressure, reaction goes in the forward direction because the number of moles decreases in the forward direction.

Addition of argon At constant volume addition of argon does not affect the equilibrium because it does not change the partial pressures of the reactants or products involved in the reaction and the equilibrium remains undisturbed.

Answer

A sparingly soluble salt having general formula $A_{x}^{p+} B_{y}^{q-}$. Its molar solubility is $S \mathrm{~mol} \mathrm{~L}^{-1}$. Then,

$$ A_{x}^{p+} B_{y}^{q-} \rightleftharpoons x A_{x}^{p+}(a q)+y B_{y}^{q-}(a q) $$

$S$ moles of $A_{x} B_{y}$ dissolve to give $x$ moles of $A^{P+}$ and $y$ moles of $B^{q-}$.

Therefore, solubility product $\left(K_{\mathrm{sp}}\right)=\left[A^{P+}\right]^{x}\left[B^{q-}\right]^{y}$

$$ \begin{aligned} & =[x S]^{x}[y S]^{y} \\ & =x^{x} y^{y} S^{x+y} \end{aligned} $$

Answer

The relation between $\Delta G$ and $Q$ is

$$ \begin{aligned} \Delta G & =\Delta G^{\ominus}+R T \ln Q \\ \Delta G & =\text { change in free energy as the reaction proceeds. } \\ \Delta G^{\ominus} & =\text { standard free energy } \\ Q & =\text { reaction quotient } \\ R & =\text { gas constant } \\ T & =\text { absolute temperature in } K \end{aligned} $$

(a) Since,

$$ \begin{aligned} \Delta G^{\ominus} & =-R T \ln K \\ \therefore \quad \Delta G & =-R T \ln K+R T \ln Q \\ \Delta G & =R T \ln \frac{Q}{K} \end{aligned} $$

If $Q<K, \Delta G$ will be negative and the reaction proceeds in the forward direction.

If $Q=K, \Delta G=0$ reaction is in equilibrium and there is no net reaction.

(b)

$$ \begin{aligned} CO(g)+3 H_{2}(g) & \rightleftharpoons CH_{4}(g)+H_{2} O(g) \\ K_{c} & =\frac{\left[CH_{4}\right]\left[H_{2} O\right]}{[CO]\left[H_{2}\right]^{3}} \end{aligned} $$

On increasing pressure, volume decreases. If we doubled the pressure, volume will be halved but the molar concentrations will be doubled. Then,

$$ Q_C = \frac{2[CH_4] . 2[H_2 O ]}{2[CO]{2[H_2 ]}^3}=\frac{1}{4} \frac{[CH_4 ][H_2 O]}{[CO][H_2 ]^3}=\frac{1}{4} K_c $$

Therefore, $Q_{C}$ is less than $K_{c}$, so $Q_{c}$ will tend to increase to re-establish equilibrium and the reaction will go in forward direction.