NCERT Exampler Questions

Hint

(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,
\(
\mathrm{NH}_4 \mathrm{Cl}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})
\)
\(
\Delta n=2-0=2
\)

Hint

(d) \(\Delta G^{\ominus}=-R T \ln K; \Delta G^{\ominus}>0\) means \(\Delta G^{\ominus}\) is +ve. This can be so only if In \(K\) is -ve, i.e., \(K<1\).

Hint

(c) All the physical processes like melting of ice and freezing of water, etc., do not stop at equilibrium.

Hint

(b)

\(
\begin{aligned}
K_{\mathrm{C}} & =\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}=\frac{1 .2 \times 10^{-3} \times 1.2 \times 10^{-3}}{0.8 \times 10^{-3}} \\
& =1.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}
\)

Hint

(b) \(\mathrm{Fe}^{3+}+\mathrm{SCN}^{-} \rightleftharpoons \mathrm{FeSCN}^{2+}(\mathrm{Red})\)
When oxalic acid is added to a solution containing iron nitrate and potassium thiocyanate, oxalic acid reacts with \(\mathrm{Fe}^{3+}\) ions to form a stable complex ion \(\left[\mathrm{Fe}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}\), thus, decreasing the concentration of free \(\mathrm{Fe}^{3+}\) ions which in \(\mathrm{mm}\) decreases the intensity of red colour.
\(
\mathrm{Fe}^{3+}+\mathrm{SCN}^{-} \rightleftharpoons[\mathrm{Fe}(\mathrm{SCN})]^{2+}(\mathrm{Blood Red})
\)

Hint

(a) Since the reaction shifts to backward direction on cooling, this means that the backward reaction is exothermic reaction. Therefore, the forward reaction is endothermic reaction and \(\Delta \mathrm{H}>0\).

Hint

(c) At \(25^{\circ} \mathrm{C},\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]=10^7\) and \(\mathrm{K}_W=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}\).
On heating, \(\mathrm{K}_w\) increases, i.e., \(\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]>10^{-}\) \({ }^{14}\) As \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right]^2>10^{-14}\) or \(\left[\mathrm{H}^{+}\right]>10^{-7} \mathrm{~M}\) or \(\mathrm{pH}<7\).

Hint

(d) \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\sqrt{K_{\mathrm{a}} \cdot \mathrm{C}}\) for the same concentration, \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \propto \sqrt{K_{\mathrm{a}}}\). But \(\mathrm{pH}=-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)
Larger the value of \(K_{\mathrm{a}}\), larger will be \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)and lower will be \(\mathrm{pH}\).

Hint

(a) For the reaction, \(\mathrm{H}_2 \mathrm{S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-}\)
\(
K_{\mathrm{a}_1}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HS}^{-}\right]}{\left[\mathrm{H}_2 \mathrm{~S}\right]}
\)
For the reaction, \(\quad \mathrm{HS}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{S}^{2-}\)
\(
K_{\mathrm{a}_2}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{S}^{2-}\right]}{\left[\mathrm{HS}^{-}\right]}
\)
When the above two reactions are added, their equilibrium constants are multiplied. Thus
\(
K_{\mathrm{a}_3}=\frac{\left[\mathrm{H}^{+}\right]^2\left[\mathrm{~S}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{~S}\right]}=K_{\mathrm{a}_1} \times K_{\mathrm{a}_2}
\)
Hence, \(K_{\mathrm{a}_3}=K_{\mathrm{a}_1} \times K_{\mathrm{a}_2}\)

Hint

(c) According to Lewis concept, a positively charged or an electron deficient species acts as Lewis acid. \(\mathrm{BF}_3\) is an electron deficient compound with \(\mathrm{B}\) having 6 electrons only.

Hint

(c) In (c), all \(\mathrm{HCl}\) will be neutralized and \(\mathrm{NH}_4 \mathrm{C} l\) will be formed. Also some \(\mathrm{NH}_4 \mathrm{OH}\) will be left unneutralized. Thus, the final solution will contain \(\mathrm{NH}_4 \mathrm{OH}\) and \(\mathrm{NH}_4 \mathrm{Cl}\) and hence will form a buffer.

Hint

(d) Silver chloride forms a soluble complex with aqueous ammonia. \(\mathrm{AgCl}+2 \mathrm{NH}_3 \rightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right] \mathrm{Cl}\)

Hint

(a)

\(\mathrm{pH}\) for weak acid is given by:
\(
\mathrm{pH}=\frac{1}{2}\left[\mathrm{pK}_{\mathrm{a}}-\log \mathrm{C}\right]
\)
\(
=\frac{1}{2}\left[-\log \left(1.74 \times 10^{-5}\right)-\log 10^{-2}\right]
\)
\(
=\frac{1}{2}[(5-0.2405)+2]=\frac{1}{2}(6.76)=3.38 \cong 3.4
\)

Alternate:

Hint: \(\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{a} \cdot} \cdot \mathrm{C}}\)
Step 1:
Calculate the concentration of \(\mathrm{H}^{+}\)ion as follows:
Given that, \(\left(\mathrm{K}_{\mathrm{a}}=1.74 \times 10^{-5}\right)\)
Concentration of \(\mathrm{CH}_3 \mathrm{COOH}=0.01 \mathrm{~mol} \mathrm{~dm}{ }^{-3}\)
\(
\begin{aligned}
& {\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}}} \\
& =\sqrt{1.74 \times 10^{-5} \times 0.01}=4.17 \times 10^{-4}
\end{aligned}
\)

Step 2 :
\(
\begin{aligned}
& \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\
& =-\log \left(4.17 \times 10^{-4}\right) \\
& =3.4
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
\mathrm{pH} & =\frac{1}{2}\left[\mathrm{p} K_{\mathrm{w}}+\mathrm{p} K_{\mathrm{a}}-\mathrm{p} K_{\mathrm{b}}\right] \\
& =\frac{1}{2}\left[14-\log \left(1.8 \times 10^{-5}\right)+\log \left(1.8 \times 10^{-5}\right)\right]=7.0
\end{aligned}
\)

Hint

(a)

\(
\Delta G^{\ominus}=-R T \ln K
\)
At the stage of half completion of reaction \([A]=[B]\),
Therefore, \(K=1\). Thus, \(\Delta G^{\ominus}=0\).

Hint

(a) According to Le Chatelier’s principle, at constant temperature, the equilibrium composition will change but \(\mathrm{K}\) will remain same. If the total pressure at which the equilibrium is established, is increased without changing the temperature, \(\mathrm{K}\) will remain same. \(\mathrm{K}\) changes only with change in temperature.

Hint

(b) Greater the boiling point, less is the vapour pressure. Hence, the correct order of vapour pressures will be:
water < acetone < ether.

Hint: Magnitude of vapour pressure and boiling point are inversely proportional to each other

Greater the boiling point, lower is the vapour pressure of the solvent.

The boiling point of water is \(100^{\circ} \mathrm{C}\), acetone boiling point is \(56^{\circ} \mathrm{C}\), and ether boiling point is \(34.6^{\circ} \mathrm{C}\).

Hint

If the given equation is multiplied by 2 , the equilibrium constant for the new equation will be squared, and on reversing the reaction the value of the equilibrium constant is reciprocated.
Thus
\(
K=5^2=25
\)
For the required reaction equation \(K=\frac{1}{25}=0.04\)

Alternate:

The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction
\(
\text { For the reaction, } \frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{I}_2(\mathrm{~g}) \rightleftharpoons \mathrm{HI}(\mathrm{g})
\)
\(
\mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{HI}]}{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}=5
\)
\(
\text { Thus, for the reaction, } 2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g})
\)
\(
\mathrm{K}_{\mathrm{c}_1}=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}=\left(\frac{1}{\mathrm{~K}_{\mathrm{c}}}\right)^2=\left(\frac{1}{5}\right)^2=\frac{1}{25}=0.04
\)

Hint

(d) The equilibrium will remain unaffected in all three cases on addition of small amount of inert gas at constant volume.

Alternate:

Hint: Inert gas doesn’t affect equilibrium at constant volume according to Le – Chatelier’s principle.
Explanation:
Step 1:
Le- Chatelier’s principle, states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

The reaction is as follows:
\(
\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})
\)
\(\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\)
\(
\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})
\)

Step 2:
According to Le chatelier’s principle, the equilibrium constant (K) remains unaffected by the addition of inert gas in constant volume. Hence, in all three cases equilibrium remain unaffected.

Hint

(a) As the value of \(\mathrm{K}\) increases with increase of temperature and \(\mathrm{K}=\mathrm{K}_{\mathrm{f}} / \mathrm{K}_{\mathrm{b}}\), this means that \(\mathrm{K}_{\mathrm{f}}\) increases, i.e., forward reaction is favoured.
Hence, reaction is endothermic.
(c). At \(400 \mathrm{~K}, Q=\frac{p_{\mathrm{NO}_2}^2}{P_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{(20)^2}{2}=200\). Thus, \(Q>K\). Equilibrium will shift backward to form more \(\mathrm{N}_2 \mathrm{O}_4\).
(d) As reaction is accompanied by increase in the number of moles, entropy increases.

Hint

(a, d) These are normal melting point and freezing point since they are measured at atmospheric pressure.

Hint

(a)

(i) \(\rightarrow\) (b), (ii) \(\rightarrow\) (d), (iii) \(\rightarrow\) (c), (iv) \(\rightarrow\) (a)
(i) Liquid \(\rightleftharpoons\) Vapour equilibrium exists at the boiling point.
(ii) Solid \(\rightleftharpoons\) Liquid equilibrium exists at the melting point.
(iii) Solid \(\rightleftharpoons\) Vapour equilibrium exists at the sublimation point.
(iv) Solute(s) \(\rightleftharpoons\) Solute (solution) equilibrium exists in saturated solution.

Hint

(c)

\(
\text { (i) } \rightarrow \text { (d), (ii) } \rightarrow \text { (c), (iii) } \rightarrow \text { (b) }
\)
For \(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})\)
\(
K_{\mathrm{C}}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}
\)
\(
\text { (i) } 2 \mathrm{~N}_2(\mathrm{~g})+6 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NH}_3(\mathrm{~g}) ; K_{\mathrm{C}}^{\prime}=K_{\mathrm{C}}^2
\)
\(
\text { (ii) } 2 \mathrm{NH}_3(\mathrm{~g}) \rightleftharpoons \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) ; K_{\mathrm{C}}^{\prime}=\frac{1}{K_{\mathrm{C}}}
\)
\(
\text { (iii) } \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g}) ; K_{\mathrm{C}}^{\prime}=K_{\mathrm{C}}^{1 / 2}
\)

Hint

(b)

(i) \(\rightarrow\) (d), (ii) \(\rightarrow\) (a), (iii) \(\rightarrow\) (b)

As we know that, \(\Delta G^{\ominus}=-\mathrm{R} T \ln K\)
(i) If \(\Delta G^{\ominus}>0\), i.e., \(\Delta G^{\ominus}\) is positive, then \(\ln K\) is negative i.e., \(K<1\)
(ii) If \(\Delta G^{\ominus}<0\), i.e., \(\Delta G^{\ominus}\) is negative then \(\ln K\) is positive i.e., \(K>1\).
(iii) If \(\Delta G^{\ominus}=0, \ln K=0\), i.e., \(K=1\)

Hint

(d) Hint: Conjugate acids are a type of acid that is formed when a base accepts a proton in solution. Explanation:

As conjugate base \(\rightarrow\) Base \(+\mathrm{H}^{+}\)
A. \(\mathrm{NH}_3+\mathrm{H}^{+} \rightarrow \mathrm{NH}_4^{+}\)
B. \(\mathrm{HCO}_3^{-}+\mathrm{H}^{+} \rightarrow \mathrm{H}_2 \mathrm{CO}_3\)
C. \(\mathrm{H}_2 \mathrm{O}+\mathrm{H}^{+} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}\)
D. \(\mathrm{HSO}_4^{-}+\mathrm{H}^{+} \rightarrow \mathrm{H}_2 \mathrm{SO}_4\)

Hint

(a)

c) Variation in Reactant conc. \(\mathrm{v} / \mathrm{s}\) time.
\(\therefore\) Reactant concentration decreases with time as it gets converted to product. So, it matches curve \((i)\).
a) Variation of product concentration with time.

Since, Concentration of Product increases with time as products will get produced from reactants. So, it matches curve (ii)
b) Reaction at equilibrium

The time from where concentration of [reactant \(]\) and [ product \(]\) becomes constant, the equilibrium is established. As at this time both rates of forward and backward reaction are equal. So, it matches curve \((\) iii \()\)

Hint

(b)

(i) \(\rightarrow\) (b) and (c) (ii) \(\rightarrow\) (d) (iii) \(\rightarrow\) (a)

Hint: The relation between \(\Delta_{\mathrm{r}} \Delta G^{\ominus}\) and \(\mathrm{K}\) is \(\Delta_{\mathrm{r}} \Delta G^{\ominus}=-\mathrm{RT} \ln \mathrm{K}\)
A. \(\Delta \mathrm{G}\) is 0 , the 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 \(\mathrm{K}>1\) 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 \(\Delta \mathrm{G}>\mathbf{0}\), then \(\mathrm{K}<1\), which implies a non-spontaneous reaction or a reaction that proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.

Hint

(a) Hint: Down the group acidity of hydrogen halide increases.
Explanation:

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

Down the group strength of \(\mathrm{HI}\) bond decreases because bond dissociation energy decreases. Hence, acidity increases.

Hint

(a) Hint: Buffer solution resist the \(\mathrm{pH}\) change.

Explanation:

A buffer solution is an acid or a base aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. Its pH changes very little when a small amount of strong acid or base is added to it.

A solution containing a mixture of acetic acid and sodium acetate acts as a buffer solution as it maintains a constant pH which is not affected by the addition of small amounts of acid or alkali.

So, option a is the correct answer.

Hint

(b) Hint: Common ion effect
\(\mathrm{H}_2 \mathrm{~S}\) is a weak acid and forms an equilibrium in the solution. The reaction is as follows:
\(
\mathrm{H}_2 \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+2 \mathrm{~S}^{2-}
\)

When a strong acid like \(\mathrm{HCl}\) is added to the solution, due to the common ion effect the ionization of \(\mathrm{H}_2 \mathrm{~S}\) is suppressed. Hence, Both assertion and reason are true and the reason is not the correct explanation of assertion.

Hint

(c)

Hint: The equilibrium constant of a reaction depends upon temperature.
Equilibrium constants are changed if we change the temperature of the system. For any chemical reaction at a particular temperature, the equilibrium constant is fixed and is a characteristic property. Hence, Assertion is true but the reason is false.

Hint

(a) Analyzing the assertion
It is correct because aqueous solution of ammonium carbonate is basic in nature.

Analyzing the reason
It is also correct because for a solution of salt of weak acid and weak base, the acidic/basic nature depends upon value of \(K_a\) for weak acid and \(K_b\) for weak base.

Solution is basic if
\(\mathrm{K}_{\mathrm{b}}\) (weak base) \(>\mathrm{K}_{\mathrm{a}}\) (weak acid).
\(
\begin{aligned}
& \mathrm{Ka}\left(\mathrm{H}_2 \mathrm{CO}_3\right)=4.3 \times 10^{-7} \\
& \mathrm{~Kb}\left(\mathrm{NH}_4 \mathrm{OH}\right)=1.8 \times 10^{-5}
\end{aligned}
\)
\(\mathrm{K}_{\mathrm{b}}\) (Ammonium hydroxide) \(>\mathrm{K}_{\mathrm{a}}\) (Hydrogen carbonate)
\(\therefore\) Hence, solution of Ammonium carbonate is basic in nature.
Both \(\mathrm{A}\) and \(\mathrm{R}\) are true and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\).
\(\therefore\) Correct answer is option (a).

Hint

(c)

Hint: An aqueous solution of ammonium acetate can act as a buffer.

Explanation:

A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. Its \(\mathrm{pH}\) changes very little when a small amount of strong acid or base is added to it.

Ammonium acetate is a salt of a weak acid \(\left(\mathrm{CH}_3 \mathrm{COOH}\right)\) and weak base \(\left(\mathrm{NH}_4 \mathrm{OH}\right)\).
Hence, both assertion and reason are true and the reason is the correct explanation of assertion.

Hint

(d)

Explanation: If the volume remains constant and an inert gas such as argon(helium) is added, the equilibrium remains undisturbed because it does not take part in the reaction.