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,

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

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., $K_{c}<1$.

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.

Answer

(b) For the reaction,

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

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

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

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.

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., $\Delta H>0$.

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 $60^{\circ} \mathrm{C}$.

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

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

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,

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

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

Answer

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

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

Answer

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

Answer

(a) Given that,

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

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

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

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

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

Answer

(a) As we know that

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

At the stage of half completion of the reaction,

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

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.

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.

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,

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

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.

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.

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.

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.