Chapter 6: Equilibrium

Q31Written answer

Predict which of the following reaction will have appreciable concentration of reactants and products:

${\text{C}}{{\text{l}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{Cl (g)}}$ ; ${K_c} = 5 \times {10^{ - 39}}$
${\text{C}}{{\text{l}}_{\text{2}}}{\text{(g}}) + 2{\text{NO (g}}) \rightleftharpoons 2{\text{NOCl (g}})$ ; ${K_c} = 3.7 \times {10^8}$
${\text{C}}{{\text{l}}_{\text{2}}}{\text{(g}}) + 2{\text{N}}{{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{\text{N}}{{\text{O}}_{\text{2}}}{\text{Cl (g}})$ ; ${K_c} = 1.8$

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Ans: If the value of ${{\text{K}}_{\text{c}}}$ lies between ${10^{ - 3}}$ and ${10^3}$ , a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration of reactants and products.

Q32Written answer

The value of Kc for the reaction ${\text{3}}{{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{{\text{O}}_{\text{3}}}{\text{(g}})$ is $2.0 \times {10^{ - 50}}$ at 25°C. If the equilibrium concentration of ${{\text{O}}_{\text{2}}}$ in air at 25°C is $1.6 \times {10^{ - 2}}$ , what is the concentration of ${{\text{O}}_{\text{3}}}$ ?

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Ans: The given reaction is:

${\text{3}}{{\text{O}}_{\text{2}}}{\text{(g}}) \rightleftharpoons 2{{\text{O}}_{\text{3}}}{\text{(g}})$

${K_C} = \dfrac{{{{\left[ {{O_{3(g)}}} \right]}^2}}}{{{{\left[ {{O_{2(g)}}} \right]}^3}}}$

${K_C} = 2.0 \times {10^{ - 50}}$ and $\left[ {{O_{2(g)}}} \right] = 1.6 \times {10^{ - 2}}$

${2.0 \times {{10}^{ - 50}} = \dfrac{{{{\left[ {{O_{3(g)}}} \right]}^2}}}{{{{\left[ {1.6 \times {{10}^{ - 2}}} \right]}^3}}}} $

${ \Rightarrow {{\left[ {{O_{3(g)}}} \right]}^2} = 2.0 \times {{10}^{ - 50}} \times {{\left( {1.6 \times {{10}^{ - 2}}} \right)}^3}} $

${ \Rightarrow {{\left[ {{O_{3(g)}}} \right]}^2} = 8.192 \times {{10}^{ - 50}}} $

${ \Rightarrow \left[ {{O_{3(g)}}} \right] = 2.86 \times {{10}^{ - 28}}M}$

Hence, the concentration of ${{\text{O}}_{\text{3}}}$ is $2.86 \times {10^{ - 28}}{\text{M}}$ .

Q33Written answer

The reaction ${\text{CO (g}}) + {\text{3}}{{\text{H}}_{\text{2}}}{\text{(g}}) \rightleftharpoons {\text{C}}{{\text{H}}_{\text{4}}}{\text{(g}}) + {{\text{H}}_{\text{2}}}{\text{O(g}})$ is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of ${\text{CO}}$ , 0.10 mol of ${{\text{H}}_{\text{2}}}$ and 0.02 mol of ${{\text{H}}_{\text{2}}}{\text{O}}$ and an unknown amount of ${\text{C}}{{\text{H}}_{\text{4}}}$ in the flask. Determine the concentration of ${\text{C}}{{\text{H}}_{\text{4}}}$ in the mixture. The equilibrium constant, ${{\text{K}}_{\text{c}}}$ for the reaction at the given temperature is 3.90.

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Ans: Let the concentration of methane at equilibrium be x.

${\text{CO (g}}) + {\text{3}}{{\text{H}}_{\text{2}}}{\text{(g}}) \rightleftharpoons {\text{C}}{{\text{H}}_{\text{4}}}{\text{(g}}) + {{\text{H}}_{\text{2}}}{\text{O(g}})$

At equilibrium, the concentration of ${\text{CO}}$ is 0.3 M , ${{\text{H}}_{\text{2}}}$ is 0.1 M and ${{\text{H}}_{\text{2}}}{\text{O}}$ is 0.02 M

Therefore,

${\dfrac{{\left[ {C{H_{4(g)}}} \right]\left[ {{H_2}{O_{(g)}}} \right]}}{{\left[ {C{O_{(g)}}} \right]{{\left[ {{H_{2(g)}}} \right]}^3}}} = {K_c}} $

${ \Rightarrow \dfrac{{x \times 0.02}}{{0.3 \times {{(0.1)}^3}}} = 3.90} $

${ \Rightarrow x = \dfrac{{3.90 \times 0.3 \times {{(0.1)}^3}}}{{0.02}}} $

= $\dfrac{0.00117}{0.02} $

$ = 0.0585M$

${ = 5.85 \times {{10}^{ - 2}}M}$

Hence, the concentration of ${\text{C}}{{\text{H}}_{\text{4}}}$ at equilibrium is $5.85 \times {10^{ - 2}}{\text{M}}$.

Q34Written answer

What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:

${\text{HN}}{{\text{O}}_2},{\text{C}}{{\text{N}}^ - },{\text{HCl}}{{\text{O}}_4},{{\text{F}}^ - },{\text{O}}{{\text{H}}^ - },{\text{CO}}_3^{2 - },{{\text{S}}^ - }$

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Ans: A conjugate acid-base pair varies from one other by only one proton. In the table below, the conjugate acid-base for each species is listed.

Species	Conjugate Acid-Base
${\text{HN}}{{\text{O}}_{\text{2}}}$	${\text{N}}{{\text{O}}_3}^ - $ (base)
${\text{C}}{{\text{N}}^ - }$	${\text{HCN}}$ (acid)
${\text{HCl}}{{\text{O}}_{\text{4}}}$	${\text{Cl}}{{\text{O}}_4}^ - $ (base)
${{\text{F}}^ - }$	${\text{HF}}$ (acid)
${\text{O}}{{\text{H}}^ - }$	${{\text{H}}_{\text{2}}}{\text{O}}$ (acid) / ${{\text{O}}^{2 - }}$ (base)
${\text{CO}}_3^{2 - }$	${\text{HC}}{{\text{O}}_3}^ - $ (acid)
${{\text{S}}^ - }$	${\text{H}}{{\text{S}}^ - }$ (acid)

Q35Written answer

Which of the followings are Lewis acids? ${{\text{H}}_{\text{2}}}{\text{O, B}}{{\text{F}}_{\text{3}}}{\text{, }}{{\text{H}}^{\text{ + }}}{\text{, N}}{{\text{H}}_4}^ + $.

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Ans: Lewis acids are those acids which can accept a pair of electrons. For example: ${\text{B}}{{\text{F}}_{\text{3}}}{\text{, }}{{\text{H}}^{\text{ + }}}{\text{, N}}{{\text{H}}_{\text{4}}}^{\text{ + }}$ are Lewis acids.

Q36Written answer

What will be the conjugate bases for the Brönsted acids: ${\text{HF, }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{, HC}}{{\text{O}}_{\text{3}}}$

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Ans: The table below lists the conjugate bases for the given Bronsted acids:

Bronsted Acid	Conjugate Base
${\text{HF}}$	${{\text{F}}^ - }$
${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$	${\text{HS}}{{\text{O}}_4}^ - $
${\text{HC}}{{\text{O}}_3}^ - $	${\text{CO}}_3^{2 - }$

Q37Written answer

Write the conjugate acids for the following Brönsted bases: ${\text{NH}}_2^ - ,\;{\text{N}}{{\text{H}}_3},\;{\text{HCO}}{{\text{O}}^ - }$

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Ans: The table below lists the conjugate acids for the given Bronsted bases:

Bronsted Base	Conjugate Acid
${\text{N}}{{\text{H}}_2}^ - $	${\text{N}}{{\text{H}}_{\text{3}}}$
${\text{N}}{{\text{H}}_{\text{3}}}$	${\text{N}}{{\text{H}}_4}^ - $
${\text{HCO}}{{\text{O}}^ - }$	${\text{HCOOH}}$

Q38Written answer

The species: ${{\text{H}}_{\text{2}}}{\text{O}},{\text{HCO}}_3^ - ,{\text{HSO}}_4^ - ,{\text{N}}{{\text{H}}_3}$ can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.

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Ans: The table below lists the conjugate acids and conjugate bases for the given species.

Species	Conjugate Acid	Conjugate Base
${{\text{H}}_{\text{2}}}{\text{O}}$	${{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}$	${\text{OH}}$
${\text{HCO}}_3^ - $	${{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$	${\text{CO}}_3^{2 - }$
${\text{HSO}}_4^ - $	${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$	${\text{SO}}_4^{2 - }$
${\text{N}}{{\text{H}}_{\text{3}}}$	${\text{NH}}_{\text{4}}^{\text{ + }}$	${\text{NH}}_2^ - $

Q39Written answer

Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:

${\text{O}}{{\text{H}}^ - }$

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Ans: ${\text{O}}{{\text{H}}^ - }$is a Lewis base since it can donate its lone pair of electrons

${{\text{F}}^ - }$

Ans: ${{\text{F}}^ - }$ is a Lewis base since it can donate a pair of electrons.

${{\text{H}}^{\text{ + }}}$

Ans: ${{\text{H}}^{\text{ + }}}$ is a Lewis acid since it can accept a pair of electrons.

${\text{BC}}{{\text{l}}_{\text{3}}}$

Ans: ${\text{BC}}{{\text{l}}_{\text{3}}}$ is a Lewis acid since it can accept a pair of electrons.

Q40Written answer

The concentration of hydrogen ion in a sample of soft drink is $3.8 \times {10^{ - 3}}{\text{ M}}$ . what is its pH?

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Ans: Given,

$[{{\text{H}}^{\text{ + }}}] = 3.8 \times {10^{ - 3}}{\text{ M}}$

Therefore, the pH value of the soft drink:

${\text{pH}} = {\text{}} - \log [{H^ + }]{\text{}}$

${\text{pH}} = {\text{}} - \log (3.8 \times {10^{ - 3}})$

${\text{pH}} = {\text{}} - \log 3.8 - \log {10^{ - 3}}$

${\text{pH}} = {\text{}} - \log 3.8 + 3$

${\text{pH}} = {\text{}} - 0.58 + 3$

${\text{pH}} = 2.42$

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