1

### AIPMT 2008

The dissociation equilibrium of a gass AB2 can be represented as :
2AB2(g) $\rightleftharpoons$ 2AB(g) + B2(g)
The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure P is
A
(2Kp/P)1/2
B
(Kp/P)
C
(2Kp/P)
D
(2Kp/P)1/3

## Explanation

2AB2(g) &#8652; 2AB(g) + B2(g)
Initial mole 2 0 0
At equilibrium 2(1 - x) 2x x

Amount of moles at equilibrium = 2(1 – x) + 2x + x = 2 + x

${K_p} = {{{{\left[ {{p_{AB}}} \right]}^2}\left[ {{p_{{B_2}}}} \right]} \over {{{\left[ {{p_{A{B_2}}}} \right]}^2}}}$

= ${{{{\left[ {{{2x} \over {2 + x}} \times P} \right]}^2}\left[ {{x \over {2 + x}} \times P} \right]} \over {{{\left[ {{{2\left( {1 - x} \right)} \over {2 + x}} \times P} \right]}^2}}}$

= ${{\left[ {{{4{x^3}} \over {2 + x}} \times P} \right]} \over {4{{\left( {1 - x} \right)}^2}}}$

$\Rightarrow$ Kp = ${{{4{x^3} \times P} \over 2} \times {1 \over 4}}$

($\because$ 1 – x ≈ 1 and 2 + x ≈ 2)

$\Rightarrow$ x = ${\left( {{{8{K_p}} \over {4P}}} \right)^{{1 \over 3}}}$

$\Rightarrow$ x = ${\left( {{{2{K_p}} \over P}} \right)^{{1 \over 3}}}$
2

### AIPMT 2007

Calculate the pOH of a solution at 25oC that contains 1 $\times$ 10$-$10 M of hydronium ions, i.e. H3O+.
A
4.000
B
9.000
C
1.000
D
7.000

## Explanation

Given, [H3O+.] = 1 $\times$ 10$-$10

$\Rightarrow$ pH = 10

Also we know, pH + pOH = 14

$\Rightarrow$ pOH = 14 - pH = 14 - 10 = 4
3

### AIPMT 2007

A weak acid, HA, has a Ka of 1.00 $\times$ 10$-$5. If 0.100 mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closest to
A
1.00%
B
99.9%
C
0.100%
D
99.0%

## Explanation

For weak acid degree of dissociation,

$\alpha$ = $\sqrt {{{{K_a}} \over C}}$

= $\sqrt {{{1 \times {{10}^{ - 5}}} \over {0.1}}} = {10^{ - 2}}$ = 1.00 %
4

### AIPMT 2007

The equilibrium constants of the following are

N2 + 3H2 $\rightleftharpoons$ 2NH3;     K1

N2 + O2 $\rightleftharpoons$ 2NO;     K2

H2 + ${1 \over 2}$O2 $\rightleftharpoons$ H2O;     K3

The equilibrium constant (K) of the reaction :

2NH3 + ${5 \over 2}$ O2 $\rightleftharpoons$ 2NO + 3H2O will be
A
K2K33/K1
B
K2K3/K1
C
K23K3/K1
D
K1K33/K2

## Explanation

2NH3 $\rightleftharpoons$ N2 + 3H2;     ${1 \over {{K_1}}}$

N2 + O2 $\rightleftharpoons$ 2NO;     K2

3H2 + ${3 \over 2}$O2 $\rightleftharpoons$ 3H2O;     (K3)3

By adding all equations, we get

2NH3 + ${5 \over 2}$ O2 $\rightleftharpoons$ 2NO + 3H2O

$\therefore$ K = ${{{K_2} \times {{\left( {{K_3}} \right)}^3}} \over {{K_1}}}$