1
MCQ (Single Correct Answer)

### AIPMT 2008

The values of for the reactions,

X $\rightleftharpoons$ Y + Z      . . . .(i)
A $\rightleftharpoons$ 2B       . . . .(ii)

are in the ratio 9 : 1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (i) and (ii) are in the ratio
A
36 : 1
B
1 : 1
C
3 : 1
D
1 : 9

## Explanation

Given

X $\rightleftharpoons$ Y + Z      . . . .(i)

A $\rightleftharpoons$ 2B       . . . .(ii)

Let the total pressure for reaction (i) and (ii) be P1 and P2 respectively, then

${{{K_{{P_1}}}} \over {{K_{{P_2}}}}} = {9 \over 1}$

X Y + Z
Initial mole 1 0 0
At equilibrium 1 - $\alpha$ $\alpha$ $\alpha$

Total number of moles at equilibrium

= 1 - $\alpha$ + $\alpha$ + $\alpha$ = 1 + $\alpha$

$\therefore$ KP1 = ${{{P_Y} \times {P_Z}} \over {{P_X}}}$ = ${{{\alpha \over {1 + \alpha }} \times {P_1} \times {\alpha \over {1 + \alpha }} \times {P_1}} \over {{{1 - \alpha } \over {1 + \alpha }} \times {P_1}}}$

A 2B
Initial mole 1 0
At equilibrium 1 - $\alpha$ 2$\alpha$

Total number of moles at equilibrium

= 1 - $\alpha$ + 2$\alpha$ = 1 + $\alpha$

$\therefore$ KP2 = ${{{{\left( {{P_B}} \right)}^2}} \over {{P_A}}}$ = ${{{{\left( {{{2\alpha } \over {1 + \alpha }} \times {P_2}} \right)}^2}} \over {{{1 - \alpha } \over {1 + \alpha }} \times {P_2}}}$

$\therefore$ ${{{K_{{P_1}}}} \over {{K_{{P_2}}}}} = {{{P_1}} \over {4{P_2}}}$

$\Rightarrow$ ${{{P_1}} \over {4{P_2}}} = {9 \over 1}$

$\Rightarrow$ ${{{P_1}} \over {{P_2}}} = {{36} \over 1}$
2
MCQ (Single Correct Answer)

### AIPMT 2008

If the concentration of OH$-$ ions in the reaction
Fe(OH)3(s) $\rightleftharpoons$ Fe3+(aq) + 3OH$-$(aq)
is decreased by 1/4 times, then equilibrium concentration of Fe3+ will increase by
A
64 times
B
4 times
C
8 times
D
16 times

## Explanation

Fe(OH)3(s) $\rightleftharpoons$ Fe3+(aq) + 3OH$-$(aq)

Equilibrium constant, Kc = ${{\left[ {F{e^{3 + }}} \right]{{\left[ {O{H^ - }} \right]}^3}} \over {\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}$

We know, equilibrium constant remains same at constant temperature.

Now, let the increase in concentration of Fe3+ be x times.

$\therefore$ Kc = ${{\left[ {x \times F{e^{3 + }}} \right]{{\left[ {{1 \over 4} \times O{H^ - }} \right]}^3}} \over {\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}$

= ${x \over {64}}{{\left[ {F{e^{3 + }}} \right]{{\left[ {O{H^ - }} \right]}^3}} \over {\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}$

$\Rightarrow$ Kc = ${x \over {64}}$Kc

$\Rightarrow$ ${x \over {64}}$ = 1

$\Rightarrow$ x = 64
3
MCQ (Single Correct Answer)

### 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 = ${\left( {{{2{K_p}} \over P}} \right)^{{1 \over 3}}}$
4
MCQ (Single Correct Answer)

### 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

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