1
MCQ (Single Correct Answer)

AIPMT 2008

The value of equilibrium constant of the reaction
HI(g) $$\rightleftharpoons$$ $${1 \over 2}$$H2(g) + $${1 \over 2}$$I2(g)
is 8.0. The The equilibrium constant of the reaction
H2(g) + I2(g) $$\rightleftharpoons$$ 2HI(g) will be
A
16
B
1/8
C
1/16
D
1/64

Explanation

HI(g) $$\rightleftharpoons$$ $${1 \over 2}$$H2(g) + $${1 \over 2}$$I2(g),

For this reaction, K = 8.0

Reversing the equation,

$$ \therefore $$$${1 \over 2}$$H2(g) + $${1 \over 2}$$I2(g) $$\rightleftharpoons$$ HI(g) ......(1)

For this reaction, K1 = $${1 \over 8}$$

H2(g) + I2(g) $$\rightleftharpoons$$ 2HI(g)

We get this equation by multiplying equation (1) by 2,

$$ \therefore $$ For this reaction, K2 = $${\left( {{1 \over 8}} \right)^2}$$ = $${1 \over {64}}$$
2
MCQ (Single Correct Answer)

AIPMT 2008

Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H+ ion concentration in the mixture?
A
3.7 $$ \times $$ 10$$-$$3 M
B
1.11 $$ \times $$ 10$$-$$3 M
C
1.11 $$ \times $$ 10$$-$$4 M
D
3.7 $$ \times $$ 10$$-$$4 M

Explanation

We know, pH = – log[H+]

For pH = 3, 3 = – log [H+]

$$ \Rightarrow $$ [H+] = 10–3 M

For pH = 4 , 4 = – log [H+]

$$ \Rightarrow $$ [H+] = 10–4 M

For pH = 5, 5 = – log [H+]

$$ \Rightarrow $$ [H+] = 10–5 M

Total concentration of [H+]

M = $${{{{10}^{ - 3}}\left( {1 + 0.1 + 0.01} \right)} \over 3}$$

M(V1 +V2 + V3) = M1V1 + M2V2 + M3V3

As V1 = V2 = V3 = V

$$ \Rightarrow $$M(3V) = (M1 + M2 + M3)V

$$ \Rightarrow $$ 3M = (10–3 + 10–4 + 10–5)

= $${{1.11 \times {{10}^{ - 3}}} \over 3}$$

= 3.7 $$ \times $$ 10$$-$$4 M
3
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
4
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) ⇌ 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 = $${\left( {{{2{K_p}} \over P}} \right)^{{1 \over 3}}}$$

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