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) ⇌ 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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