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VITEEE 2010 Paper

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Question : 53 of 120

Marks: +1, -0

For the reaction,

2\mathrm{A}(g)+\mathrm{B_2}(g)\rightleftharpoons2\mathrm{AB_2}(g)

The equilibrium constant,

K_{p}

at 300 K is 16.0. The value of

K_{p}

\mathrm{AB_2}(g)\rightleftharpoons\mathrm{A}(g)+\frac{1}{2}\;\;\mathrm{B_2}(g)

is

8
0.25
0.125
32

Solution:

For the given reaction,

2\mathrm{A}(g)+\mathrm{B_2}(g)\rightleftharpoons2\mathrm{AB_2}(g)

The equilibrium constant,

K_{p}=\frac{P^{2}_{\mathrm{AB_2}}}{P^{2}_{\mathrm{A}}\cdot P_{\mathrm{B_2}}}=16

...(i)

For the other given reaction,

\mathrm{AB_2}(g)\rightleftharpoons\mathrm{A}(g)+\frac{1}{2}\mathrm{B_2}(g)

The equilibrium constant,

K'_{p}=\frac{P_{\mathrm{A}}\cdot P_{\mathrm{B_2}}^{1/2}}{P_{\mathrm{AB_2}}}

...(ii)

On squaring Eq.(ii), we obtain,

(K'_{p})^{2}=\frac{P^{2}_{\mathrm{A}}\cdot P_{\mathrm{B_2}}}{P^{2}_{\mathrm{AB_2}}}

...(iii)

Now, from Eq.(i) and (ii), we obtain,

K_{p}\cdot (K'_{p})^{2}=1

\Rightarrow 16\cdot (K'_{p})^{2}=1\qquad (\because K_{p}=16.0)

\Rightarrow (K'_{p})=\left[\frac{1}{16}\right]^{1/2}

\Rightarrow K'_{p}=\frac{1}{4}=0.25

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