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Q.

The equilibrium constant for $2 H_{2} S (g) ⇌ 2 H_{2} (g) + S_{2} (g)$ is 0.0118 at 1300 $K$ while the heat of dissociation is 597.4 $kJ$ . The standard equilibrium constant of the action at 1200 $K$ is :

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a

$1.180 X 10^{- 4}$

b

11.80

c

118.0

d

cannot be calculated from given data

answer is A.

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Detailed Solution

$- \log \frac{K_{1}}{K_{2}} = \frac{Δ H}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$- \log [\frac{K_{1}}{0.0118}] = \frac{597.4 \times 10^{3}}{2.303 \times 8.314} [\frac{1}{1200} - \frac{1}{1300}]$

$K_{1} = 1.18 \times 10_{-}^{- 4}$

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The equilibrium constant for 2H2S(g)⇌2H2(g)+S2(g) is 0.0118 at 1300 K while the heat of dissociation is 597.4 kJ. The standard equilibrium constant of the action at 1200 K is :