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Equilibrium constant K(p) for H(2)S(g)...

Equilibrium constant `K_(p)` for
`H_(2)S(g) hArr 2H_(2)(g)+S_(2)(g)`
is `0.0118` atm at `1065^(@)C` and heat of dissociation is `42.4` Kcal. Find equilibrium constant at `1132^(@)C`.

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The correct Answer is:
A, B
`2.303 "log"K_(p_(2))/K_(p_(1))=(DeltaH)/(R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`2.303"log"K_(p_(2))/K_(p_(1))=(42.4xx10^(3))/(2)[(1405-1338)/(1405xx1338)]`
`:. K_(p_(2))/K_(p_(1))=2.129`
`:. K_(p_(2))=2.129xx0.0118=0.025 "atm"`
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