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`.

Equilibrium constant `K_(p)` for `H_(2)S(g) hArr 2H_(2)(g)+S_(2)(g)` i

1.	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`.
Answer» Correct Answer - 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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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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