Question

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 1

`2.303log_(10)"(K_(2))/(K_(1))=(DeltaH)/(K)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`2.303log_(10)"(K_(p_(2)))/(K_(p_(1)))=(42.4xx10^(3))/(2)[(1405-1338)/(1405xx1338)]`
`:. (K_(p_(2)))/(K_(p_(1)))=2.129`
`:. K_(p_(2))=2.129xx0.469`
`=0.998~~1 atm`