The standard enthalpy and entropy changes for the reaction in equilibrium for the forward direction are given below:
`CO_((g))+H_(2)O_((g))hArrCO_(2(g))+H_(2(g))`
`DeltaH_(300K)^(@)=-41.16 kJ mol^(-1)`
`DeltaS_(300 K)^(@)=-4.24xx10^(-2) kJ mol^(-1)`
`DeltaH_(1200 K)^(@)=-32.93 K J mol^(-1)`
`DeltaS_(1200 K)^(@)=-2.96xx10^(-2) k J mol^(-1)`
Calculate `K_(p)` at each temperature and predict the direction of reaction at `300 K` and `1200 K`, when `P_(CO)=P_(CO_(2))=P_(H_(2))=P_(H_(2)O)=1` atm at initial state.

Answer

`K_(P) = 8.94 xx 10^(4)`, at 1200 K. `K_(P) = 0.77`

Explanation

At equilibrium `Delta G^(0) = Delta H^(0) - T Delta S^(0)`
`Delta G_(300 K)^(0) = -41.16 - 300 xx (-4.24 xx 10^(-2)) = - 28.44 kJ`
and `Delta G_(1200 K)^(0) = -32.93 - 1200 xx (-2.96 xx 10^(2)) - + 2.59 kJ`
Thus, at 300 K, reaction will process in forword direction and 1000 K back ward direction also,
`Delta g^(0) = -2.303 RT log K_(p) , :. K_(P) = 8.94 xx 10^(4)`
Similarly, at 1200 K, `K_(p) = 0.77`