Question

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 1

Correct Answer - At `300 K,DeltaG^(@)=-28.44 kJ,K_(p)=8.94xx10^(4)`
`at 1200 K,DeltaG^(@)=+2.59 kJ,K_(p)=0.77`