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

`DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
`=41.16-300xx(-4.24xx10^(-2))`
`=-28.44kJ`
since, `DeltaG^(@)` is negative hence reaction is spontaneous in forward direction,
`DeltaG^(@)=-2.303RT" "log" "K_(P)`
`-28.44=-2.303xx8.314xx10^(-3)xx300log" "K_(P)`
`K_(P)=8.93xx10^(4)`
At 1200K: `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
`=-32.93-1200(-2.96xx10^(-2))=+2.59kJ`
Positive value of `DeltaG^(@)` shows that the reaction is spontaneous in backward direction ltbr. `DeltaG^(@)=-2.303RT" "log" "K_(P)`
`2.59=-2.303xx8.314xx10^(-3)xx1200log" "K_(P)`
`K_(P)=0.77`