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) hArr CO_(2)(g) +H_(2)(g)`
`DeltaH^(Theta)underset(300K). =- 41.16 kJ mol^(-1)`
`DeltaS^(Theta)underset(300K). =- 4.14 xx 10^(-2) kJ mol^(-1)`
`DeltaH^(Theta)underset(1200K). =- 31.93 kJ mol^(-1)`
`DeltaH^(Theta)underset(1200K). =- 2.96 xx 10^(-2) kJ mol^(-1)`
Calculate `K_(p)` at each temperature and predict the direction of reaction at `300K` and `1200k`, when `P_(CO) = P_(CO_(2)) =P_(H_(2)) = P_(H_(2)O) =1` atm at initial state.

Answer 1

At `300K: DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)`
`=- 41.16 - 300 xx (-4.24xx10^(-2))`
`=- 28.44 kJ`
Since, `DeltaG^(Theta)` is negative, hence recaiton is spontaneous is forward direction.
`DeltaG^(Theta) =- 2.3030 RT log K_(p)`
`-28.44 = - 2.303 xx 8.314 xx 10^(-3) xx 300 log_(10) K_(p)`
`K_(p) = 8.93 xx 10^(4)`
At `1200K: DeltaG^(Theta) = DeltaH^(Theta) - T DeltaS^(Theta)`
`=- 32.93 - 1200 (-2.96 xx 10^(-2))`
` =+ 2.59 kJ`
Positive value of `DeltaG^(Theta)` shows that the reaction is sponteneous in backward direction
`Deltag^(Theta) =- 2.303 RT log_(10) K_(p)`
`2.59 =- 2.303 xx 8.314 xx 1200 log K_(p)`
`K_(p) = 0.77`