It is planned to carry out the reaction:
`CaCO_(3)(g) hArr CaO(s) +CO_(2)(g)` at `1273K` and `1`bar pressure.
`Delta_(r)G^(Theta) = 176 kJ mol^(-1)` and `Delta_(r)S^(Theta) = 157.2 kJ mol^(-1)`
a. Is the reaction spontaneous at this temperature and pressure ?
b. Calculate the value of
i. `K_(p)` at `1273K` for the reaction
ii. partial pressure of `CO_(2)` at equilibrium

Explanation

a. `Delta_(r)G^(Theta)` for the reaction at `1273 K`
`Delta_(r)G^(Theta) = Delta_(r)H^(Theta) - T Delta_(r)S^(Theta)`
`Delta_(r)G^(Theta) = 176 kJ mol^(-1), Delta_(r)S^(Theta) = 157.2 J mol^(-1)` and `T = 1273 K`
`:. Delta_(r)H^(Theta) = 176 kJ mol^(-1) - 1273 xx 157.2 xx 10^(-3) kJ mol^(-1)`
`=176 - 200.1 =- 24.10 kJ`
Since `Delta_(r)G^(Theta)` is negative, the reaction is spontaneous.
b. i. `Delta_(r)^(Theta) =- 2.303 RT log K_(P)`
or `logK_(p) =- (Delta_(r)G^(Theta))/(2.303 RT)`
`Delta_(r)G^(Theta) =- 24.10 kJ mol^(-1), R = 8.314 xx 10^(-3) kJ mol^(-1), T = 1273 K`
`:. log K_(P) =- (-24.10)/(2.303 xx 8.314 xx 10^(-3)xx1273) = 0.9887`
`:. K_(p) = Antilog (0.9887) = 9.74`
ii. For the reaction,
`K_(P) = P_(CO_(2))`
`:. P_(CO_(2)) = 9.74 bar`