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). =- 4.24 xx 10^(-2) kJ mol^(-1)`
`DeltaH^(Theta)underset(1200K). =- 32.93 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.

Explanation

To solve the problem, we need to calculate the equilibrium constant \( K_p \) at two different temperatures (300 K and 1200 K) and predict the direction of the reaction at these temperatures. We will use the given values of standard enthalpy change (\( \Delta H^\Theta \)) and standard entropy change (\( \Delta S^\Theta \)) for the reaction.

### Step 1: Calculate \( \Delta G^\Theta \) at 300 K

The Gibbs free energy change (\( \Delta G^\Theta \)) can be calculated using the formula:

\[
\Delta G^\Theta = \Delta H^\Theta - T \Delta S^\Theta
...