For the reaction:
`CO(g) +H_(2)O(g) hArr CO_(2)(g) +H_(2)(g)`
`(Delta_(r)H)_(300K) = - 41.2 kJ mol^(-1)`
`(Delta_(r)H)_(1200K) =- 33.0 kJ mol^(-1)`
`(Delta_(r)S)_(300K) = - 4.2 xx 10^(-2) kJ mol^(-1)`
`(Delta_(r)S)_(1200K) =- 3.0 xx10^(-2) kJ mol^(-1)`
Predict the direction of spontaneity of the reaction at `300K` and `1200K`. also calculated `log_(10)K_(p)` at `300K` and `1200K`.

Explanation

To solve the problem, we need to determine the spontaneity of the reaction at two different temperatures (300 K and 1200 K) and calculate the logarithm of the equilibrium constant (log Kp) at these temperatures. We'll use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

### Step 1: Calculate ΔG at 300 K

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