Answer
$$K_p = 2.15 \times 10^4$$
Work Step by Step
$T_2/K = 0 + 273.15 = 273.15$
The reaction produces 1 mol of $CH_3OH$
$$\Delta H^o_{rxn} = -128 \space kJ/mol = -1280 \space J/mol$$
Van't Hoff's equation:
$$ln \space \frac{K_2}{K_1} = -\frac{\Delta H^o_{rxn}}{R}\Bigg(\frac {1}{T_2} - \frac{1}{T_
1} \Bigg)$$
$$ln \space \frac{K_2}{2.25 \times 10^4} = -\frac{-1280}{8.314}\Bigg(\frac {1}{273.15} - \frac{1}{298} \Bigg)$$
$$ln \space \frac{K_2}{2.25 \times 10^4} = -0.047$$
$$\frac{K_2}{2.25 \times 10^4} = e^{-0.047} = 0.954$$
$$K_2 = 2.25 \times 10^4 \times 0.954 = 2.15 \times 10^4$$
