Answer Questions 1 -4 about the following reaction. DON'T FORGET TO INCUDE UNITS WHEREAPPROPRIATE \( \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \Leftrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \quad \mathrm{Kp} \) at \( 675 \mathrm{~K}=\mathbf{1 1 . 8} \) 1.) If the initial pressures of carbon monoxide and steam are 2.00 atm , will the reaction run forwards or backwards... explain.

To determine whether the reaction will run forwards or backwards, we need to compare the reaction quotient \( Q_p \) to the equilibrium constant \( K_p \). The reaction is: \[ \mathrm{CO}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \Leftrightarrow \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2}(\mathrm{g}) \] The expression for the equilibrium constant \( K_p \) is given by: \[ K_p = \frac{P_{\mathrm{CO}_2} \cdot P_{\mathrm{H}_2}}{P_{\mathrm{CO}} \cdot P_{\mathrm{H}_2O}} \] Where: - \( P_{\mathrm{CO}} \) is the partial pressure of carbon monoxide, - \( P_{\mathrm{H}_2O} \) is the partial pressure of steam, - \( P_{\mathrm{CO}_2} \) is the partial pressure of carbon dioxide, - \( P_{\mathrm{H}_2} \) is the partial pressure of hydrogen. Given that the initial pressures of carbon monoxide and steam are both 2.00 atm, we can calculate \( Q_p \) at the start of the reaction. Initially, the pressures of the products \( P_{\mathrm{CO}_2} \) and \( P_{\mathrm{H}_2} \) are both 0 atm since the reaction has not yet proceeded. Thus, the expression for \( Q_p \) at the start is: \[ Q_p = \frac{P_{\mathrm{CO}_2} \cdot P_{\mathrm{H}_2}}{P_{\mathrm{CO}} \cdot P_{\mathrm{H}_2O}} = \frac{0 \cdot 0}{2.00 \cdot 2.00} = 0 \] Now we compare \( Q_p \) to \( K_p \): - \( Q_p = 0 \) - \( K_p = 11.8 \) Since \( Q_p < K_p \), the reaction will proceed in the forward direction to reach equilibrium. This means that the reaction will produce more products (CO₂ and H₂) until the equilibrium is established. **Conclusion:** The reaction will run forwards because \( Q_p < K_p \).