To calculate the equilibrium constant Kp for the given reaction, we need to know the equilibrium partial pressures of all the species involved in the reaction. We are given the partial pressures of H2 and O2 at equilibrium, but we need to find the partial pressure of H2O at equilibrium as well.Let's assume that initially, there is no H2O present. As the reaction proceeds, some amount of H2 and O2 will react to form H2O. Let's denote the change in the partial pressures of H2, O2, and H2O as x.At equilibrium, the partial pressures will be:H2: 0.15 - 2xO2: 0.10 - xH2O: 2xNow, we can write the expression for Kp:Kp = [H2O]^2 / [H2]^2 * [O2] Plugging in the equilibrium partial pressures:Kp = 2x ^2 / 0.15 - 2x ^2 * 0.10 - x However, we still need the value of Kp at 298 K to solve for x. We can find this value using the standard Gibbs free energy change G and the following equation:Kp = exp -G / RT Where R is the gas constant 8.314 J/molK and T is the temperature in Kelvin 298 K . For the given reaction, the standard Gibbs free energy change G is -228.6 kJ/mol. Converting this to J/mol, we get -228600 J/mol.Now, we can find Kp at 298 K:Kp = exp - -228600 J/mol / 8.314 J/molK * 298 K Kp 1.22 10^83Now we have the value of Kp, we can plug it back into the equation we derived earlier:1.22 10^83 = 2x ^2 / 0.15 - 2x ^2 * 0.10 - x This equation is quite difficult to solve analytically, but we can make an approximation. Since Kp is extremely large, the reaction is heavily favored towards the formation of H2O. Therefore, we can assume that x is very close to the initial partial pressures of H2 and O2, meaning that 2x 0.15 and x 0.10.With this approximation, we can calculate the equilibrium partial pressures:H2: 0.15 - 2x 0O2: 0.10 - x 0H2O: 2x 0.15Now, we can calculate the approximate equilibrium constant using these partial pressures:Kp [H2O]^2 / [H2]^2 * [O2] Kp 0.15 ^2 / 0 ^2 * 0 Since the denominator is approximately zero, Kp is extremely large, which is consistent with our earlier calculation of Kp 1.22 10^83. This confirms that the reaction is heavily favored towards the formation of H2O at 298 K.