At \(327^{\circ} \mathrm{C},\) the equilibrium concentrations are \(\left[\mathrm{CH}_{3} \mathrm{OH}\right]=\) \(0.15 M,[\mathrm{CO}]=0.24 M,\) and \(\left[\mathrm{H}_{2}\right]=1.1 M\) for the reaction $$\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)$$.Calculate \(K_{\mathrm{p}}\) at this temperature.

We are given the following equilibrium concentrations: - [CH3OH] = 0.15 M - [CO] = 0.24 M - [H2] = 1.1 M We also know the temperature, which is 327°C, but we should convert it to Kelvin (K) for our calculations: Temperature = 327 + 273.15 = 600.15 K

To find the partial pressures, we can use the Ideal Gas Law given by: PV = nRT Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 Latm/molK), and T is the temperature in Kelvin. Also, note that for our equilibrium constant, it doesn't matter what the volume of the container is because it is constant for all the species involved so it would cancel out. Calculating partial pressures using the Ideal Gas Law and replacing the concentrations with n/V, we get: P(CH3OH) = [CH3OH]RT P(CO) = [CO]RT P(H2) = [H2]RT Plug in the given concentrations and the temperature: P(CH3OH) = (0.15 M)(0.0821 Latm/molK)(600.15 K) = 7.37 atm P(CO) = (0.24 M)(0.0821 Latm/molK)(600.15 K) = 11.81 atm P(H2) = (1.1 M)(0.0821 Latm/molK)(600.15 K) = 54.05 atm

The expression for Kp for the given reaction is: Kp = (P(CO) * P(H2)^2) / P(CH3OH) Plug in the calculated partial pressures: Kp = (11.81 atm * (54.05 atm)^2) / 7.37 atm = 56207.13 So, the equilibrium constant Kp for the reaction CH3OH(g) ⇌ CO(g) + 2 H2(g) at 327°C is approximately 56207.13.