Methanol is produced commercially by the catalysed reaction of carbon monoxide and hydrogen: CO(g) + 2H2(g)  CH3OH(g). An equilibrium mixture in a 2.00 L vessel is found to contain 0.0406 mol CH3OH, 0.170 mol CO, and 0.302 mol H2 at 500 K. Calculate Kp at this temperature.

(Answer = 6.22  10-3)

(CO) = mols/L = ?

(H2) = mols/L = ?
(CH3OH) = mols/L = ?

Substitue the concentrations into the Kc expression and sole for Kc. Convert to Kp by Kp = Kc(RT)^delta n. 6.22E-3 is the right answer.

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To calculate Kp at a given temperature, we need to know the partial pressures of the reactants and products in the equilibrium mixture.

First, let's calculate the partial pressures of CO, H2, and CH3OH in the equilibrium mixture.

Given:
- Volume of the vessel (V) = 2.00 L
- Moles of CH3OH (nCH3OH) = 0.0406 mol
- Moles of CO (nCO) = 0.170 mol
- Moles of H2 (nH2) = 0.302 mol

To calculate the partial pressure of a gas, we can use the ideal gas law:
PV = nRT

Where:
- P is the partial pressure of the gas
- V is the volume of the vessel
- n is the number of moles of the gas
- R is the ideal gas constant (0.0821 L·atm/(K·mol))
- T is the temperature in Kelvin

Let's calculate the partial pressures of CO, H2, and CH3OH using the given values:

Partial pressure of CO (PCO) = (nCO * R * T) / V
Partial pressure of H2 (PH2) = (nH2 * R * T) / V
Partial pressure of CH3OH (PCH3OH) = (nCH3OH * R * T) / V

Substituting the given values:
PCO = (0.170 mol * 0.0821 L·atm/(K·mol) * 500 K) / 2.00 L
PH2 = (0.302 mol * 0.0821 L·atm/(K·mol) * 500 K) / 2.00 L
PCH3OH = (0.0406 mol * 0.0821 L·atm/(K·mol) * 500 K) / 2.00 L

Calculating the values:
PCO = 6.025 atm
PH2 = 10.764 atm
PCH3OH = 1.211 atm

Now, we can calculate Kp by using the expression:
Kp = (PCH3OH) / (PCO * (PH2)^2)

Substituting the calculated partial pressures:
Kp = 1.211 atm / (6.025 atm * (10.764 atm)^2)

Calculating the value:
Kp ≈ 6.22 × 10^(-3)

Therefore, the calculated value of Kp at 500 K is approximately 6.22 × 10^(-3).