Question

Methanol, ch3oh, is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen. co(g) + 2 h2(g) equilibrium reaction arrow 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 1

Answer:

6.0588 x 10⁻⁷.

Explanation:

For the reaction: CO + 2H₂ ↔ CH₃OH,

Kc = [products]/[reactants] = [CH₃OH]/[CO][H₂]².

[CH₃OH] = no. of moles/V = (0.0406 mol)/(2.0 L) = 0.0203 mol/L.

[CO] = no. of moles/V = (0.170 mol)/(2.0 L) = 0.085 mol/L.

[H₂] = no. of moles/V = (0.302 mol)/(2.0 L) = 0.151 mol/L.

∴ Kc = [CH₃OH]/[CO][H₂]² = (0.0203 mol/L)/(0.085 mol/L)(0.151 mol/L)² = 10.47.

∵ Kp =Kc(RT)∧Δn.

R is the general gas constant = 8.314 J/mol.K.

T is the temperature of the vessel = 500.0 K.

Δn =  no. of moles of gases in products - no. of moles of gases in the reactants = 1 -3 = - 2.

∴ Kp =Kc(RT)∧Δn = (10.47)[(8.314)(500.0)]⁻² = 6.0588 x 10⁻⁷.