Methanol (CH3OH) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen:

CO (g) + 2H2 (g) ⇌CH3OH (g).

An equilibrium mixture in a 1.50 L vessel is found to contain 0.0679 mol CH3OH, 0.160 mol CO, and 0.302 mol H2 at 500 K.

Required:

Calculate Kc at this temperature.

Question

Darnell Crane

Chemistry

26 August, 10:40

Methanol (CH3OH) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen:

CO (g) + 2H2 (g) ⇌CH3OH (g).

An equilibrium mixture in a 1.50 L vessel is found to contain 0.0679 mol CH3OH, 0.160 mol CO, and 0.302 mol H2 at 500 K.

Required:

Calculate Kc at this temperature.

+4

Answers (1)

Know the Answer?

Christopher Henry · Answer 1

Answer: Kc = 4.653

Explanation:

From The equation of reaction

CO (g) + 2H2 (g) ⇌CH3OH (g)

From the question

[CH3OH] = 0.0679 moles

[CO] = 0.160moles

[H2] = 0.302

Kc = [CH3OH]/[CO] [H2]^2

Kc=[0.069]/[0.16]*[0.302]^2

Kc = 4.653

Methanol (CH3OH) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen:CO (g) + 2H2 (g) ⇌CH3OH (g).An equilibrium mixture in a 1.50 L vessel is found to contain 0.0679 mol CH3OH, 0.160 mol CO, and 0.302 mol H2 at 500 K.Required:Calculate Kc at this temperature.