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1.6 mol of CH3OH(g) are injected into a 4.0 L container and the following equilibrium becomes established.
2H2(g) + CO(g) <=====> CH3OH(g) + 92 kJ
If at equilibrium 0.80 mol of H2 is found in the container the Ke must be which of the following?

a) 25
b) 0.78
c) 6.25
d) 4.69
e) 75.0

Respuesta :

nuralfaqih

2H2(g) + CO(g) ←→ CH3OH

At equilibrium :

n H2 = 0.8 mol

n CO = ½ • 0.8 = 0.4 mol

n CH3OH = 1.6 - (½ • 0.8) = 1.2 mol

Equilibriun constant :

K = [CH3OH] / [H2]²[CO]

K = (1.2) / (0.8)²(0.4)

K = 4.69

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