Question

the reaction AB2C (g) <---> B2 (g) + AC (g) reached equilibrium at 900 K in a 5.00 L vessel. At equilibrium 0.0840 mol of AB2C, 0.0350 mol of B2, and 0.0590 mol of AC were detected. What is the equilibrium constant at this temperature for this system?

Answer 1

Answer:

k = 4,92x10⁻³

Explanation:

For the reaction:

AB₂C (g) ⇄ B₂(g) + AC(g)

The equilibrium constant, k is defined as:

$k = \frac{[B_{2}][AC]}{[AB_2C]}$ (1)

Molar concentration of the species are:

[AB₂C]: 0,0840mol / 5L = 0,0168M

[B₂]: 0,0350mol / 5L = 0,0070M

[AC]: 0,0590mol / 5L = 0,0118M

Replacing this values in (1):

$k = \frac{[0,0070][0,0118]}{[0,0168]}$

k = 4,92x10⁻³

I hope it helps!