What is the value of the equilibrium constant?

Question

For the reaction: 2 A (g) + B (s) ⇌ 2 C (s) + D (g) At 298 K in a 10.0 L vessel, the equilibrium values are as follows: 0.721 atm of A, 4.18 mol of B, 6.25 mol of C, and 2.93 atm of D.

Answer 1

2 A (g) + B (s) ⇌ 2 C (s) + D (g)

Keq = [D] / [A]²

First, we'll determine the [A] using the ideal gas law:

PV = nRT

n = PV/RT = (0.721 atm)(10 L) / (0.0821 Latm/Kmol)(298K)

n = 0.295 moles

[A] = 0.295 mols / 10 L = 0.0295 M

Now we do the same for D:

PV = nRT

n = PV/RT = (2.39 atm)(10 L) / (0.0821 Latm/Kmol)(298K)

n = 0.977 mols

[D] = 0.977 mols / 10 L = 0.0977 M

Now plug these concentrations into the Keq expression:

Keq = 0.0977 / (0.0295)² = 0.0977 / 0.000870

Keq = 112

(be sure to check the math)