Question

Consider the following equilibrium system:

• PCl₅(g) -->PCl₃(g) + Cl₂(g)

A 10.00 L evacuated flask is filled with 0.4334 mol PCl₅(g) at 297.3 K. The temperature is then raised to 510.0 K, where the decomposition of PCl₅ gas takes place to an appreciable extent. When equilibrium is established, the total pressure in the flask is 2.665 atm.

What is the value of the equilbrium constant in terms of concentrations, K_c, at 510.0 K?

Answer 1

K_c = 0.0180

Explanation

Total pressure = 2.665 atm

Total concentration = (total pressure) / [(R) * (T)]

Total concentration = (2.665 atm) / [(0.0821 L-atm/mol-K) * (510.0 K)]

Total concentration = 0.06368 M

initial moles PCl₅ = 0.4334 mol

initial concentration PCl₅ = (initial moles PCl₅) / (volume)

initial concentration PCl₅ = (0.4334 mol) / (10.00 L)

initial concentration PCl₅ = 0.04334 M

ICE table	PCl5 (g)	$\rightleftharpoons$	PCl3 (g)	Cl2 (g)
Initial conc.	0.04334 M		0	0
Change	-x		+x	+x
Equilibrium conc.	0.04334 M - x		+x	+x

Total concentration = [PCl₅]_eq + [PCl₃]_eq + [Cl₂]_eq

0.06368 M = (0.04334 M - x) + (x) + (x)

0.06368 M = 0.04334 M + x

x = 0.06368 M - 0.04334 M

x = 0.02034 M

[PCl₅]_eq = 0.04334 M - x

[PCl₅]_eq = 0.04334 M - 0.02034 M

[PCl₅]_eq = 0.023 M

[PCl₃]_eq = x = 0.02034 M

[Cl₂]_eq = x = 0.02034 M

K_c = [PCl₃]_eq[Cl₂]_eq / [PCl₅]_eq

K_c = (0.02034 M) * (0.02034 M) / (0.023 M)

K_c = 0.0180