Question

Consider the reaction below. 3.2 mol of A and 6.7 mol of B are added to a 7 L container. At equilibrium, the concentration of A is 0.187 M. What is the value of the equilibrium constant?

1 A + 2 B ↔ 5 C

Answer 1

The given reaction is

$1 \ A + 2\ B \rightleftharpoons 5\ C$

The initial concentrations of A and B can be calculated by dividing the number of moles of A and B by the volume of the container assuming they are gases.

Hence,

$[A] = \frac{3.2\ mol}{7 \ L } \approx 0.457 \ mol$

$[B] = \frac{6.7\ mol}{7 \ L } \approx 0.957 \ mol$

Now, we can create an ICE table to calculate the equilibrium concentrations of the molecules,

		2 B	$\rightleftharpoons 5\ C$
Initial, M	0.457	0.957	0
Change, M	-x	-2x	+5x
Equilibrium, M	0.457-x	0.957-2x	5x

Note that according to the balanced equation, if x moles of A and 2x moles of B are consumed we get 5x moles of C.

Hence, from the ICE table, we can write the expression of equilibrium constant K_c as follows:

$K_c = \frac{[C]_{(eq)}^5}{[A]_{(eq)} \times [B]^2_{(eq)}} = \frac{(5x)^5}{(0.457-x) \times (0.957-2x)^2}$

We are given that the equilibrium concentration of A is 0.187 M.

Hence, from the ICE table, we can write

$[A]_{(eq)} = 0.457 \ M - x = 0.187 \ M \\ \Rightarrow x = 0.457 \ M - 0.187 \ M = 0.270 \ M$

Hence, the value of K_c can be calculated as

$K_c = \frac{(5x)^5}{(0.457-x) \times (0.957-2x)^2} =\frac{( 5\times 0.270) ^5 }{0.187 \times (0.957-2\times 0.270)^2} \approx 137.8 \\ \Rightarrow K_c \approx 138$

Hence, the equilibrium constant of the reaction is 138 approximately. (rounded to three significant figures).