Question

The reaction, 2A${}_{\left(g\right)}$ + B${}_{\left(g\right)}$ $\rightleftharpoons$ 3C${}_{\left(g\right)}$ + D${}_{\left(g\right)}$ is begun with the concentrations of A and B both at an initial value of 1.00 M.

When equilibrium is reached the concentration of D is measured and found to be 0.25 M.

What is the value of the equilibrium constant for the given reaction?

• A. $\frac{\left(0.75{)}^3\right(0.25)}{\left(1.0{)}^2\right(1.0)}$
• B. $\frac{\left(0.75{)}^3\right(0.25)}{\left(0.5{)}^2\right(0.75)}$
• C. $\frac{\left(0.75{)}^3\right(0.25)}{\left(0.5{)}^2\right(0.25)}$
• D. $\frac{\left(0.75{)}^3\right(0.25)}{\left(0.75{)}^2\right(0.25)}$

Answer 1

Answer: (B)

$\frac{\left(0.75{)}^3\right(0.25)}{\left(0.5{)}^2\right(0.75)}$

The initial and equilibrium concentrations are given below.

	2A${}_{\left(g\right)}$	+	B${}_{\left(g\right)}$	$\rightleftharpoons$	3C${}_{\left(g\right)}$	+	D${}_{\left(g\right)}$
Initial molarity (M)	1		1		0		0
Equilibrium molarity (M)	1 - (2 x 0.25) = 0.5		1 - 0.25 = 0.75		3 x 0.25 = 0.75		0.25

The expression for the equilibrium constant is as shown below.

K = $\frac{\left[C{]}^3\right[D]}{\left[A{]}^2\right[B]}$

Substitute values in the above expression.

$K=\frac{\left(0.75{)}^3\right(0.25)}{\left(0.5{)}^2\right(0.75)}$.

This expression gives the value of the equilibrium constant.

So, the correct option is $B$