Solving Equilibrium Constant: 2A(g)⇌B(g)+C(g)

[jdstokes]

Samples of A (2.5 mol) and B (1.0 mol) are placed in a 5.0 L contained and the following reaction takes place

\rm 2A(g) \rightleftharpoons B(g) + C(g)

At equilibrium, the concentration of A is 0.20 M. What is the vlaue of K_\mathrm{c}?

n_0(\mathrm{A}) = 2.5\thinspace\mathrm{mol}
n_0(\mathrm{B}) = 1.0\thinspace\mathrm{mol}
n_\mathrm{eq}(\mathrm{A}) = 1.0\thinspace\mathrm{mol}
\Delta n(\mathrm{A}) = -1.5\thinspace\mathrm{mol}
\Delta n(\mathrm{B}) = -0.75\thinspace\mathrm{mol} = \Delta n(\mathrm{C})

Now,

K_\mathrm{c} = \frac{c(\mathrm{B})c(\mathrm{C})}{c(\mathrm{A})^2}

Putting

c(\mathrm{A}) = 0.2\thinspace\mathrm{mol}
c(\mathrm{B}) = \frac{1.0-0.75}{5.0}\thinspace\mathrm{mol} = 0.050\thinspace\mathrm{mol}
c(\mathrm{C}) = \frac{0.75}{5.0}\thinspace\mathrm{mol} = 0.15\thinspace\mathrm{mol}

I compute

K_\mathrm{c} = 0.1875\thinspace\mathrm{mol}.

But the correct answer is 1.3 mol.

 

[siddharth]

c(B) must be (1.0 + 0.75)/5 mol/L

 

[jdstokes]

Thanks for the tip.