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Equilibrium Constan

  • Thread starter jdstokes
  • Start date
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

[itex]\rm 2A(g) \rightleftharpoons B(g) + C(g)[/itex]

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

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

Now,

[itex]K_\mathrm{c} = \frac{c(\mathrm{B})c(\mathrm{C})}{c(\mathrm{A})^2}[/itex]

Putting

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

I compute

[itex]K_\mathrm{c} = 0.1875\thinspace\mathrm{mol}[/itex].

But the correct answer is 1.3 mol.
 

siddharth

Homework Helper
Gold Member
1,110
0
c(B) must be (1.0 + 0.75)/5 mol/L
 
Thanks for the tip.
 

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