# Question #87759

Aug 29, 2016

Trying to present an answer assuming slight change in the given reaction to keep consistency with the unit of equilibrium constant.

The reaction and ICE table

$\text{ "A" "+" "2 B" " rightleftharpoons" } C$

I$\text{ "x" mol "+" "4""mol " 0 " mol}$

C$\text{-1" mol "+" "-2""mol " +1 " mol}$

E$\text{(x-1)" mol "+" "2""mol " 1 " mol}$

Where x mole is the amount of A to be mixed with 4 moles of B initially

The volume of the reaction mixture given is $V = 5 {\mathrm{dm}}^{3}$.

So at equilibrium the concentration of the components will be

$\left[A\right] = \frac{x - 1}{V} {\text{mol"/"dm}}^{3}$

$\left[B\right] = \frac{2}{V} {\text{mol"/"dm}}^{3}$

$\left[C\right] = \frac{1}{V} {\text{mol"/"dm}}^{3}$

Now equilibrium constant

${K}_{c} = \frac{\left[C\right]}{\left[A\right] {\left[B\right]}^{2}} = \frac{\frac{1}{V}}{\frac{x - 1}{V} \cdot {\left(\frac{2}{V}\right)}^{2}}$

$\implies {K}_{c} = {V}^{2} / \left(4 \left(x - 1\right)\right) {\mathrm{dm}}^{6} m o {l}^{-} 2$

Inserting the value of ${K}_{c} \mathmr{and} V$ we get

$0.25 = {5}^{2} / \left(4 \left(x - 1\right)\right)$

$\implies x - 1 = 25 \implies x = 26$

So the amount of A to be added to 4 moles of B is 26 moles