Question

For the reaction $2A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right);K_c={10}^{12}$. If the initial moles of $A,B,C,D$ are $2,1,7,3$ moles respectively in a one litre vessel. What is the equilibrium concentration of $A$?

• A. $4\times {10}^{-4}$
• B. $2\times {10}^{-4}$
• C. ${10}^{-4}$
• D. $8\times {10}^{-4}$

Answer 1

The correct option is A $4\times {10}^{-4}$

Since the volume of the vessel is 1 L, the number of moles correspond to the molar concentration.

	A	B	C	D
Initial concentration (M)	2	1	7	3
Change (M)	-2x	-x	x	x
Equilibrium concentration (M)	2-2x	1-x	7+x	3+x

The equilibrium constant expression is
$K_c=\frac{\left[C\right]\left[D\right]}{\left[A{]}^2\right[B]}$
${10}^{12}=\frac{(7+x)(3+x)}{(2-2x{)}^2(1-x)}$
$x=0.9998$
Hence, the equilibrium concentration of A is $\left[A\right]=2-2x=2-2\left(0.9998\right)=4\times {10}^{-4}$ M