Question

One mole of A and 2 moles of B are allowed to react in a 0.5 lit flask. What is the value of $K_c$? if at equilibirum, 0.4 moles of C is formed in the reaction.
$A+2B\iff C+2D$

• A. 4/9
• B. 9/4
• C. 8/27
• D. 27/8

Answer 1

Answer: (C)

The expression for the equilibrium constant is $A+2B\iff C+2D$.
The expression for the equilibrium constant is $\frac{\left[C\right][D{]}^2}{\left[A\right]\left[B\right]}$.
The following table lists the equilibrium concentration of various species.

	A	B	C	D
Initial	1	2	0	0
Change	-x	-2x	x	2x
Equilibrium	1-x	2-2x	x	2x

But $\left[C\right]=0.4=x$.
Hence, the number of moles of A,B,C and D in 0.5 L are 06,1.2,0.4 and 0.8 respectively.
The molar concentrations of A,B,C and D are 1.2, 2.4, 0.8 and 1.6 respectively.
Substitute the values in the expression for the equilibrium constant.
$K_c=\frac{0.8\times (1.6{)}^2}{1.2\times (2.4{)}^2}=\frac{8}{27}$.