Question

In the reaction, $A+2B\rightleftharpoons 2C$, if $2$ moles of A, $3.0$ moles of B and $2.0$ moles of C are placed in a $2L$ flask and the equilibrium concentration of C is $0.5mol/L$. The equilibrium constant $\left(K_c\right)$ for the reaction is:

• A. $0.073$
• B. $0.1470$
• C. $0.0500$
• D. $0.0080$

Answer 1

The correct option is D $0.0080$

$A$ $+$ $2B$ $\leftrightharpoons$ $C$

2 3 2 initial moles

(2-x) (3-2x) (2-2x) at eqbm.

But for C eqbm. moles = 0.5

$2-2x=0.5$

$x=0.75$

Equilibrium moles: A = (2 + 0.75) = 2.75 mol

B = (3 + 2 $\times$ 0.75) = 4.5 mol

C = 0.5 mol

Equilibrium volume: [A] = 2.75/2 M

[B] = 4.5/2 M

[C] = 0.5/2 M

$K=\frac{(0.5/2{)}^2}{\left(4.5/2{)}^2\right(2.75/2)}=0.0080$