Question

When $A_2$ and $B_2$ are allowed to react, the equilibrium constant of the reaction at 27${}^{\circ }$C is found to be 4 ($K_c$= 4).
$A_{2\left(g\right)}+B_{2\left(g\right)}\leftrightharpoons 2A{B}_{\left(g\right)}$
What will be the equilibrium concentration of AB?

• A. 1.33 M
• B. 2.66 M
• C. 1.6 M
• D. 0.33 M

Answer 1

The correct option is C 1.6 M

The given reaction is :-

$A_{2\left(g\right)}+B_{2\left(g\right)}\rightleftharpoons 2AB\left(g\right);K_C=4$

Initial ${reac}^n$ : $2$ $4/3$ $0$ $(\because concentration=\frac{No.ofmoles}{volume})$

At ${eq}^m$ : $(2-x)$ $(\frac{4}{3}-x)$ $2x$ (let)

$\because K_C=4$

$\Rightarrow \frac{{\left[AB\right]}^2}{\left[A_2\right]\left[B_2\right]}=4\Rightarrow \frac{{\left(2x\right)}^2}{(2-x)\left(\frac{4-3x}{3}\right)}=4$

$\Rightarrow {3x}^2=(2-x)(4-3x)$

$\Rightarrow {3x}^2=8-6x-4x+3x^2$

$\Rightarrow 10x=8\Rightarrow x=\frac{4}{5}=0.8$

$\therefore$ concentration of $AB=2\times 0.8=1.6$.