Question

If a mixture 0.4 mole $H_2$ and 0.2 mole $B{r}_2$ is heated at $700\text{K}$ at equilibrium, the value of equilibrium constant is $0.25\times {10}^{10}$ then find out the ratio of concentrations of $\left(B{r}_2\right)$ and $\left(HBr\right)$ (Report your answer as $\frac{B{r}_2}{HBr}\times {10}^{11}$)

• A. 70
• B. 80
• C. 90
• D. 100

Answer 1

The correct option is B 80

$\begin{array}{cccccc}& H_2& +& B{r}_2& \rightleftharpoons & 2HBr\\ t=0& 0.4& & 0.2& & -\\ t=t_{\text{eq}}& 0.2& & y& & 0.4\\ & & & =\text{negligible}& & =y\end{array}$
$\because \frac{1}{4}\times {10}^{10}=\frac{0.4\times 0.4}{0.2\times y}\Rightarrow y=3.2\times {10}^{-10}\Rightarrow \frac{B{r}_2}{HBr}\times {10}^{11}=\frac{3.2}{0.2}\times {10}^{-10}\times {10}^{11}=80$