Question

$2NOBr\left(g\right)\rightleftharpoons 2NO\left(g\right)+B{r}_2\left(g\right)$

If nitrosyl bromide ($NOBr)$ is $40\%$ dissociated at certain temperature and a total pressure of $0.30$ atm, then $K_p$ for the reaction,

$2NO\left(g\right)+B{r}_2\left(g\right)\rightleftharpoons 2NOBr\left(g\right)$ is:

• A. $45$
• B. $25$
• C. $0.022$
• D. $0.25$

Answer 1

The correct option is A $45$

$2NOBr\left(g\right)\rightleftharpoons 2NO\left(g\right)+B{r}_2\left(g\right)$
At eqm. $1-0.4=0.6$ $0.4$ $0.2$

So, $P_{NOBr}=0.6\times \frac{0.3}{1.2}=\frac{0.6}{4}=0.15atm$

$P_{NO}=0.4\times \frac{0.3}{1.2}=\frac{0.4}{4}=0.1atm$

$P_{B{r}_2}=0.2\times \frac{0.3}{1.2}=\frac{0.2}{4}=0.05atm$

So, for $2NO\left(g\right)+B{r}_2\left(g\right)\rightleftharpoons 2NOBr\left(g\right)$

$K_p=\frac{(0.15{)}^2}{{0.1}^2\times 0.05}=45$