Question

$NOBr\left(g\right)\rightleftharpoons 2NO\left(g\right)+B{r}_2\left(g\right)$. If nitrosyl bromide $\left(NOBr\right)$ is $34\%$ dissociated at ${25}^{\circ }C$ & total pressure of $0.25$atm. Calculate $K_p$ for the dissociation at this temperature.

Answer 1

Consider the following reaction.

$2NOBr\rightleftharpoons 2NO+B{r}_2$

If the initial pressure of $NOBr$ is $p$, then at equilibrium, since the percent dissociation of $NOBr$ is $34$,

$P\left(NOBr\right)=0.66p$

$P\left(NO\right)=0.34p$

$P\left(B{r}_2\right)=\frac{0.34p}{2}=0.17p$

Total pressure $=0.66p+0.34p+0.17p=0.25atm$

$p=0.214atm$

The equilibrium partial pressures are then,

$P\left(NOBr\right)=0.66p=0.141atm$

$P\left(NO\right)=0.34p=0.073atm$

$P\left(B{r}_2\right)=0.17p=0.036atm$

Now, from the original balance equation,

$K_p=\frac{{\left(P_{NO}\right)}^2\left(P_{B{r}_2}\right)}{{\left(P_{NOBr}\right)}^2}$

$K_p=\frac{{\left(0.073\right)}^2\times 0.036}{{\left(0.141\right)}^2}$

$K_p=0.000147atm$

Hence, this is the required result.