Question

$2NOBr\left(g\right)\rightleftharpoons 2NO\left(g\right)+B{r}_2\left(g\right)$. If nitrosyl bromide $\left(NOBr\right)$ is 33.33% dissociated at ${25}^o$ C and a total pressure of 0.28 atm. Calculate $K_p$ for the dissociation at this temperature.

Answer 1

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

At $t=0:-$ $P$ $0$ $0$

At equilibrium:- $P(1-\alpha )$ $2P\alpha$ $P\alpha$

$P_T=P-P\alpha +2P\alpha +P\alpha =0.28atm$ and $\alpha =0.3333$

$P(1+2\alpha )=0.28atm$

$P(1+0.6666)=0.28atm$

$\therefore P=0.168atm.$

$\therefore$ $K_P=\frac{{\left(2P\alpha \right)}^{2}P\alpha }{P-P\alpha }=\frac{{4P}^3{\alpha }^3}{P(1-\alpha )}=\frac{{4P}^2{\alpha }^3}{(1-\alpha )}=\frac{4\times (0.168{)}^2\times (0.3333{)}^3}{(1-0.3333)}=6.27\times {10}^{-3}$

$\therefore K_P=6.27\times {10}^{-3}$