Question

Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below:
$2N{O}_{\left(g\right)}+B{r}_{2\left(g\right)}\rightleftharpoons 2NOB{r}_{\left(g\right)}$

When $0.087$ mole of $NO$ and $0.0437$ mole of $B{r}_2$ are mixed is a closed container at constant temperature, $0.0518$ mole of $NOBr$ is obtained at equilibrium. Calculate the equilibrium amount of nitric oxide and bromine.

Answer 1

$2N{O}_{\left(g\right)}+B{r}_{2\left(g\right)}\rightleftharpoons 2NOB{r}_{\left(g\right)}$

Initial moles of $NO=0.087$

Initial moles of $B{r}_2=0.0437$

Final moles of $NO=2a$

Final moles of $B{r}_2=a$

$\therefore$ number of moles of $NO$ at equilibrium $=(0.087-2a)$

Number of moles of $B{r}_2$ at equilibrium $= (0.0437 - a)

Given:

$2a=0.0518$

$\therefore a=\frac{0.0518}{2}=0.0259$

At equilibrium,

Number of moles (amount) of $NO=0.0870-0.0518=0.0352$ mole
Number of moles (amount) of $B{r}_2=0.0437-0.0259=0.0178$ mole