Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below:
$2 N O_{(g)} + B r_{2 (g)} ⇌ 2 N O B r_{(g)}$
When $0.087$ mole of $N O$ and $0.0437$ mole of $B r_{2}$ are mixed is a closed container at constant temperature, $0.0518$ mole of $N O B r$ is obtained at equilibrium. Calculate the equilibrium amount of nitric oxide and bromine.

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Solution

$2 N O_{(g)} + B r_{2 (g)} ⇌ 2 N O B r_{(g)}$
Initial moles of $N O = 0.087$
Initial moles of $B r_{2} = 0.0437$
Final moles of $N O = 2 a$
Final moles of $B r_{2} = a$

$∴$ number of moles of $N O$ at equilibrium $= (0.087 - 2 a)$
Number of moles of $B r_{2}$ at equilibrium $= (0.0437 - a)

Given:
$2 a = 0.0518$
$∴ a = \frac{0.0518}{2} = 0.0259$
At equilibrium,
Number of moles (amount) of $N O = 0.0870 - 0.0518 = 0.0352$ mole
Number of moles (amount) of $B r_{2} = 0.0437 - 0.0259 = 0.0178$ mole

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