Question

Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below:
\(2NO (g) + Br_2 (g) ⇋ 2NOBr (g)\)
When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br₂ .

Answer 1

Given:

Reaction:
2NO(g) + Br₂(g) ⇌ 2NOBr(g)

Initial moles of NO = 0.087 mol
Initial moles of Br₂ = 0.0437 mol
Moles of NOBr formed at equilibrium = 0.0518 mol

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Step 1: Assume extent of reaction

From the balanced equation:
2 mol NO → 2 mol NOBr
1 mol Br₂ → 2 mol NOBr

Let the amount of Br₂ reacted be x mol.

Then:
NOBr formed = 2x

Given:
2x = 0.0518

x = 0.0259 mol

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Step 2: Calculate moles consumed

Br₂ consumed = 0.0259 mol
NO consumed = 2x = 0.0518 mol

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Step 3: Calculate equilibrium amounts

Equilibrium moles of NO:

= Initial NO − NO consumed
= 0.087 − 0.0518
= 0.0352 mol

Equilibrium moles of Br₂:

= Initial Br₂ − Br₂ consumed
= 0.0437 − 0.0259
= 0.0178 mol

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Final Answer:

Equilibrium amount of NO = 0.0352 mol
Equilibrium amount of Br₂ = 0.0178 mol