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Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:

2NO(g) + Br2(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.

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2NO(g) + Br2(g) ⇌ 2NOBr(g)

According to this equation, 2 moles of NO(g) react with 1 mole of Br2(g) to form 2 moles of NOBr(g). The composition of the equilibrium mixture can be calculated as follows:

No. of moles of NOBr(g) formed at equilibrium = 0.0518 mol

No. of moles of NO(g) taking part in reaction = 0.0518 mol

No. of moles of NO(g) left at equilibrium = 0.087 - 0.0518 = 0.0352 mol

No. of moles of Br2(g) taking part in reaction = \(\frac{1}{2}\) x 0.0518 = 0.0259 mol

No. of moles of Br2(g) left at equilibrium = 0.0437 - 0.0259 = 0.0178 mol

Hence 0.0352 mol of NO and 0.0178 mol of Br2

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