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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 9 of 73
Marks: +1, -0
Nitric oxide reacts with Br2\mathrm{Br}_2 and gives nitrosyl bromide as reaction given below :
2NO(g)+Br2(g)2\mathrm{NO}_{(g)} + \mathrm{Br}_{2(g)}2NOBr(g)2\mathrm{NOBr}_{(g)}
When 0.087 mol of NO and 0.0437 mol of Br2\mathrm{Br}_2 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\mathrm{Br}_2.
Solution:  
2NO(g)+Br2(g)2\mathrm{NO}_{(g)} + \mathrm{Br}_{2(g)}2NOBr(g)2\mathrm{NOBr}_{(g)}
Initial moles0.0870.04370Moles at eqm.(0.0872x)(0.0437x)2x\begin{array}{c|c|c|c} \text{Initial moles} & 0.087 & 0.0437 & 0 \\ \text{Moles at eqm.} & (0.087-2x) & (0.0437-x) & 2x \end{array}
But 2x = 0.0518 ∴ x = 0.0259
(nNO)eq(n_{\mathrm{NO}})_{\mathrm{eq}} = 0.087 – 0.0518 = 0.0352 mol
(nBr2)eq(n_{\mathrm{Br}_2})_{\mathrm{eq}} = 0.0437 – 0.0259 = 0.0178 mol
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