question_answer 37)

Nitric oxide reacts with bromine and gives nitrosyl bromide as per the reaction given below: \[2NO(g)+B{{r}_{2}}(g)\rightleftharpoons 2NOBr(g)\] when 0.087 mol of NO and 0.0437 mol of\[B{{r}_{2}}\] are mixed in a closed container at constant temperature, 0.0518 mol of\[NOBr\]is obtained at equilibrium. Calculate the equilibrium amount of nitric oxide and bromine.

Answer:

\[\underset{{{t}_{\begin{smallmatrix} 0 \\ {{t}_{eq}} \end{smallmatrix}}}}{\overset{{}}{\mathop{{}}}}\,\,\,\,\,\underset{\begin{smallmatrix} 0.087 \\ (0.087-2x) \end{smallmatrix}}{\mathop{2NO(g)}}\,\,+\underset{\begin{smallmatrix} 0.0437 \\ (0.0437-x) \end{smallmatrix}}{\mathop{B{{r}_{2}}(g)}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ 2x \end{smallmatrix}}{\mathop{2NOBr(g)}}\,\] Given, \[2x=0.0518\] x = 0.0259 mol Equilibrium amount of \[B{{r}_{2}}\,=(0.0437-0.0259)mol\] = 0.0178 mol Equilibrium amount of \[NO\text{ }=\text{ }0.087\text{ }-2x\] = 0.087 - 0.0518 = 0.0352 mol