question_answer 36)

Reaction between \[{{N}_{2}}\] and \[{{O}_{2}}\] takes place as follows: \[2{{N}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2{{N}_{2}}O(g)\] If a mixture of 0.482 mol \[{{N}_{2}}\] and 0.933 mol of \[{{O}_{2}}\] is placed in a 10L reaction vessel and allowed to form \[{{N}_{2}}O\]at a temperature for which\[{{K}_{c}}=2\times {{10}^{-37}}\], determine the composition of equilibrium mixture.

Answer:

\[\underset{{{t}_{\begin{smallmatrix} 0 \\ _{\begin{smallmatrix} {{t}_{eq}} \\ Equilibrium \\ concentration \end{smallmatrix}} \end{smallmatrix}}}}{\overset{{}}{\mathop{{}}}}\,\underset{\underset{\begin{smallmatrix} (0.482-x) \\ \frac{(0.482-x)}{10} \end{smallmatrix}}{\mathop{0.482}}\,}{\mathop{2{{N}_{2}}(g)}}\,\,\,\,+\,\,\,\,\,\,\,\underset{\underset{\begin{smallmatrix} (0.933-x/2) \\ \frac{(0.933-x/2)}{10} \end{smallmatrix}}{\mathop{0.933}}\,}{\mathop{{{O}_{2}}(g)}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ x \\ \frac{x}{10} \end{smallmatrix}}{\mathop{2{{N}_{2}}O(g)}}\,\]   The value of is very small, i.e., extent of reaction will also be very small. \[\therefore \] \[[{{N}_{2}}]=\left[ \frac{0.482-x}{10} \right]=0.0482M\] \[[{{O}_{2}}]=\left[ \frac{0.933-x/2}{10} \right]=0.0933M\] \[[{{N}_{2}}O]=0.1x\] \[{{K}_{c}}=\frac{{{[{{N}_{2}}O]}^{2}}}{{{[{{N}_{2}}]}^{2}}[{{O}_{2}}]}=\frac{{{(0.1x)}^{2}}}{{{(0.0482)}^{2}}(0.0933)}\] \[2\times {{10}^{-37}}=\frac{0.01{{x}^{2}}}{{{(0.0482)}^{2}}\times (0.0933)}\] On solving \[x=6.6\times {{10}^{-20}}\] \[\therefore \] \[[{{N}_{2}}O]=0.1x\] \[=0.1\times 6.6\times {{10}^{-20}}\] \[=6.6\times {{10}^{-21}}M.\]