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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 8 of 73
Marks: +1, -0
Reaction between N2\mathrm{N}_2 and O2\mathrm{O}_2 takes place as follows :
2N2(g)+O2(g)2\mathrm{N}_{2(g)} + \mathrm{O}_{2(g)} ⇌ 2N2O(g)2\mathrm{N}_2\mathrm{O}_{(g)}
If a mixture of 0.482 mol of N2\mathrm{N}_2 and 0.933 mol of O2\mathrm{O}_2 is placed in a 10 L reaction vessel and allowed to form N2O\mathrm{N}_2\mathrm{O} at a temperature for which KcK_c = 2.0 × 10−3710^{-37}, determine the composition of equilibrium mixture.
Solution:  
2N2(g)+O2(g)2\mathrm{N}_{2(g)} + \mathrm{O}_{2(g)} ⇌ 2N2O(g)2\mathrm{N}_2\mathrm{O}_{(g)}
Initial no. of moles :0.4820.9330At eqm. no. of moles :(0.482−x)(0.933−x/2)xMolar conc.0.482−x100.933−(x/2)10x10\begin{array}{lccc} \text{Initial no. of moles :} & 0.482 & 0.933 & 0 \\ \text{At eqm. no. of moles :} & (0.482-x) & (0.933-x/2) & x \\ \text{Molar conc.} & \frac{0.482-x}{10} & \frac{0.933-(x/2)}{10} & \frac{x}{10} \end{array}
As KcK_c = 2.0 × 10−3710^{-37} is very small, this means that the amount of N2\mathrm{N}_2 and O2\mathrm{O}_2reacted (x) is very very small. Hence, at equilibrium, we have[N2][\mathrm{N}_2] = 0.0482 mol L−1,[O2]\text{mol L}^{-1}, [\mathrm{O}_2] = 0.0933 mol L−1,[N2O]\text{mol L}^{-1}, [\mathrm{N}_2\mathrm{O}] = 0.1x
∴ KcK_c = (0.1x)2(0.0482)2(0.0933)\frac{(0.1x)^2}{(0.0482)^2(0.0933)} = 2.0 × 10−3710^{-37} (given)
On solving this gives, x = 6.6 × 10−2010^{-20}
∴ [N2O][\mathrm{N}_2\mathrm{O}]= 0.1x = 6.6 × 10−21 mol L−110^{-21}\,\text{mol L}^{-1}
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