A) 1.5 mol
B) 2.0 mol
C) 0.5 mol
D) 1.0 mol
Correct Answer: B
Solution :
\[{{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g)\] \[Q=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}{{O}_{4}}]}=\frac{{{[3]}^{2}}}{[1]}=9>{{K}_{c}}\] Thus, equilibrium is displaced in backward side. \[\underset{\begin{smallmatrix} 3 \\ (3-2x) \end{smallmatrix}}{\mathop{2N{{O}_{2}}}}\,\rightleftharpoons \underset{\begin{smallmatrix} \,\,\,1 \\ (1+x) \end{smallmatrix}}{\mathop{{{N}_{2}}{{O}_{4}}}}\,\] \[K{{}_{c}}=\frac{1}{{{K}_{c}}}=\frac{1}{0.67}=\frac{1+x}{{{(3-2x)}^{2}}}=1.5\] Only permissible value of x that satisfies above equation is \[x=0.5\]Thus, \[N{{O}_{2}}\]at equilibrium \[=3-2x=2\,mol.\]
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