Question

For the following equilibrium, $ {{N}_{2}}{{O}_{4}}2N{{O}_{2}};{{K}_{c}}=0.67. $ If we start with 3 moles of $ N{{O}_{2}} $ and 1 mole of $ {{N}_{2}}{{O}_{4}} $ in 1L flask, then $ N{{O}_{2}} $ present at equilibrium is

• A. 1.5 mol
• B. 2.0 mol
• C. 0.5 mol
• D. 1.0 mol

Answer 1

Answer: (B)

$ {{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. $