d
\(\mathop {\,\,\,}\limits_{\mathop {Initial\,moles}\limits_{Moles\,of\,equil.} } \mathop {\mathop {{N_2}{O_4}(g)}\limits_1 }\limits_{1 - \alpha }  \leftrightarrow \mathop {\mathop {2N{O_2}(g)}\limits_0 }\limits_{2\alpha } \)

(\(\alpha =\) degree of dissociation)

Total number of moles at equil.

\( = (1 - \alpha ) + 2\alpha \)

\( = (1 + \alpha )\)

\({P_{{N_2}{O_4}}} = \frac{{(1 - \alpha )}}{{(1 + \alpha )}} \times P\)

\({P_{N{O_2}}} = \frac{{2\alpha }}{{(1 + \alpha )}} \times P\)

\({K_P} = \frac{{{{({P_{N{O_2}}})}^2}}}{{{P_{{N_2}{O_4}}}}} = \frac{{{{\left( {\frac{{2\alpha }}{{(1 + \alpha )}} \times P} \right)}^2}}}{{\left( {\frac{{1 - \alpha }}{{(1 + \alpha )}}} \right) \times P}} = \frac{{4{\alpha ^2}P}}{{1 - {\alpha ^2}}}\)

Given, \({K_P} = 2,\,P = 0.5\,atm\)

\(\therefore \,{K_P} = \frac{{4{\alpha ^2}P}}{{1 - {\alpha ^2}}}\)

\( = \frac{{4{\alpha ^2} \times 0.5}}{{1 - {\alpha ^2}}}\)

\(\alpha  = 0.707 \approx 0.71\)

\(\therefore \) Percentage dissociation

\( = 0.71 \times 100 = 71\)