Where $\alpha =$ Degree of dissociation.
Mol. wt. of mixture
$ = \frac{{(1 - \alpha ) \times {M_{{N_2}{O_4}}} + 2\alpha \times {M_{N{O_2}}}}}{{(1 + \alpha )}}$
$ = \frac{{(1 - 0.2)92 + 2 \times 0.2 \times 46}}{{(1 + 0.2)}} = 76.66$
Now, As per ideal gas equation
$PV = nRT$
$P{M_{mixture}} = dRT$
$\therefore d = \frac{{P{M_{mix.}}}}{{RT}} = \frac{{1 \times 76.66}}{{0.0821 \times 300}} = 3.11\,g/L$
$2S{O_2} + {O_2}$ $\rightleftharpoons$ $2S{O_3} + Q$ $Cal$
$2NH_3 + 5/2O_2 $ $\rightleftharpoons$ $ 2NO + 3H_2O, K_1, K_2$ અને $K_3$ ના સંદર્ભમાં ...... થાય.
$NO_{(g)} + \frac{1}{2}{O_2} \rightleftharpoons N{O_2}_{(g)}$
$2N{O_2}_{(g)} \rightleftharpoons 2NO_{(g)} + {O_2}_{(g)}$