$4{\text{P}}\quad {\text{P}}\quad 0\quad 0\,\,\,lnitially$
$2\,{\text{P}}\quad 0.5\,{\text{P}}\quad 1.5\,{\text{P}}\,\,2\,{\text{P}}\,\,At\,equilibrium$
${\mathrm{K}_{\mathrm{p}}=\frac{(1.5 \mathrm{P})^{3}(2 \mathrm{P})^{2}}{(2 \mathrm{P})^{2}(0.5 \mathrm{P})}=6.75 \mathrm{P}^{2}}$
${\mathrm{P}=\sqrt{\frac{\mathrm{K}_{\mathrm{P}}}{6.75}}=0.4 \sqrt{\mathrm{K}_{\mathrm{P}}}}$
$(B)$ is correct answer.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| List $-I$ (Atomic Number) | List $-II$ (Block of periodic table) |
| $A$ $37$ | $I$ $p-$block |
| $B$ $78$ | $II$ $d-$block |
| $C$ $52$ | $III$ $f-$block |
| $D$ $65$ | $IV$ $s-$block |
નીચે આપેલા વિકલ્પોમાંથી યોગ્ય ઉત્તરની પસંદગી કરો.
$(C= 3 \times 10^8 \,ms^{-1}$ and $N_A =6.02 \times mol^{-1}).$