\(\frac{{d(N{H_3})}}{{dt}}\,\, = \,\,2.5\,\, \times \,{10^{ - 4}}\)
\(r=\frac{1}{2}\times 2.5\times {{10}^{-4}}\) \(=\frac{2.5}{2}\times {{10}^{-4}}\)
\(\frac{d({{N}_{2}})}{dt}=\) \(\frac{2.5}{2}\times {{10}^{-4}}=1.25\times {{10}^{-4}}\)
$\frac{d[NH_3]}{dt} = 2 \times 10^{-4} \, mol \,L^{-1} \, s^{-1}$ હોય, તો $\frac{-d[H_2]}{dt}$ ની કિંમત ............. $mol \,L^{-1} \, s^{-1}$ થશે.
${O_3} \rightleftharpoons {O_2} + \left[ O \right]$
${O_3} + \left[ O \right] \to 2{O_2}$ (slow)
તો $2{O_3} \to 3{O_2}$ પ્રક્રિયાનો કમ જણાવો.