$\frac{{0.693}}{T} = \frac{{2.303}}{t}\log \frac{{100}}{{100 - 30}}$
$\therefore \,\,T = 58.2\,min.$
પ્રક્રિયા $P \to Q$ માટે ${K_2} = {10^{10}}\,{e^{ - 8000/8.34\,\,T}}$ હોય તો ....... $K$ તાપમાને $K_1 = K_2$ થશે.
${{H}_{2}}+C{{l}_{2}}\xrightarrow{\text{Sunlight}}2HCl$
| No | $[NH_4^+]$ | $[NO_2^-]$ | rate of reaction |
| $1.$ | $0.24\, M$ | $0.10\, M$ | $7.2 \times {10^{ - 6}}$ |
| $2.$ | $0.12\, M$ | $0.10\, M$ | $3.6 \times {10^{ - 6}}$ |
| $3.$ | $0.12\, M$ | $0.15\, M$ | $5.4 \times {10^{ - 6}}$ |