$a=100,$ $a-x=0.1$
${{t}_{1}}=\frac{2.303}{K}\log \frac{100}{0.1}$
$=\frac{2.303}{K}\log {{10}^{3}}$,
${{t}_{2}}=\frac{2.303}{K}\log \frac{100}{50}$
$=\frac{2.303}{K}\log \,2$
$\frac{{{t_1}}}{{{t_2}}}\, = \,\frac{3}{{0.3010}}\, = \,10$
| No | $[NH_4^+]$ | $[NO_2^-]$ | rate of reaction |
| $1.$ | $0.24\, M$ | $0.10\, M$ | $7.2 \times {10^{ - 6}}$ |
| $2.$ | $0.12\, M$ | $0.10\, M$ | $3.6 \times {10^{ - 6}}$ |
| $3.$ | $0.12\, M$ | $0.15\, M$ | $5.4 \times {10^{ - 6}}$ |
(નજીકનાં પૂર્ણાકમાં રાઉન્ડ ઑફ) $\left[ R =8.314\, J \,K ^{-1} \,mol ^{-1}\right]$
