so rate $=2^{n}=2^{1}=2$
When temperature is increased from $10\,^{o} \mathrm{C}$ to $100\,^{o} \mathrm{C},$ change in temperature
${=100-10=90\,^{o} \mathrm{C}}$
${n=9}$
So, rate $=2^{9}=512$ times
Alternate method with every $10^{o}$ rise in temperature, rate becomes double,
so $\frac{r^{\prime}}{r}=2^{\left(\frac{100-10}{10}\right)}=2^{9}=512$ times.
$T$ (in, $K$) $- 769$ , $1/T$ (in, $K^{-1}$ ) $- 1.3\times 10^{-3},$
$\log_{10}K - 2.9\,T$ (in, $K$) $- 667$, $1/T$ (in, $K^{-1}) - 1.5\times 10^{-3}$, $\log_{10}\,K - 1.1$