In a zero order reaction for every $${10^ \circ }C$$  rise of temperature, the rate is doubled. If the temperature is increased from $${10^ \circ }C$$  to $${100^ \circ }C,$$  the rate of the reaction will become

Solution :
For $${10^ \circ }$$ rise in temperature, $$n = 1$$
$${\text{so rate}} = {2^n} = {2^1} = 2$$
When temperature is increased from $${10^ \circ }C$$  to $${100^ \circ }C,$$  change in temperature
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 100 - 10 = {90^ \circ }C \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,n = 9 \cr & {\text{so,}}\,{\text{rate}} = {2^9} = 512\,{\text{times}} \cr} $$
Alternate method with every $${10^ \circ }$$ rise in temperature, rate becomes double,
So $$\frac{{r'}}{r} = {2^{\left( {\frac{{100 - 10}}{{10}}} \right)}} = {2^9} = 512\,{\text{times}}{\text{.}}$$