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In a zero-order reaction for every $10^\circ$ 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:
A
128 times
B
256 times
C
512 times
D
64 times
Detailed Solution
$\frac{r_2}{r_1} = 2$ for every $10^\circ$ rise
Rise in temperature = 100 − 10 = $90^\circ$, i.e. nine intervals of $10^\circ$
$\frac{r_n}{r_1} = 2^{\Delta T/10} = 2^9 = 512$
So the rate becomes 512 times.
Rise in temperature = 100 − 10 = $90^\circ$, i.e. nine intervals of $10^\circ$
$\frac{r_n}{r_1} = 2^{\Delta T/10} = 2^9 = 512$
So the rate becomes 512 times.