Two $1$ order reaction having same reactant concentration proceed at $25^{\circ} C$ at the same rate. The temperature coefficient of the rate of the first reaction is $2$ and that of second reaction is $3$ . Find the ratio of the rates of there reactions at $75^{\circ} C$ .

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Solution

For first order reaction, $r_{1} = k {[A]}^{1}$
$∴ r_{1} / r_{2} = k_{1} / k_{2} =$ temperature coefficient
Let the rate of reaction for Ist at $25^{o} C$ be $R_{1}$ and the rate of reaction for IInd at $25^{o} C$ be $R_{2}$ .
Also, $R_{1} = R_{2}$
Rates of reaction
$∴$ At $25^{o} C$ $R_{1}$ $R_{2}$
$35^{o} C$ $2 R_{1}$ $3 R_{2}$
$45^{o} C$ ${(2)}^{2} R_{1}$ ${(3)}^{2} R_{2}$
$55^{o} C$ ${(2)}^{3} R_{1}$ ${(3)}^{3} R_{2}$
$65^{o} C$ ${(2)}^{4} R_{1}$ ${(3)}^{4} R_{2}$
$75^{o} C$ ${(2)}^{5} R_{1}$ ${(3)}^{5} R_{2}$
$∴$ Temperature coefficient for 1st reaction $= \frac{k_{35}}{k_{25}} = \frac{R_{35}}{R_{25}} = 2$
i.e., for each $10^{o} C$ rise in temperature, rate becomes $2$ times.
Similarly, for $2$ nd reaction it becomes $3$ times,
$∴$ At $75^{o} C$
$\frac{r a t e o f r e a c t i o n f o r 2 n d}{r a t e o f r e a c t i o n f o r 1 s t} = \frac{{(3)}^{5} R_{2}}{{(2)}^{5} R_{1}} = 7.5937 (∵ R_{1} = R_{2})$

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