Question

Two reactions $r_1$ and $r_2$ proceed at ${25}^{o}C$ at the same rate. The temperature coefficient of the second reaction is $2.5$ times of temperature coefficient of the first reaction. The ratio of rates of these reactions at ${96}^{o}C$ will be $\left(r_2/r_1\right)$, approx.

• A. $10:1$
• B. $5:1$
• C. $15:1$
• D. $25:1$

Answer 1

The correct option is A $10:1$

We know $k=A{e}^{\frac{E_a}{RT}}$ where $\frac{E_a}{R}$ is temperature coefficient and k is rate constant:

${\vartheta }_2=k_2{C}^n;{\vartheta }_1=k_1{C}_0^n$

$\therefore \frac{{\vartheta }_2}{{\vartheta }_1}{\mid }_{{25}^{\circ }C}=\frac{k_2}{k_1}=e^{-(\frac{E_a^1}{R}-\frac{E_a^2}{R})\frac{1}{298}}=1\to$eq 1

also

$\frac{{\vartheta }_2}{{\vartheta }_1}{\mid }_{8^{\circ }C}=\frac{k_2}{k_1}=e^{-(\frac{E_a^1}{R}-\frac{E_a^2}{R})\frac{1}{359}}\to$eq 2

substuting $(\frac{E_a^1}{R}-\frac{E_a^2}{R})$ from eq 1 into eq 2 we get

$\frac{{\vartheta }_2}{{\vartheta }_1}{\mid }_{{96}^{\circ }C}=10=10:1$