Chemistry - Chemical Kinetics Question with Solution | TestHub

To find the temperature coefficient ($\mu$), we use the formula:

$$\frac{r_2}{r_1}=(\mu {)}^{\Delta T/10}$$

Given $r_1=2.6\times 10^{-3}mol{L}^{-1}{s}^{-1}$ at $T_1=50^{\circ }C$ and $r_2=7.02\times 10^{-2}mol{L}^{-1}{s}^{-1}$ at $T_2=80^{\circ }C$.

$\Delta T=T_2-T_1=80^{\circ }C-50^{\circ }C=30^{\circ }C$.

Substitute the values:

$$\frac{7.02\times 10^{-2}}{2.6\times 10^{-3}}=(\mu {)}^{30/10}$$

$$27={\mu }^3$$

Taking the cube root of both sides:

$$\mu =\sqrt[3]{27}$$

$$\mu =3$$