Answer the following: The rate constant of a first order reaction increases five times when the…

Formula: $\log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2\cdot303 R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)$ Substitution: $\log(5) = \frac{E_a}{2\cdot303 \times 8\cdot314\text{ J K}^{-1}\text{mol}^{-1}}\left(\frac{500 - 350}{350 \times 500}\right)$ $0\cdot6990 = \frac{E_a}{19\cdot147} \times \left(\frac{150}{175000}\right)$ $E_a = \frac{0\cdot6990 \times 19\cdot147 \times 175000}{150} = 15614\text{ J mol}^{-1} = 15\cdot61\text{ kJ mol}^{-1}$ Final answer: 15.61 kJ mol^-1

Zero order. Reason: The rate of reaction has the unit $\text{mol L}^{-1}\text{s}^{-1}$. The general unit of rate constant for an $n^{\text{th}}$ order reaction is $(\text{mol L}^{-1})^{1-n}\text{s}^{-1}$. Equating the units gives $1 - n = 1$, which yields $n = 0$ (zero order).