the rate constant of a certain reaction is known to obey the arrhenius equation, and to have an activation…

Explanation:

Step1: Convert temperatures to Kelvin

$T_1 = 6.0 + 273.15=279.15\ K$ $T_2=- 64.0+273.15 = 209.15\ K$

Step2: Write the Arrhenius - equation in two - point form

The two - point form of the Arrhenius equation is $\ln\frac{k_2}{k_1}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$, where $k_1$ and $k_2$ are the rate constants at temperatures $T_1$ and $T_2$ respectively, $E_a$ is the activation energy, and $R = 8.314\ J/(mol\cdot K)$. First, convert $E_a$ to $J/mol$: $E_a=67.0\times10^{3}\ J/mol$.

Step3: Calculate $\frac{1}{T_1}-\frac{1}{T_2}$

$\frac{1}{T_1}-\frac{1}{T_2}=\frac{1}{279.15}-\frac{1}{209.15}=\frac{209.15 - 279.15}{279.15\times209.15}=\frac{-70}{279.15\times209.15}\approx - 1.19\times10^{-3}\ K^{-1}$

Step4: Calculate $\ln\frac{k_2}{k_1}$

$\ln\frac{k_2}{k_1}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})=\frac{67.0\times10^{3}\ J/mol}{8.314\ J/(mol\cdot K)}\times(-1.19\times10^{-3}\ K^{-1})\approx - 9.61$

Step5: Solve for $k_2$

We know that $\ln\frac{k_2}{k_1}=-9.61$, so $\frac{k_2}{k_1}=e^{-9.61}$. Given $k_1 = 2.9\times10^{-4}\ M^{-1}\ s^{-1}$, then $k_2=k_1\times e^{-9.61}=2.9\times10^{-4}\ M^{-1}\ s^{-1}\times e^{-9.61}$. $e^{-9.61}\approx6.9\times10^{-5}$, so $k_2=2.9\times10^{-4}\times6.9\times10^{-5}\ M^{-1}\ s^{-1}\approx2.0\times10^{-8}\ M^{-1}\ s^{-1}$

Answer:

$2.0\times10^{-8}\ M^{-1}\ s^{-1}$