Chemistry - Chemical Kinetics Question with Solution | TestHub

For the first reaction, $A\to B$:

Given that 50% of the reaction completes in 2 hours at 500 K. For a first-order reaction, the half-life $t_{1/2}=\frac{\ln 2}{k}$.

So, $k_{1,500K}=\frac{\ln 2}{t_{1/2}}=\frac{\ln 2}{2}hou{r}^{-1}$.

Given that $k_{1,500K}=2\times k_{1,300K}$.

Using the Arrhenius equation: $\ln \left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$.

$\ln \left(\frac{k_{1,500K}}{k_{1,300K}}\right)=\ln (2)=\frac{E_{a1}}{R}\left(\frac{1}{300}-\frac{1}{500}\right)=\frac{E_{a1}}{R}\left(\frac{500-300}{300\times 500}\right)=\frac{E_{a1}}{R}\left(\frac{200}{150000}\right)=\frac{E_{a1}}{R}\left(\frac{1}{750}\right)$.

So, $E_{a1}=750R\ln 2$.

For the second reaction, $C\to D$:

Given $E_{a2}=\frac{E_{a1}}{2}=\frac{750R\ln 2}{2}=375R\ln 2$.

Given $k_{2,500K}=2\times k_{1,500K}=2\times \frac{\ln 2}{2}=\ln 2hou{r}^{-1}$.

Now, we need to find $k_{2,300K}$ for the second reaction.

Using the Arrhenius equation for the second reaction:

$\ln \left(\frac{k_{2,500K}}{k_{2,300K}}\right)=\frac{E_{a2}}{R}\left(\frac{1}{300}-\frac{1}{500}\right)$.

$\ln \left(\frac{\ln 2}{k_{2,300K}}\right)=\frac{375R\ln 2}{R}\left(\frac{1}{750}\right)$.

$\ln \left(\frac{\ln 2}{k_{2,300K}}\right)=\frac{375\ln 2}{750}=\frac{\ln 2}{2}$.

$\frac{\ln 2}{k_{2,300K}}=e^{\frac{\ln 2}{2}}=(e^{\ln 2}{)}^{1/2}=2^{1/2}=\sqrt{2}$.

$k_{2,300K}=\frac{\ln 2}{\sqrt{2}}$.

$k_{2,300K}=\frac{0.693}{1.414}\approx 0.490hou{r}^{-1}$.

In the required format, $k_{2,300K}=4.9\times 10^{-1}hou{r}^{-1}$.

The nearest integer is 5.