$$A \rightarrow B$$ (first reaction)
$$C \rightarrow D$$ (second reaction)
Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At 500 K, 50% of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. lf the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is______$$\times 10^{-1}hour^{-1}$$ (nearest integer).

We consider two first-order reactions, $$A \rightarrow B$$ and $$C \rightarrow D$$. The rate constant for the first reaction at 500 K is double its rate constant at 300 K, so $$k_{1,500} = 2 k_{1,300}$$. Since 50% of the first reaction completes in 2 hours at 500 K and the half-life for a first-order reaction is given by $$t_{1/2} = \frac{\ln 2}{k}$$, it follows that $$2 = \frac{\ln 2}{k_{1,500}}$$ and hence $$k_{1,500} = \frac{\ln 2}{2}\ \text{hour}^{-1}$$. The activation energy of the second reaction is half that of the first reaction, so $$E_{a2} = \frac{1}{2} E_{a1}$$, and at 500 K the rate constant of the second reaction is twice that of the first reaction, giving $$k_{2,500} = 2k_{1,500}\,. $$

We need to find the rate constant of the second reaction at 300 K, denoted $$k_{2,300}$$, and express it as an integer multiplied by $$10^{-1}\ \text{hour}^{-1}$$. We begin by using the Arrhenius equation, $$k = A e^{-E_a / (R T)}\,, $$ where $$k$$ is the rate constant, $$A$$ is the pre-exponential factor, $$E_a$$ is the activation energy, $$R$$ is the gas constant, and $$T$$ is the absolute temperature.

Since for the first reaction at 300 K and 500 K we have $$k_{1,300} = A_1 e^{-E_{a1}/(R\cdot 300)}$$ and $$k_{1,500} = A_1 e^{-E_{a1}/(R\cdot 500)}$$ and it is given that $$k_{1,500} = 2 k_{1,300}$$, we set $$A_1 e^{-E_{a1}/(500R)} = 2 A_1 e^{-E_{a1}/(300R)}$$. Dividing by $$A_1$$ and taking natural logarithms gives $$-\frac{E_{a1}}{500R} = \ln 2 \;-\;\frac{E_{a1}}{300R}\,. $$ Rearranging yields $$\frac{E_{a1}}{R}\Bigl(-\tfrac{1}{500}+\tfrac{1}{300}\Bigr)=\ln 2,\quad -\tfrac{1}{500}+\tfrac{1}{300}=\tfrac{1}{750},$$ so $$\frac{E_{a1}}{R}=750\ln 2\,. $$

Substituting the half-life condition at 500 K, $$k_{1,500}=\frac{\ln 2}{2}$$, into the Arrhenius form $$k_{1,500}=A_1 e^{-E_{a1}/(500R)}$$ and using $$\frac{E_{a1}}{R}=750\ln 2$$ gives $$\frac{\ln 2}{2}=A_1 e^{-(750\ln 2)/(500)}=A_1 e^{-(3/2)\ln 2}=A_1\cdot 2^{-3/2}=\frac{A_1}{2\sqrt2}\,. $$ Solving for $$A_1$$ yields $$A_1=\ln 2\cdot\sqrt2\,. $$

Since $$k_{2,500}=2k_{1,500}$$ and $$k_{1,500}=\frac{\ln 2}{2}$$, it follows that $$k_{2,500}=2\cdot\frac{\ln 2}{2}=\ln 2\ \text{hour}^{-1}\,. $$

For the second reaction the Arrhenius equation at 300 K and 500 K gives $$k_{2,300}=A_2 e^{-E_{a2}/(300R)},\quad k_{2,500}=A_2 e^{-E_{a2}/(500R)},$$ so $$\frac{k_{2,500}}{k_{2,300}}=e^{E_{a2}/R\bigl(-\frac{1}{500}+\frac{1}{300}\bigr)}=e^{(E_{a2}/R)(1/750)}\,. $$ Since $$E_{a2}=\tfrac12E_{a1}$$ and $$E_{a1}/R=750\ln 2$$, we have $$E_{a2}/R=375\ln 2$$ and therefore $$k_{2,300}=k_{2,500}\,e^{-(375\ln 2)/(750)}=\ln 2\;e^{-(1/2)\ln 2}=\ln 2\cdot2^{-1/2}=\frac{\ln 2}{\sqrt2}\,. $$

Using $$\ln 2\approx0.693147$$ and $$\sqrt2\approx1.414214$$ gives $$k_{2,300}=\frac{0.693147}{1.414214}\approx0.4900\ \text{hour}^{-1}=4.900\times10^{-1}\ \text{hour}^{-1},$$ and the nearest integer to 4.900 is 5.

Therefore, the rate constant for the second reaction at 300 K is $$5\times10^{-1}\ \text{hour}^{-1}\,.$$