Question

The reactions \(A \xrightarrow{k_1} B\) and \(C \xrightarrow{k_2} D\) follow first order kinetics. At \(500\,\text{K}\), rate constants are \(k_1\) and \(k_2\) respectively, where \(k_2 = 2k_1\). Activation energies \(E_{a1}\) and \(E_{a2}\) are related as \(E_{a2} = \dfrac{E_{a1}}{2}\). The rate constant for the first reaction at \(300\,\text{K}\) is half of its value at \(500\,\text{K}\). Given: half-life of first reaction is \(2\) hrs at \(500\,\text{K}\). Find the value of \(10 \times (k_2)_{300\,\text{K}\). }

Hint:

If activation energy is halved, the temperature dependence of the rate constant becomes the {square root} of the original relation.

Answer 1

Correct Answer: 5

Concept:


For first order reaction: \[ t_{1/2} = \frac{0.693}{k} \]
Temperature dependence of rate constant is given by Arrhenius equation: \[ \ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

Step 1: Calculate \(k_1\) at \(500\,\text{K}\)
Given: \[ t_{1/2} = 2\,\text{hrs} \] \[ k_{1,500} = \frac{0.693}{2} = 0.3465\,\text{hr}^{-1} \]
Step 2: Rate Constant of Reaction 1 at \(300\,\text{K}\)
Given: \[ k_{1,300} = \frac{1}{2} k_{1,500} \] \[ k_{1,300} = 0.17325\,\text{hr}^{-1} \]
Step 3: Relation Between \(k_1\) and \(k_2\)
At \(500\,\text{K}\): \[ k_{2,500} = 2k_{1,500} \] Using Arrhenius relation: \[ \ln\frac{k_{2,300}}{k_{2,500}} = \frac{E_{a2}}{R}\left(\frac{1}{500}-\frac{1}{300}\right) \] But since: \[ E_{a2} = \frac{E_{a1}}{2} \quad \text{and} \quad \ln\frac{k_{1,300}}{k_{1,500}} = \frac{E_{a1}}{R}\left(\frac{1}{500}-\frac{1}{300}\right) \] Therefore: \[ \ln\frac{k_{2,300}}{k_{2,500}} = \frac{1}{2}\ln\frac{k_{1,300}}{k_{1,500}} \] \[ \frac{k_{2,300}}{k_{2,500}} = \sqrt{\frac{k_{1,300}}{k_{1,500}}} = \sqrt{\frac{1}{2}} \] \[ k_{2,300} = 2k_{1,500} \times \frac{1}{\sqrt{2}} = \sqrt{2}\,k_{1,500} \]
Step 4: Numerical Value
\[ k_{2,300} \approx \sqrt{2} \times 0.3465 \approx 0.49 \] \[ 10 \times k_{2,300} \approx 4.9 \approx 5 \] \[ \boxed{10 \times (k_2)_{300\,\text{K}} = 5} \]