The activation energies of two reactions are `E_(a1)` and `E_(a2)` with `E_(a1) gt E_(a2)`. If the temperature of the reacting systems is increased from `T` to `T'`, which of the following is correct?

Options

A. `(k1')/(k_(1))=(k'_(2))/(k_(2))`
B. `(k'_(1))/(k_(1)) lt 2(k'_(2))/(k_(2))`
C. `(k'_(1))/(k_(1)) gt (k'_(2))/(k_(2))`
D. `(k'_(1))/(k_(1)) lt (k'_(2))/(k_(2))`

Answer

To solve the problem, we need to analyze the effect of temperature on the rate constants of two reactions with different activation energies. Let's break it down step by step.

### Step 1: Understand the relationship between rate constant and activation energy
The rate constant \( k \) of a reaction is related to the activation energy \( E_a \) and temperature \( T \) by the Arrhenius equation:
\[ k = A e^{-\frac{E_a}{RT}} \]
where \( A \) is the pre-exponential factor, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.

### Step 2: Write the expression for the change in rate constant with temperature
When the temperature changes from \( T \) to \( T' \), the new rate constants \( k_1' \) and \( k_2' \) for the two reactions can be expressed as:
\[ \log \frac{k_1'}{k_1} = \frac{E_{a1}}{R} \left( \frac{1}{T} - \frac{1}{T'} \right) \]
\[ \log \frac{k_2'}{k_2} = \frac{E_{a2}}{R} \left( \frac{1}{T} - \frac{1}{T'} \right) \]

### Step 3: Compare the two reactions
Since it is given that \( E_{a1} > E_{a2} \), we can analyze the implications:
- The term \( \frac{E_{a1}}{R} \left( \frac{1}{T} - \frac{1}{T'} \right) \) will be larger than \( \frac{E_{a2}}{R} \left( \frac{1}{T} - \frac{1}{T'} \right) \) when \( T' > T \).
- This means that the increase in the rate constant \( k_1' \) for the reaction with higher activation energy \( E_{a1} \) will be less pronounced compared to the increase in \( k_2' \) for the reaction with lower activation energy \( E_{a2} \).

### Step 4: Conclusion
From the above analysis, we can conclude that:
\[ \frac{k_1'}{k_1} < \frac{k_2'}{k_2} \]
This implies that the rate constant for the reaction with the lower activation energy increases more significantly with the increase in temperature than that of the reaction with the higher activation energy.

### Final Answer
Thus, the correct conclusion is that the reaction with lower activation energy (reaction 2) will have a greater increase in its rate constant compared to the reaction with higher activation energy (reaction 1) when the temperature is increased from \( T \) to \( T' \).