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. `K_1/K_1=K_2/K_2`
B. `K_1/K_1 gt K_2/K_2`
C. `K_1/K_1 lt K_2/K_2`
D. `K_1/K_2 lt 2 K_2/K_2`

Answer

To solve the problem regarding the activation energies of two reactions and the effect of temperature on their rate constants, we can follow these steps:

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

### Step 2: Compare the activation energies
Given that \( E_{a1} > E_{a2} \), we can infer that the reaction with the higher activation energy (reaction 1) will have a lower rate constant at a given temperature compared to the reaction with the lower activation energy (reaction 2).

### Step 3: Analyze the effect of temperature increase
When the temperature of the reacting systems is increased from \( T \) to \( T' \), both reactions will experience an increase in their rate constants. However, the increase will be more significant for the reaction with the lower activation energy \( E_{a2} \) because the exponential factor \( e^{-\frac{E_a}{RT}} \) will change more dramatically for smaller values of \( E_a \).

### Step 4: Compare the rate constants at the new temperature
Let’s denote the rate constants at the original temperature \( T \) as \( k_1 \) and \( k_2 \), and at the new temperature \( T' \) as \( k_1' \) and \( k_2' \). Since \( E_{a1} > E_{a2} \):
\[ \frac{k_1'}{k_1} > \frac{k_2'}{k_2} \]
This indicates that the rate constant for the reaction with lower activation energy (reaction 2) will increase more significantly than that of the reaction with higher activation energy (reaction 1).

### Conclusion
Thus, after the temperature increase, the reaction with the lower activation energy will have a higher rate constant compared to the reaction with the higher activation energy. Therefore, the correct conclusion is that the rate constant for the reaction with \( E_{a2} \) will be greater than that of \( E_{a1} \) after the temperature increase.

### Final Answer
The correct option is that the rate constant for the reaction with \( E_{a2} \) will be greater than that for \( E_{a1} \) after the temperature increase.

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