The activation energies of two reactions are `E_(1)` & `E_(2)` with `E_(1)gtE_(2)`. If temperature of reacting system is increased from `T_(1)` (rate constant are `k_(1)` and `k_(2)`) to `T_(2)` (rate constant are `k_(1)^(1)` and `k_(2)^(1)`) predict which of the following alternative is correct.

Options

A. `(k_(1)^(1))/(k_(1))=(k_(2)^(1))/(k_(2))`
B. `(k_(1)^(1))/(k_(1))gt(k_(2)^(1))/(k_(2))`
C. `(k_(1)^(1))/(k_(1))lt(k_(2)^(1))/(k_(2))`
D. `k_(1)^(1)ltk_(2)^(1)`

Explanation

To solve the problem, we need to analyze the effect of temperature on the rate constants of two reactions with different activation energies. Here’s a step-by-step solution:

### Step 1: Understand the Arrhenius Equation
The Arrhenius equation relates the rate constant (k) of a reaction to its activation energy (Ea) and temperature (T):
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor
- \( E_a \) = activation energy
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Write the Rate Constants for Two Temperatures
For two reactions with activation energies \( E_1 \) and \( E_2 \) (where \( E_1 > E_2 \)), we can express the rate constants at two different temperatures \( T_1 \) and \( T_2 \):
1. For the first reaction:
\[ k_1' = A_1 e^{-\frac{E_1}{RT_2}} \]
\[ k_1 = A_1 e^{-\frac{E_1}{RT_1}} \]

2. For the second reaction:
\[ k_2' = A_2 e^{-\frac{E_2}{RT_2}} \]
\[ k_2 = A_2 e^{-\frac{E_2}{RT_1}} \]

### Step 3: Formulate the Ratios of Rate Constants
We can form the ratios of the rate constants at the two temperatures:
1. For the first reaction:
\[ \frac{k_1'}{k_1} = \frac{e^{-\frac{E_1}{RT_2}}}{e^{-\frac{E_1}{RT_1}}} = e^{-\frac{E_1}{RT_2} + \frac{E_1}{RT_1}} = e^{\frac{E_1}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)} \]

2. For the second reaction:
\[ \frac{k_2'}{k_2} = \frac{e^{-\frac{E_2}{RT_2}}}{e^{-\frac{E_2}{RT_1}}} = e^{-\frac{E_2}{RT_2} + \frac{E_2}{RT_1}} = e^{\frac{E_2}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)} \]

### Step 4: Compare the Ratios
Now we can compare the two ratios:
\[ \frac{k_1'}{k_1} \text{ and } \frac{k_2'}{k_2} \]
Taking the natural logarithm of both ratios:
\[ \ln\left(\frac{k_1'}{k_1}\right) = \frac{E_1}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
\[ \ln\left(\frac{k_2'}{k_2}\right) = \frac{E_2}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

### Step 5: Analyze the Effect of Activation Energies
Since \( E_1 > E_2 \), it follows that:
\[ \frac{E_1}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) > \frac{E_2}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
This indicates that:
\[ \ln\left(\frac{k_1'}{k_1}\right) > \ln\left(\frac{k_2'}{k_2}\right) \]
Thus, we conclude:
\[ \frac{k_1'}{k_1} > \frac{k_2'}{k_2} \]

### Conclusion
This means that the increase in the rate constant for the reaction with the higher activation energy \( E_1 \) is greater than that for the reaction with the lower activation energy \( E_2 \).

### Final Answer
The correct alternative is that \( \frac{k_1'}{k_1} > \frac{k_2'}{k_2} \).

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