The rate of reaction increases isgnificantly with increase in temperature. Generally, rate of reactions are doubled for every `10^(@)C` rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every `10^(@)C` change in temperature.
`"Temperature coefficient" (mu) = ("Rate constant of" (T + 10)^(@)C)/("Rate constant at" T^(@)C)`
Arrhenius gave an equation which describes aret constant `k` as a function of temperature
`k = Ae^(-E_(a)//RT)`
where `k` is the rate constant, `A` is the frequency factor or pre-exponential factor, `E_(a)` is the activation energy, `T` is the temperature in kelvin, `R` is the universal gas constant.
Equation when expressed in logarithmic form becomes
`log k = log A - (E_(a))/(2.303 RT)`
Activation energies of two reaction are `E_(a)` and `E_(a)'` with `E_(a) gt E'_(a)`. If the temperature of the reacting systems is increased form `T_(1)` to `T_(2)` (`k'` is rate constant at higher temperature).

Options

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

Explanation

To solve the problem, we need to analyze the relationship between the rate constants \( k \) and \( k' \) of two reactions with different activation energies \( E_a \) and \( E_a' \) as the temperature changes from \( T_1 \) to \( T_2 \).

### Step-by-Step Solution:

1. **Understand the Arrhenius Equation**:
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where \( k \) is the rate constant, \( A \) is the frequency factor, \( E_a \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

2. **Write the Rate Constants for Both Temperatures**:
For the first reaction at temperature \( T_1 \):
\[ k = A e^{-\frac{E_a}{RT_1}} \]
For the second reaction at temperature \( T_2 \):
\[ k' = A e^{-\frac{E_a'}{RT_2}} \]

3. **Form the Ratio of Rate Constants**:
To find the ratio \( \frac{k'}{k} \):
\[ \frac{k'}{k} = \frac{A e^{-\frac{E_a'}{RT_2}}}{A e^{-\frac{E_a}{RT_1}}} \]
The \( A \) cancels out:
\[ \frac{k'}{k} = e^{-\frac{E_a'}{RT_2}} \cdot e^{\frac{E_a}{RT_1}} = e^{\left(\frac{E_a}{RT_1} - \frac{E_a'}{RT_2}\right)} \]

4. **Simplify the Exponent**:
We can express the exponent as:
\[ \frac{k'}{k} = e^{\left(\frac{E_a}{RT_1} - \frac{E_a'}{RT_2}\right)} \]

5. **Analyze the Activation Energies**:
Given that \( E_a > E_a' \), we need to analyze how this affects the ratio. Since \( T_2 > T_1 \), the term \( \frac{1}{RT_2} \) will be smaller than \( \frac{1}{RT_1} \). Thus:
\[ \frac{E_a}{RT_1} - \frac{E_a'}{RT_2} \text{ will be greater for the reaction with higher } E_a. \]

6. **Conclusion**:
Since \( E_a > E_a' \), it follows that:
\[ \frac{k'}{k} \text{ for the reaction with } E_a \text{ will be greater than that for } E_a'. \]
Therefore, the ratio \( \frac{k'}{k} \) is larger for the reaction with greater activation energy.

### Final Result:
Thus, we conclude that:
\[ \frac{k'}{k} \text{ (for } E_a \text{) } > \frac{k'}{k} \text{ (for } E_a' \text{)} \]