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)`
For which of the following reactions `k_(310)//k_(300)` would be maximum?

Options

A. `P + Q rarr R, E_(a) = 10 kJ`
B. `E + F rarr D, E_(a) = 21 kJ`
C. `A + B rarr C, E_(a) = 10.5 kJ`
D. `L+M rarr N, E_(a) = 5 kJ`

Explanation

To solve the question regarding which reaction will have the maximum ratio of rate constants \( k_{310}/k_{300} \), we will use the Arrhenius equation and analyze the activation energies of the given reactions.

### 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 at Two Temperatures**:
For two temperatures, \( T_1 = 300 \, K \) and \( T_2 = 310 \, K \):
\[ k_{300} = A e^{-\frac{E_a}{R \cdot 300}} \]
\[ k_{310} = A e^{-\frac{E_a}{R \cdot 310}} \]

3. **Calculate the Ratio \( \frac{k_{310}}{k_{300}} \)**:
The ratio of the rate constants at these two temperatures can be expressed as:
\[ \frac{k_{310}}{k_{300}} = \frac{A e^{-\frac{E_a}{R \cdot 310}}}{A e^{-\frac{E_a}{R \cdot 300}}} \]
The \( A \) cancels out:
\[ \frac{k_{310}}{k_{300}} = e^{-\frac{E_a}{R \cdot 310} + \frac{E_a}{R \cdot 300}} \]

4. **Simplify the Exponent**:
This can be rewritten as:
\[ \frac{k_{310}}{k_{300}} = e^{\frac{E_a}{R} \left( \frac{1}{300} - \frac{1}{310} \right)} \]

5. **Calculate the Difference**:
The term \( \frac{1}{300} - \frac{1}{310} \) can be simplified:
\[ \frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = \frac{1}{9300} \]

6. **Final Expression for the Ratio**:
Thus, we have:
\[ \frac{k_{310}}{k_{300}} = e^{\frac{E_a}{R \cdot 9300}} \]
This shows that the ratio \( \frac{k_{310}}{k_{300}} \) is directly proportional to the activation energy \( E_a \).

7. **Conclusion**:
To determine which reaction has the maximum ratio \( \frac{k_{310}}{k_{300}} \), we need to identify the reaction with the highest activation energy \( E_a \). According to the information given, the reaction with the highest activation energy is the second reaction with \( E_a = 21 \, \text{kJ/mol} \).

### Final Answer:
The reaction for which \( \frac{k_{310}}{k_{300}} \) would be maximum is the second reaction with an activation energy of 21 kJ/mol.

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