For a particular reaction taking place at \(T = 300K\), \(E_a = 300 kJ/mol\) and \(log k = -3.45\); calculate the temperature at which the rate constant becomes 4 times that of \(k\).

304 K

The correct answer is option 2. 304 K.

Given:

Initial temperature: \( T_1 = 300 \, \text{K} \)

Activation energy: \( E_a = 300 \, \text{kJ/mol} = 300 \times 10^3 \, \text{J/mol} \)

Rate constant at \( T_2 \): \( k_2 = 4k_1 \)

The Arrhenius equation relates the rate constant \(k\) to the temperature \(T\) as follows:

\(k = A \exp\left(-\frac{E_a}{RT}\right)\)

Where \(A\) is a pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature.

For two different temperatures \( T_1 \) and \( T_2 \), the ratio of rate constants \( k_1 \) and \( k_2 \) is:

\(\frac{k_2}{k_1} = \exp\left[\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right]\)

Given that \( k_2 = 4k_1 \):

\(4 = \exp\left[\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right]\)

Taking the natural logarithm of both sides:

\(\ln(4) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)

Substitute the given values:

\(\ln(4) = \frac{300 \times 10^3}{8.314} \left(\frac{1}{300} - \frac{1}{T_2}\right)\)

Now, rearrange to solve for \( \frac{1}{T_2} \):

\(\frac{1}{T_2} = \frac{1}{300} - \frac{\ln(4) \times 8.314}{300 \times 10^3}\)

Plugging in the numbers:

\(\frac{1}{T_2} = \frac{1}{300} - \frac{1.386 \times 8.314}{300 \times 10^3}\)

This simplifies to:

\(T_2 \approx 303.5 \, \text{K}\)

Since \( T_2 \) is approximately \( 303.5 \, \text{K} \), the closest answer among the provided options is 304 K.