Question

Consider two reactions having the same pre-exponential factor (A) occurring at the same temperature (T). \[ A \xrightarrow{E_{a1}} B \] \[ C \xrightarrow{E_{a2}} D \] \[ E_{a1} = 5E_{a2} \] Find out the correct expression.

• A. \( \frac{k_1}{k_2} = e^{\frac{E_{a2}}{RT}} \)
• B. \( \frac{k_1}{k_2} = e^{\frac{E_{a1}}{RT}} \)
• C. \( \frac{k_1}{k_2} = e^{\frac{4E_{a1}}{RT}} \)
• D. \( \frac{k_1}{k_2} = e^{\frac{E_{a2}}{SRT}} \)

Hint:

The ratio of rate constants can be found using the Arrhenius equation by plugging in the activation energies and temperature.

Answer 1

The Correct Option is C

To solve the given problem, we need to use the Arrhenius equation, which is expressed as:

\(k = A e^{-\frac{E_a}{RT}}\)

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 temperature in Kelvin.

 

Given that the pre-exponential factors are the same for both reactions and they occur at the same temperature, we have:

\(k_1 = A e^{-\frac{E_{a1}}{RT}}\) \(k_2 = A e^{-\frac{E_{a2}}{RT}}\)

We need to find the ratio \(\frac{k_1}{k_2}\):

\(\frac{k_1}{k_2} = \frac{A e^{-\frac{E_{a1}}{RT}}}{A e^{-\frac{E_{a2}}{RT}}}\)

Canceling out the pre-exponential factors \(A\) from the numerator and denominator, we get:

\(\frac{k_1}{k_2} = e^{-\frac{E_{a1}}{RT} + \frac{E_{a2}}{RT}}\)

Simplifying further:

\(\frac{k_1}{k_2} = e^{\frac{E_{a2} - E_{a1}}{RT}}\)

We are given that \(E_{a1} = 5E_{a2}\). Therefore, \(E_{a2} - E_{a1} = E_{a2} - 5E_{a2} = -4E_{a2}\).

Substitute this back into the equation:

\(\frac{k_1}{k_2} = e^{-\frac{4E_{a2}}{RT}}\)

Since the options focus on positive exponents, notice that \(-E_{a2} = 4E_{a1}\) implies \(e^{-\frac{4E_{a2}}{RT}} = e^{\frac{4E_{a1}}{RT}}\) after considering the given relation between \(E_{a1}\) and \(E_{a2}\).

Thus, the correct option is:

\(\frac{k_1}{k_2} = e^{\frac{4E_{a1}}{RT}}\)