Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Physical: Chemical Kinetics

Question:

The activation energy for a reaction at a temperature \(T K\) was found to be \(2.303RT \text{ J/mol}\). The ratio of the rate constant to Arrhenius factor is

Options:

\(10^{-1}\)

\(10^{-2}\)

\(2 × 10^{-2}\)

\(2 × 10^{-3}\)

Correct Answer:

\(10^{-1}\)

Explanation:

The correct answer is (1) \(10^{-1}\).

The Arrhenius equation relates the rate constant (\(k\)) of a reaction to the temperature (\(T\)), the Arrhenius factor (\(A\)), and the activation energy (\(E_a\)) as follows:
\[k = A \cdot e^{-\frac{E_a}{RT}}\]
Where:
- \(k\) is the rate constant
- \(A\) is the Arrhenius factor (pre-exponential factor)
- \(E_a\) is the activation energy
- \(R\) is the universal gas constant (8.314 J/mol·K)
- \(T\) is the temperature in Kelvin
In this case, you are given that the activation energy (\(E_a\)) is \(2.303RT\) J/mol. Now, let's calculate the ratio of the rate constant (\(k\)) to the Arrhenius factor (\(A\)).
\[ \frac{k}{A} = e^{-\frac{E_a}{RT}} = e^{-\frac{2.303RT}{RT}} = e^{-2.303} \]
Now, you can calculate this value:
\[ \frac{k}{A} \approx e^{-2.303} \approx 0.1 \]
So, the ratio of the rate constant to the Arrhenius factor is approximately \(0.1\), which is the same as \(10^{-1}\).
Therefore, the correct option is (1) \(10^{-1}\).