The rate constant of a reaction at temperature 200 is 10 times less than the rate constant at 400 K. What is the activation energy (E_a) of the reaction? (R = gas constant):

921.2 R

The correct answer is option 2. 921.2 R

To determine the activation energy \((E_a)\) of the reaction, we can use the Arrhenius equation:

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

where:

\(k\) is the rate constant,

\(A\) is the pre-exponential factor,

\(E_a\) is the activation energy,

\(R\) is the gas constant, and

\(T\) is the temperature in Kelvin.

We are given that the rate constant at 200 K is 10 times less than the rate constant at 400 K. Mathematically, we can write this as:

\( k_{200} = \frac{1}{10} k_{400} \)

Now, we can substitute these values into the Arrhenius equation and solve for Ea.

For 200 K:

\(k_{200} = Ae^{-\frac{E_a}{RT_{200}}}\)

For 400 K:

\(k_{400} = Ae^{-\frac{E_a}{RT_{400}}}\)

Taking the ratio of these two equations:

\(\frac{k_{200}}{k_{400}} = \frac{Ae^{-\frac{E_a}{RT_{200}}}}{Ae^{-\frac{E_a}{RT_{400}}}} \)

Simplifying:

\(\frac{k_{200}}{k_{400}} = e^{-\frac{E_a}{R}\left(\frac{1}{T_{200}} - \frac{1}{T_{400}}\right)} \)

Given that \( k_{200} = \frac{1}{10} k_{400} \), we can substitute this into the equation:

\(\frac{1}{10} = e^{-\frac{E_a}{R}\left(\frac{1}{T_{200}} - \frac{1}{T_{400}}\right)} \)

Taking the natural logarithm of both sides:

\(\ln\left(\frac{1}{10}\right) = -\frac{E_a}{R}\left(\frac{1}{T_{200}} - \frac{1}{T_{400}}\right) \)

Simplifying further:

\(\frac{E_a}{R} = -\frac{\ln\left(\frac{1}{10}\right)}{\frac{1}{T_{200}} - \frac{1}{T_{400}}} \)

Now, we can substitute the values for \( T_{200} = 200 \) K and \( T_{400} = 400 \) K, as well as the value of the gas constant R, to calculate Ea:

\(\frac{E_a}{R} = -\frac{\ln\left(\frac{1}{10}\right)}{\frac{1}{200} - \frac{1}{400}}\)

Calculating this expression:

\(\frac{E_a}{R} \approx 921.2 \)

Hence, the activation energy (Ea) of the reaction is approximately 921.2 times the gas constant (R).

Therefore, the correct answer is (2) 921.2 R.