Question

For an elementary chemical reaction, the Arrhenius plot is given below.

If the energy of activation is $6.64 \text{ kJ mol}^{-1}$ and $R = 8.3 \text{ J K}^{-1}\text{ mol}^{-1}$, the temperature at which the rate constant becomes $e^2 \text{ min}^{-1}$, is

• A. $250\text{ K}$
• B. $125\text{ K}$
• C. $150\text{ K}$
• D. $200\text{ K}$

Hint:

Always double-check your units in chemical kinetics calculations! Activation energy ($E_a$) is usually given in kJ/mol, while the gas constant $R$ is given in $\text{J/(K}\cdot\text{mol)}$. Multiplying the kilojoules by $10^3$ is a vital step to avoid being off by a factor of 1000!

Answer 1

The Correct Option is D

Step 1: Write the Arrhenius equation in log form.
The rate constant depends on temperature as \[ \ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T} \] This is a straight line of \(\ln k\) versus \(1/T\) with intercept \(\ln A\).
Step 2: Read the intercept from the graph.
The line cuts the \(\ln k\) axis at \(6\), so \(\ln A = 6\).
Step 3: Convert the data to consistent units.
\(E_a = 6.64\ kJ\,mol^{-1} = 6640\ J\,mol^{-1}\) and \(R = 8.3\ J\,K^{-1}mol^{-1}\). Now the joules cancel cleanly.
Step 4: Put in the target rate constant.
We want \(k = e^2\), so \(\ln k = 2\). Substitute into the equation: \[ 2 = 6 - \frac{6640}{8.3\,T} \]
Step 5: Solve for the temperature.
Rearrange: \(\dfrac{6640}{8.3\,T} = 6-2 = 4\), so \[ T = \frac{6640}{8.3\times 4} = \frac{6640}{33.2} \]
Step 6: Compute and conclude.
Since \(33.2\times 200 = 6640\), we get \(T = 200\ K\).
\[ \boxed{T = 200\ \text{K}} \]