Question

The rate constant $ k $ of a reaction doubles when the temperature is raised from 300 K to 310 K. Calculate the activation energy $ E_a $ of the reaction. (Use $ R = 8.314 \, J/mol\cdot K $)

• A. \( 52 \, kJ/mol \)
• B. \( 55 \, kJ/mol \)
• C. \( 53 \, kJ/mol \)
• D. \( 60 \, kJ/mol \)

Hint:

Tip: Use the logarithmic form of the Arrhenius equation and careful unit conversions.

Answer 1

The Correct Option is C

To calculate the activation energy \(E_a\) of the reaction, we use the Arrhenius equation in its logarithmic form: \(\ln\left(\frac{k_2}{k_1}\right)=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\).

Given that the rate constant \(k\) doubles from \(k_1\) to \(k_2\) as temperature increases from \(T_1=300\,K\) to \(T_2=310\,K\), we have \(\frac{k_2}{k_1}=2\).

Substitute the known values into the equation: \(\ln(2)=-\frac{E_a}{8.314}\left(\frac{1}{310}-\frac{1}{300}\right)\).

Calculate the temperature difference: \(\frac{1}{310}-\frac{1}{300}=\frac{300-310}{310\times300}=-\frac{10}{93,000}\approx-0.0001075\).

So, \(\ln(2)\approx0.693\) can be substituted into the equation:\[0.693=-\frac{E_a}{8.314}\times-0.0001075\].

Solving for \(E_a\):\[E_a=\frac{0.693\times8.314}{0.0001075}\approx53,000\ J/mol=53\ kJ/mol\].

Thus, the activation energy \(E_a\) of the reaction is \({53\ kJ/mol}\), which matches the correct answer.

Answer 2

Step 1: Use Arrhenius equation ratio form 
\[ \frac{k_2}{k_1} = e^{\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)} \] Given \( \frac{k_2}{k_1} = 2 \), \( T_1 = 300\,K \), \( T_2 = 310\,K \).

Step 2: Take natural logarithm 
\[ \ln 2 = \frac{E_a}{R} \left(\frac{1}{300} - \frac{1}{310}\right) \]

Step 3: Calculate difference in reciprocals 
\[ \frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} \approx 1.075 \times 10^{-4} \]

Step 4: Calculate \( E_a \) 
\[ \ln 2 \approx 0.693 \] \[ 0.693 = \frac{E_a}{8.314} \times 1.075 \times 10^{-4} \implies E_a = \frac{0.693}{1.075 \times 10^{-4}} \times 8.314 \approx 53642 \, J/mol = 53.6 \, kJ/mol \]