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Step-by-Step Solution
Step 1: Recall the Arrhenius Equation
The dependence of the rate constant k on temperature T is given by the Arrhenius equation:
k = A e^{-\frac{E_a}{RT}}
where
A is the frequency factor,
E_a is the activation energy,
R is the gas constant, and
T is the temperature in Kelvin.
Step 2: Use the Logarithmic Form
Taking the natural logarithm and comparing the rate constants at two different temperatures gives:
\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)
Step 3: Identify Known Quantities
The rate of the reaction decreases by 3.555 times going from 40^\circ C (i.e., 313\,K ) to 30^\circ C (i.e., 303\,K ). Hence,
\frac{k_1}{k_2} = 3.555
\quad\Longrightarrow\quad
\ln \left(\frac{k_1}{k_2}\right) = \ln(3.555) = 1.268.
Gas constant, R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}.
Temperatures:
T_1 = 313\,K,\;
T_2 = 303\,K.
We will use the above form in which
\ln \left(\frac{k_1}{k_2}\right) = \frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)
or its equivalent arrangement (the sign depends on which ratio is placed in the numerator). In the provided solution reference, it is taken as
\ln(3.555) = \frac{E_a}{R}\left(\frac{1}{303} - \frac{1}{313}\right).
Step 4: Plug In the Data
The difference in reciprocals of the temperatures is:
\left(\frac{1}{303} - \frac{1}{313}\right) \approx \frac{313 - 303}{303 \times 313} = \frac{10}{303 \times 313}.
The provided solution simplifies this difference in a way that leads to the final approximate value for E_a . Following that reference directly:
\ln(3.555) = 1.268 = \frac{E_a}{8.314}\left(\frac{1}{303} - \frac{1}{313}\right).
Step 5: Final Calculation and Approximation
Using the reference approach shown:
E_a
= \frac{1.268 \times 8.314 \times 3.3 \times 313}{10}
\approx 99980.7\ \mathrm{J\,mol^{-1}}
= 99.98\ \mathrm{kJ\,mol^{-1}}
\approx 100\ \mathrm{kJ\,mol^{-1}}.
The final (rounded) activation energy is:
\boxed{100\ \mathrm{kJ\,mol^{-1}}}
