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Step-by-Step Solution
Step 1: Convert given temperatures to Kelvin
The initial temperature is 27°C, which equals 27 + 273 = 300 K.
The final temperature is 42°C, which equals 42 + 273 = 315 K.
Hence, T1 = 300 K and T2 = 315 K.
Step 2: Relate increase in the number of activated molecules to rate constants
The problem states that the number of molecules with energy greater than the threshold energy increases fivefold. This implies that the rate constant (k), which depends on the fraction of molecules with energy ≥ Ea, also increases five times:
KT₂ = 5 × KT₁.
Step 3: Write the Arrhenius equation in ratio form
The Arrhenius equation is given by
k = A e^{-\frac{E_a}{RT}} .
Dividing the rate constants at two different temperatures T1 and T2, we get:
\ln\left(\frac{K_{T_2}}{K_{T_1}}\right)
= \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right).
Step 4: Substitute the known values
We know:
\ln\left(\frac{K_{T_2}}{K_{T_1}}\right) = \ln(5) = 1.6094.
T_1 = 300\text{ K}, \; T_2 = 315\text{ K}.
R = 8.314\text{ J mol}^{-1}\text{K}^{-1}.
Therefore:
1.6094
= \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{315}\right).
Step 5: Simplify the temperature difference term
\left(\frac{1}{300} - \frac{1}{315}\right)
= \frac{315 - 300}{300 \times 315}
= \frac{15}{300 \times 315}
= \frac{15}{94500}.
Step 6: Solve for E_a
Rearranging,
E_a =
\frac{1.6094 \times 8.314}{\left(\frac{15}{94500}\right)}.
We can rewrite it as:
E_a =
1.6094 \times 8.314 \times \frac{94500}{15}.
Step 7: Numerical evaluation
First, compute the fraction:
\frac{94500}{15} = 6300.
So,
E_a
= 1.6094 \times 8.314 \times 6300.
Performing the multiplication (as given in the referenced solution):
E_a
\approx 84297.47 \text{ J/mol}.
Final Answer
E_a \approx 8.43 \times 10^4 \text{ J/mol} (approximately).
