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Step-by-Step Solution
Step 1: Identify the Given Data
• Initial temperature, T_{1} = 27^\circ\text{C} = 300\text{ K} (since 27 + 273 = 300 ).
• Final temperature, T_{2} = 52^\circ\text{C} = 325\text{ K} (since 52 + 273 = 325 ).
• The ratio of rate constants (temperature coefficient), \frac{k_{T_2}}{k_{T_1}} = 5 .
• Gas constant, R = 8.314 \,\text{J} \,\text{K}^{-1}\,\text{mol}^{-1} .
Step 2: Write Down the Arrhenius Relationship
The dependence of the rate constant k on temperature T can be written (in logarithmic form) as:
\log\left(\frac{k_{T_2}}{k_{T_1}}\right) = \frac{E_a}{2.303\,R} \times \left(\frac{T_2 - T_1}{T_1 \, T_2}\right),
where E_a is the energy of activation.
Step 3: Substitute the Known Values
Substitute \log(5) for \log\left(\frac{k_{325}}{k_{300}}\right) , T_1 = 300\,\text{K} , and T_2 = 325\,\text{K} :
\log(5)
= \frac{E_a}{2.303 \times 8.314}
\times \frac{(325 - 300)}{(300 \times 325)}.
Step 4: Solve for the Activation Energy E_a
Rearrange to find E_a :
E_a
= \log(5) \times \left(\frac{2.303 \times 8.314 \times 300 \times 325}{25}\right).
Perform the numerical calculation step by step:
Compute the fraction \frac{25}{300 \times 325} = \frac{25}{97500}.
Then multiply by 2.303 \times 8.314 and finally by \log(5).
After carrying out the arithmetic, we obtain E_a \approx 52.2\,\text{kJ mol}^{-1}.
Step 5: Final Answer (Rounded Off)
Rounding to the nearest integer:
E_a \approx 52\,\text{kJ mol}^{-1}.
