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Step-by-Step Solution
Step 1: Identify the Given Data
• Rate constant at 327°C: $k_1 = 2.5 \times 10^{-4}\,\text{dm}^3\,\text{mol}^{-1}\,\text{s}^{-1}$
• Rate constant at 527°C: $k_2 = 1.0\,\text{dm}^3\,\text{mol}^{-1}\,\text{s}^{-1}$
• Universal gas constant: $R = 8.314\,\text{J K}^{-1}\,\text{mol}^{-1}$
• Temperatures (in Kelvin):
$T_1 = 327 + 273 = 600\,\text{K}$
$T_2 = 527 + 273 = 800\,\text{K}$
Step 2: Write the Arrhenius Equation
The Arrhenius equation relating two rate constants $k_1$ and $k_2$ at temperatures $T_1$ and $T_2$ is:
$$
\log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303\,R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)
$$
where $E_a$ is the activation energy.
Step 3: Plug in the Known Values
The ratio of the rate constants is
$$
\frac{k_2}{k_1} = \frac{1.0}{2.5 \times 10^{-4}} = 4 \times 10^3.
$$
Taking the logarithm (base 10):
$$
\log \left( 4 \times 10^3 \right) = \log(4000) \approx 3.6.
$$
Substitute into the Arrhenius equation:
$$
\log \left( 4 \times 10^3 \right) = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{600} - \frac{1}{800} \right).
$$
Step 4: Compute the Temperature Difference Term
Calculate the difference inside the parenthesis:
$$
\frac{1}{600} - \frac{1}{800} = \frac{4}{2400} - \frac{3}{2400} = \frac{1}{2400}.
$$
Thus,
$$
\log(4000) = \frac{E_a}{2.303 \times 8.314} \times \frac{1}{2400}.
$$
Step 5: Solve for $E_a$
Rearrange to isolate $E_a$:
$$
E_a = \log(4000) \times 2.303 \times 8.314 \times 2400.
$$
Numerically,
$$
E_a \approx 3.6 \times 2.303 \times 8.314 \times 2400.
$$
After calculation and converting to kJ/mol (by dividing by 1000 if needed),
$$
E_a \approx 166 \,\text{kJ mol}^{-1}.
$$
Final Answer:
The activation energy for the reaction is approximately $166\,\text{kJ mol}^{-1}$.
