The rate of a reaction decreased by 3.555 times when the temperature was changed from 40ºC to 30ºC. — Chemical Kinetics Chemistry Question
Question
The rate of a reaction decreased by 3.555 times when the temperature was changed from 40ºC to 30ºC. The activation energy (in KJ mol ) of the reaction is .......[Take; R = 8.314 J mol K In 3.555 = 1.268] Round off the answer to the nearest integer. -1 -1 -1
💡 Solution & Explanation
**Step 1: Identify the Arrhenius equation relationship** Use the Arrhenius equation in logarithmic form: $$\ln\left(\frac{k_1}{k_2}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$ **Step 2: Determine the rate ratio and temperatures** - Rate decreased by 3.555 times when going from 40°C to 30°C - This means: k₁/k₂ = 3.555 (at T₁ = 40°C and T₂ = 30°C) - T₁ = 40°C = 313.15 K - T₂ = 30°C = 303.15 K **Step 3: Calculate the natural logarithm** $$\ln(3.555) = 1.268$$ (given) **Step 4: Calculate the temperature term** $$\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{313.15} - \frac{1}{303.15}$$ $$= 0.003193 - 0.003298 = -0.000105 \text{ K}^{-1}$$ **Step 5: Solve for activation energy** $$1.268 = \frac{E_a}{8.314} \times (-0.000105)$$ $$E_a = \frac{1.268 \times 8.314}{0.000105} = \frac{10.54}{0.000105}$$ $$E_a = 100,381 \text{ J/mol} ≈ 100 \text{ kJ/mol}$$ Therefore, the answer is **100.00**.