Question

The rate of a gaseous reaction triples when temperature is increased from 17°C to 27°C. Calculate the energy of activation for this reaction. $[Given : 2 303 R = 19 15 JK^{-1} mol^{-1}, log 3 = 0 48]$

Hint:

- The Arrhenius equation is useful to calculate the energy of activation when the rate constant changes with temperature. - The rate of reaction is highly sensitive to temperature, and small temperature changes can significantly impact the rate.

Answer 1

The Arrhenius equation is: \[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \quad \text{or} \quad \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right] \] Substituting the known values for the rate constants and temperatures into the equation. \(k_2/k_1 = 3\), \(T_1 = 290\,K\), and \(T_2 = 300\,K\): \[ \log \left( \frac{3k_1}{k_1} \right) = \frac{E_a}{19.15} \left( \frac{1}{290} - \frac{1}{300} \right) \] we get: \[ 0.48 = \frac{E_a}{19.15} \left( \frac{10}{290 \times 300} \right) \] Activation energy, \(E_a\), by multiplying both sides of the equation: \[ E_a = 0.48 \times 19.15 \times 290 \times 300 \quad \Rightarrow \quad E_a = \frac{0.48 \times 19.15 \times 290 \times 300}{10} \] Finally, calculate the value of \(E_a\). The result gives the activation energy in J/mol. After performing the calculation, we find: \[ E_a = 79970 \, \text{J mol}^{-1} \quad \text{or} \quad E_a = 79.970 \, \text{kJ mol}^{-1} \]