Question Bank • QB_DFE_034 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Preservation of Food

Subtopic: Kinetics of microbial death – pasteurization and sterilization of milk and other liquid foods

Question Statement:

Thermal destruction of bacterial spores exhibits first-order kinetics. The reaction rate constants are $k_1 = 0.046\text{ s}^{-1}$ at $T_1 = 383.15\text{ K}$ ($110^\circ\text{C}$) and $k_2 = 0.460\text{ s}^{-1}$ at $T_2 = 393.15\text{ K}$ ($120^\circ\text{C}$). Taking the universal gas constant $R = 8.314\text{ J/(mol}\cdot\text{K)}$, the Arrhenius activation energy ($E_a$) in $\text{kJ/mol}$ is ________ (round off to 1 decimal place).

Answer Key:

288.4

Step-by-Step Solution & Explanation

From the Arrhenius equation: $\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$ $\ln\left(\frac{0.460}{0.046}\right) = \ln(10) = 2.302585$ $\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{383.15} - \frac{1}{393.15} = 0.00260994 - 0.00254356 = 6.6384 \times 10^{-5}\text{ K}^{-1}$ $2.302585 = \frac{E_a}{8.314} \times 6.6384 \times 10^{-5}$ $E_a = \frac{2.302585 \times 8.314}{6.6384 \times 10^{-5}} = 288378\text{ J/mol} \approx 288.4\text{ kJ/mol}$
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