Question

If $T_m$ is the average absolute temperature, $R$ is the universal gas constant, then the quantitative relationship between activation energy $E$ and $Z$-value is:

• A. $E = \dfrac{2.303RT_m}{Z}$
• B. $E = \dfrac{2.303RT_m^2}{Z}$
• C. $E = \dfrac{2.303RZ}{T_m}$
• D. $E = \dfrac{2.303RZ}{T_m^2}$

Hint:

Remember key relation: $E \propto \dfrac{T^2}{Z}$. Higher $Z$ → lower temperature sensitivity.

Answer 1

The Correct Option is B

Concept: The relationship between activation energy ($E$) and $Z$-value arises from thermal destruction kinetics and the Arrhenius equation. The $Z$-value represents the temperature change required to bring about a tenfold change in the reaction rate (or D-value), while activation energy quantifies the sensitivity of the reaction rate to temperature.

Step 1: Starting from Arrhenius equation.
The Arrhenius equation is: \[ k = A e^{-E/(RT)} \] Taking logarithm (base 10): \[ \log k = \log A - \frac{E}{2.303RT} \]

Step 2: Relation between D-value and temperature.
In thermal processing: \[ \log D = -\frac{T}{Z} + \text{constant} \] This shows that the logarithm of D-value varies linearly with temperature.

Step 3: Comparing both expressions.
From Arrhenius relation and thermal death time relation:
• Slope from Arrhenius form = $\frac{E}{2.303RT^2}$
• Slope from $Z$-value definition = $\frac{1}{Z}$

Step 4: Equating slopes.
\[ \frac{E}{2.303RT_m^2} = \frac{1}{Z} \]

Step 5: Rearranging the equation.
Multiplying both sides: \[ E = \frac{2.303RT_m^2}{Z} \]

Step 6: Matching with options.
• Option (A) missing $T_m^2$
• Option (B) matches derived expression
• Option (C) incorrect arrangement
• Option (D) incorrect form Final Conclusion:
The correct relationship between activation energy and $Z$-value is: \[ E = \frac{2.303RT_m^2}{Z} \] Hence, the correct answer is option (2).