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Q3.7:
What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?

Solution :

1. Effect of Temperature on the Rate Constant

The rate constant ($k$) of a chemical reaction is highly sensitive to changes in temperature. For most chemical reactions, an increase in temperature leads to an exponential increase in the rate constant, typically resulting in a doubling or tripling of the reaction rate for every $10^\circ\text{C}$ rise in temperature.

Qualitative Justification: According to the Maxwell-Boltzmann distribution of molecular energies, an increase in temperature shifts the distribution curve to the right and increases the area under the curve representing molecules with energy equal to or greater than the threshold activation energy ($E_a$). Consequently, the frequency of effective collisions increases, directly elevating the rate constant [$Per\ Arrhenius\ collision\ theory$].

Kinetic Energy (E) Fraction of molecules Threshold Energy (Ea) T1 T2 (T2 > T1)

2. Quantitative Representation: The Arrhenius Equation

The quantitative relationship between the rate constant ($k$) and temperature ($T$) is defined by the Arrhenius equation:

$k = A e^{-\frac{E_a}{RT}}$

Where:

  • $k$: Rate constant of the reaction.
  • $A$: Arrhenius factor or frequency factor (pre-exponential factor).
  • $E_a$: Activation energy ($\text{J mol}^{-1}$).
  • $R$: Universal gas constant ($8.314 \text{ J K}^{-1} \text{ mol}^{-1}$).
  • $T$: Temperature in Kelvin.

3. Mathematical Derivation for Temperature Variation

To evaluate the effect of changing temperature from $T_1$ to $T_2$, we take the natural logarithm ($\ln$) of the Arrhenius equation:

$\ln k = \ln A - \frac{E_a}{RT}$

For two different temperatures $T_1$ and $T_2$ with corresponding rate constants $k_1$ and $k_2$:

$\ln k_1 = \ln A - \frac{E_a}{RT_1}$

$\ln k_2 = \ln A - \frac{E_a}{RT_2}$

Subtracting the first equation from the second yields the linear form used for calculation [$By\ algebraic\ subtraction$]:

$\ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$

Converting to base-10 logarithm ($\log$):

$\log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$

ParameterEffect of Increasing Temperature
Rate Constant ($k$)Increases exponentially
Fraction of molecules with $E \ge E_a$Increases
Frequency of collisionsIncreases

Final Solution: The effect of temperature on the rate constant is quantitative expressed via the Arrhenius equation $k = A e^{-E_a/RT}$. This shows that $k$ increases exponentially with temperature because a higher fraction of reactant molecules possess sufficient kinetic energy to overcome the activation energy barrier.


More Questions from Class 12 Chemistry Chemical Kinetics Exercises


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