Question

The rate constant of a reaction at 27°C is 3 x 10$^{-3}$ min$^{-1}$. The fraction of molecule crossing the energy barrier if the reaction at that temperature is 10$^{-3}$%. If temperature is increased, the maximum value of rate constant K can be attained is x x 100 min$^{-1}$. The value of x is

Answer 1

3

Answer 2

The fraction of molecules crossing the energy barrier is given by the Boltzmann factor, $e^{-E_a/RT}$.

Given fraction at $T = 27^\circ\text{C} = 27 + 273 = 300 \text{ K}$ is $10^{-3}\%$.

Fraction $= \frac{10^{-3}}{100} = 10^{-5}$.

So, at $T = 300 \text{ K}$, $e^{-E_a/RT} = 10^{-5}$.

The Arrhenius equation relates the rate constant ($k$) to the pre-exponential factor ($A$), activation energy ($E_a$), gas constant ($R$), and absolute temperature ($T$):

$k = A e^{-E_a/RT}$

We are given the rate constant at $27^\circ\text{C}$ is $k = 3 \times 10^{-3} \text{ min}^{-1}$.

Substituting the values into the Arrhenius equation:

$3 \times 10^{-3} = A \times (e^{-E_a/RT})$

$3 \times 10^{-3} = A \times 10^{-5}$

Now, we can solve for the pre-exponential factor $A$:

$A = \frac{3 \times 10^{-3}}{10^{-5}} = 3 \times 10^{-3} \times 10^5 = 3 \times 10^{(-3+5)} = 3 \times 10^2 \text{ min}^{-1}$.

The pre-exponential factor $A$ represents the theoretical maximum value of the rate constant that can be achieved when all collisions are effective, which occurs as the temperature approaches infinity ($T \to \infty$). In this limit, $e^{-E_a/RT} \to e^0 = 1$, so $k \to A$.

Thus, the maximum value of the rate constant $K$ is equal to $A$.

$K_{\text{max}} = A = 3 \times 10^2 \text{ min}^{-1}$.

The problem states that the maximum value of the rate constant is $x \times 100 \text{ min}^{-1}$.

We have $K_{\text{max}} = 3 \times 10^2 \text{ min}^{-1} = 3 \times 100 \text{ min}^{-1}$.

Comparing $3 \times 100 \text{ min}^{-1}$ with $x \times 100 \text{ min}^{-1}$, we find $x = 3$.