Question

Calculate rate constant of a first order reaction having pre-exponential factor $1.6\times10^{-13}\text{s}^{-1}$. ($E_{a}/2.303RT=21$)}

• A. $1.6\times10^{-13}$
• B. $3.2\times10^{-13}$
• C. $3.2\times10^{-8}$
• D. $1.6\times10^{-34}$

Hint:

When dealing with Arrhenius equations, remember that $e^{-x}$ is mathematically equivalent to $10^{-x / 2.303}$. Therefore, $e^{-E_a/RT} = 10^{-E_a/2.303RT}$. You can plug the given exponent directly onto a base 10!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We must calculate the rate constant ($k$) using the Arrhenius equation, given the pre-exponential factor ($A$) and an explicitly pre-calculated logarithmic exponent term ($E_a / 2.303RT = 21$).

Step 2: Detailed Explanation:
The standard Arrhenius equation is:
$k = A \cdot e^{-\frac{E_a}{RT}}$
This equation is often converted to base-10 to make manual calculations easier. Taking the common logarithm (base 10) of both sides yields:
$\log_{10}(k) = \log_{10}(A) - \frac{E_a}{2.303 RT}$
We are provided with the exact numerical value for the exponential term:
$\frac{E_a}{2.303 RT} = 21$
Substitute this into the logarithmic equation:
$\log_{10}(k) = \log_{10}(A) - 21$
To solve for $k$, convert this equation back into its exponential form (base 10):
$k = 10^{\log_{10}(A) - 21}$
$k = 10^{\log_{10}(A)} \times 10^{-21}$
Because $10^{\log_{10}(A)}$ simplifies identically to $A$, the equation becomes:
$k = A \times 10^{-21}$
We are given the pre-exponential factor $A = 1.6 \times 10^{-13} \text{ s}^{-1}$. Substitute this value:
$k = (1.6 \times 10^{-13}) \times 10^{-21}$
Combine the exponents using exponent rules ($10^a \times 10^b = 10^{a+b}$):
$k = 1.6 \times 10^{-13 - 21}$
$k = 1.6 \times 10^{-34} \text{ s}^{-1}$

Step 3: Final Answer:
The rate constant is $1.6\times10^{-34}$, matching option (d).