Understanding the Problem

We are dealing with a first-order reaction: A → B + C. At 27°C, 3.8 × 10^–16% of the reactant molecules are in the activated state. We need to find the activation energy (E_a).

Key Concept: Activated Complex Theory

According to the Arrhenius equation and transition state theory, the fraction of molecules in the activated state is related to the activation energy. For a first-order reaction, the rate constant k is given by:

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

where A is the pre-exponential factor, E_a is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

The fraction of molecules in the activated state (f) is proportional to the Boltzmann factor:

$f\propto e^{-\frac{E_a}{RT}}$

Thus, we can write:

$f=C·e^{-\frac{E_a}{RT}}$

where C is a constant. For the purpose of finding E_a, we can compare the given fraction to a reference, but a more direct approach is to use the formula:

$f=\frac{N_{\text{activated}}}{N_{\text{total}}}=e^{-\frac{E_a}{RT}}$

This is because the probability of a molecule having energy equal to or greater than E_a is given by the Boltzmann factor.

Step-by-Step Solution

Step 1: Convert Temperature to Kelvin

Given temperature T = 27°C.

$T=27+273=300\text{K}$

Step 2: Express the Given Fraction in Decimal Form

Given: 3.8 × 10^–16% of molecules are activated.

First, convert percentage to decimal:

$f=\frac{3.8\times {10}^{-16}}{100}=3.8\times {10}^{-18}$

Step 3: Set Up the Equation

Using the formula:

$f=e^{-\frac{E_a}{RT}}$

So,

$3.8\times {10}^{-18}=e^{-\frac{E_a}{RT}}$

Step 4: Take Natural Logarithm on Both Sides

$\ln (3.8\times {10}^{-18})=-\frac{E_a}{RT}$

Calculate the left side:

$\ln (3.8\times {10}^{-18})=\ln \left(3.8\right)+\ln \left({10}^{-18}\right)=1.335-41.4465=-40.1115$

(Note: ln(10^–18) = –18 * ln(10) = –18 * 2.302585 = –41.44653; ln(3.8) ≈ 1.335; so sum ≈ –40.1115)

Step 5: Solve for E_a

So,

$-\frac{E_a}{RT}=-40.1115$

Therefore,

$\frac{E_a}{RT}=40.1115$

$E_a=40.1115\times R\times T$

R = 8.314 J mol^–1 K^–1, T = 300 K

$E_a=40.1115\times 8.314\times 300$

First, calculate 8.314 × 300 = 2494.2

Then, 40.1115 × 2494.2 ≈ 100,000 J/mol = 100 kJ/mol

(More precisely: 40.1115 * 2494.2 = 100,000. (approx))

Final Answer: 100 kJ/mol

Related Topics

• Arrhenius Equation
• Transition State Theory
• Boltzmann Distribution
• Chemical Kinetics
• Activation Energy

Important Formulae

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

Fraction of Activated Molecules: $f=e^{-\frac{E_a}{RT}}$

Gas Constant: R = 8.314 J mol^–1 K^–1