Q.The activation energy for the reaction
$2HI(g) \rightarrow H_2 + I_2(g)$
is $209.5\ \text{kJ mol}^{-1}$ at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

The key idea is the Arrhenius equation, where the fraction of molecules with energy $\ge E_a$ is given by the exponential factor $e^{-E_a/RT}$.

Step 1: Write the expression for the fraction $f$:

Step 2: Insert the given values. $E_a = 209.5 \times 10^3\ \text{J mol}^{-1}$, $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$, $T = 581\ \text{K}$.

Step 3: Compute the exponent.

Step 4: Evaluate the fraction (via base-10 logs, as NCERT does). …

The Arrhenius equation tells us that not all collisions lead to a reaction — only those molecules that possess energy equal to or greater than the activation energy $E_a$ can overcome the energy barrier. The fraction of such molecules in a gas at temperature $T$ follows the Boltzmann distribution: it is proportional to $e^{-E_a/RT}$. This exponential factor is the heart of the temperature dependence of reaction rates.

Here, we are given $E_a = 209.5\ \text{kJ mol}^{-1}$ and $T = 581\ \text{K}$. The fraction $f$ is simply:

But careful: $E_a$ is in kJ, while $R$ is usually in J. So we must convert units first.

1. Convert activation energy to J/mol

   $E_a = 209.5\ \text{kJ mol}^{-1} = 209.5 \times 10^3\ \text{J mol}^{-1}$

2. Write the gas constant

   $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$

3. Compute the exponent

First, denominator: $8.314 \times 581 = 4830.434$

Then, numerator: $209500$

So exponent = $-\frac{209500}{4830.434} = -43.371$

4. Find the fraction

Evaluating via base-10 logarithms (NCERT's route):

$\log f = -\frac{43.371}{2.303} = -18.832$ …

Method: Direct Application of the Arrhenius Exponential Factor

This method uses the exponential term from the Arrhenius equation to find the fraction of molecules with energy ≥ activation energy.

Concept (Why this works)

The Arrhenius equation is:

The factor $e^{-E_a / RT}$ represents the fraction of molecules that have energy equal to or greater than the activation energy $E_a$. This comes from the Maxwell–Boltzmann distribution of molecular energies.

Steps

Step 1: Identify the given data

• Activation energy, $E_a = 209.5\ \text{kJ mol}^{-1}$
• Temperature, $T = 581\ \text{K}$
• Gas constant, $R = 8.314\ \text{J mol}^{-1} \text{K}^{-1}$

Step 2: Convert units to match $R$

Since $R$ is in Joules, convert $E_a$ from kJ to J:

Step 3: Compute the exponent

First calculate denominator:

Then:

Step 4: Calculate the fraction …

Here are the common mistakes students make when solving this Arrhenius equation problem, along with how to avoid each.

Mistake 1: Forgetting to convert activation energy to the same units as $R$

The error:

Students plug $E_a = 209.5\ \text{kJ mol}^{-1}$ directly into the formula but use $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$. This gives a completely wrong exponent.

How to avoid:

Always check units before substituting.

• $E_a$ must be in J mol$^{-1}$ when $R$ is in J mol$^{-1}$ K$^{-1}$.
• Convert:

Mistake 2: Using the wrong formula for "fraction of molecules"

The error:

Some students try to use the full Arrhenius equation $k = A e^{-E_a/RT}$ and then solve for $k$, but the question asks only for the fraction of molecules with energy $\ge E_a$, which is not $k$.

How to avoid:

The fraction of molecules with energy $\ge E_a$ is given directly by the Boltzmann factor:

This is not the rate constant — it is the exponential part alone.

Mistake 3: Sign error in the exponent

The error:

Writing $e^{+E_a/RT}$ instead of $e^{-E_a/RT}$, or forgetting the negative sign entirely.

How to avoid:

The Arrhenius equation has a negative exponent because higher activation energy means fewer molecules have enough energy. Remember:

A quick sanity check: if $E_a$ is large, the fraction should be very small (much less than 1).

Mistake 4: Incorrect temperature unit

The error:

Using temperature in Celsius instead of Kelvin, or forgetting to convert.

How to avoid:

The problem gives $T = 581\ \text{K}$ — no conversion needed here. But always verify: if temperature is in °C, add 273.15.

Mistake 5: Calculation errors in the exponent value

The error:

Mishandling the large numbers when computing $\frac{E_a}{RT}$.

How to avoid:

Do the calculation step-by-step:

1. $E_a = 209.5 \times 10^3 = 209500\ \text{J mol}^{-1}$
2. $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$
3. $T = 581\ \text{K}$
4. Compute denominator: $RT = 8.314 \times 581$
   • $8.314 \times 500 = 4157$
   • $8.314 \times 81 = 673.434$
   • Total = $4157 + 673.434 = 4830.434$
5. Compute exponent:

6. Fraction = $e^{-43.371}$

Use a calculator carefully — this is a very small number. Evaluated via logs: $\log f = -43.371/2.303 = -18.832$, so $f = 1.471 \times 10^{-19}$.

Mistake 6: Reporting the fraction as a percentage without stating it

The error: …