Q.The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate $E_a$.

Concept: Arrhenius Equation — the temperature dependence of the rate constant is given by

$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$.

Step 1: The rate doubles, so $\frac{k_2}{k_1} = 2$.

$T_1 = 298\ \text{K}$, $T_2 = 308\ \text{K}$, and $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$.

Step 2: Substitute into the equation:

$\log 2 = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{298} - \frac{1}{308} \right)$.

Step 3: Compute the temperature difference:

$\frac{1}{298} - \frac{1}{308} = \frac{308 - 298}{298 \times 308} = \frac{10}{91784} \approx 1.089 \times 10^{-4}$. …

The Arrhenius equation tells us how the rate constant $k$ depends on temperature:

Here, $A$ is the pre-exponential factor (frequency factor), $E_a$ is the activation energy, $R$ is the gas constant (8.314 J mol⁻¹ K⁻¹), and $T$ is the absolute temperature. The key idea: when temperature increases, the fraction of molecules with energy ≥ $E_a$ grows exponentially, so the rate constant rises.

When the problem says "the rate doubles," it means the rate constant $k$ doubles (assuming concentration terms cancel). So we have $k_2 = 2 k_1$ for $T_1 = 298$ K and $T_2 = 308$ K.

We can avoid knowing $A$ by taking a ratio. For two temperatures:

Taking natural logs gives the working form:

This is the two-point Arrhenius equation. Notice the sign: $\frac{1}{T_1} - \frac{1}{T_2}$ is positive when $T_2 > T_1$, matching the positive $E_a$.

Now plug in the numbers.

1. Identify the given data.

   $T_1 = 298$ K, $T_2 = 298 + 10 = 308$ K, $\frac{k_2}{k_1} = 2$, $R = 8.314$ J mol⁻¹ K⁻¹.

2. Compute the temperature reciprocal difference.

Find a common denominator:

Calculate $298 \times 308$: $300 \times 308 = 92400$, minus $2 \times 308 = 616$, gives $91784$. So:

3. Apply the formula.

$\ln(2) \approx 0.6931$. So:

4. Solve for $E_a$. Multiply both sides by $8.314$: …

Method: Arrhenius Equation (Two-Point Form)

This problem uses the two-point form of the Arrhenius equation, which relates rate constants at two different temperatures.

Why this method works

The Arrhenius equation tells us how reaction rate depends on temperature through activation energy. When the rate doubles, the rate constant $k$ also doubles (since concentration terms cancel). So we know $k_2 = 2k_1$.

Steps

Step 1: Write the two-point Arrhenius equation

Where:

• $k_1$ = rate constant at $T_1 = 298 \text{ K}$
• $k_2$ = rate constant at $T_2 = 308 \text{ K}$ (10 K increase)
• $R = 8.314 \text{ J mol}^{-1}\text{K}^{-1}$
• $E_a$ = activation energy (what we want)

Step 2: Substitute the known ratio

Since the rate doubles: $\frac{k_2}{k_1} = 2$

Step 3: Calculate the temperature difference term

$\frac{1}{298} - \frac{1}{308} = \frac{308 - 298}{298 \times 308} = \frac{10}{298 \times 308}$ …

Common Mistakes with the Arrhenius Equation (Rate Doubling Problem)

This is a classic numerical from chemical kinetics. Let's break down the most frequent errors students make and how to avoid each.

✗ Mistake 1: Forgetting to Convert Temperature to Kelvin

What students do:

They plug in 10°C directly as $T_2 - T_1$ without realising the problem already gives absolute temperature in Kelvin.

Why it's wrong:

The Arrhenius equation uses absolute temperature (K). Here, $T_1 = 298\ \text{K}$ and $T_2 = 308\ \text{K}$. The 10 K rise is already in Kelvin — no conversion needed.

✓ How to avoid:

Always check the unit of temperature in the question. If it says "10 K", it's already Kelvin. If it says "10°C", convert: $T(K) = T(°C) + 273$.

✗ Mistake 2: Misinterpreting "Rate Doubles"

What students do:

They write $k_2 = 2 \times k_1$ but then incorrectly set $\frac{k_2}{k_1} = \frac{1}{2}$.

Why it's wrong:

"Rate doubles" means the rate constant becomes twice — so $\frac{k_2}{k_1} = 2$, not $\frac{1}{2}$.

✓ How to avoid:

Read carefully:

• "Doubles" → multiply by 2 → $\frac{k_2}{k_1} = 2$
• "Halves" → divide by 2 → $\frac{k_2}{k_1} = \frac{1}{2}$

✗ Mistake 3: Using the Wrong Form of the Arrhenius Equation

What students do:

They use $\ln k = \ln A - \frac{E_a}{RT}$ for two different temperatures but forget to subtract the two equations.

Why it's wrong:

That single-equation form doesn't directly give $E_a$ unless you know $A$. You need the two-point form:

✓ How to avoid:

Memorise the two-point form explicitly. When given two temperatures and the ratio of rate constants, this is the only direct route.

✗ Mistake 4: Sign Error in the Temperature Term

What students do:

They write $\frac{1}{T_2} - \frac{1}{T_1}$ instead of $\frac{1}{T_1} - \frac{1}{T_2}$.

Why it's wrong:

Since $T_2 > T_1$, $\frac{1}{T_1} - \frac{1}{T_2}$ is positive. The correct formula is:

✓ How to avoid:

Always write the smaller temperature first inside the bracket. Or remember: "Start with the initial temperature minus the final temperature".

✗ Mistake 5: Using $\log_{10}$ Instead of $\ln$

What students do:

They use $\log_{10}$ but forget the conversion factor $2.303$.

Why it's wrong:

The standard Arrhenius equation uses natural log ($\ln$). If you use $\log_{10}$, you must multiply by $2.303$:

✓ How to avoid: …