Q.The first order rate constant for the decomposition of ethyl iodide by the reaction
$C_2H_5I(g) \rightarrow C_2H_4(g) + HI(g)$
at 600 K is $1.60\times10^{-5}\ \text{s}^{-1}$. Its energy of activation is $209\ \text{kJ/mol}$. Calculate the rate constant of the reaction at 700 K.

The key idea is the Arrhenius equation in its two-point logarithmic form, which relates rate constants at two temperatures through the activation energy.

Step 1: Write the two-point Arrhenius equation

Step 2: Insert the known values

$k_1 = 1.60 \times 10^{-5}\ \text{s}^{-1}$, $T_1 = 600\ \text{K}$, $T_2 = 700\ \text{K}$, $E_a = 209\ \text{kJ/mol} = 209000\ \text{J/mol}$, $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$.

Step 3: Simplify the calculation (keep full precision — rounding early here shifts the final digit)

Working with natural logarithms is equivalent and keeps the arithmetic clean: …

Equation

Data: $k_1 = 1.60\times10^{-5}\ \text{s}^{-1}$, $T_1 = 600\ \text{K}$, $T_2 = 700\ \text{K}$, $E_a = 209\,000\ \text{J mol}^{-1}$, $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$.

Step 1 — Temperature term.

Step 2 — Substitute.

$\log\frac{k_2}{k_1} = \frac{209000}{2.303 \times 8.314}\,(2.381\times10^{-4}) = (10915)(2.381\times10^{-4}) = 2.599$ …

Method: Two-Point Form of the Arrhenius Equation

This method is used when you know the rate constant at one temperature and the activation energy, and need to find the rate constant at another temperature.

Steps

Step 1: Write the two-point Arrhenius equation

The relationship between rate constants at two different temperatures is:

Where:

• $k_1$ = rate constant at temperature $T_1$
• $k_2$ = rate constant at temperature $T_2$
• $E_a$ = activation energy (in J/mol)
• $R$ = gas constant = $8.314\ \text{J mol}^{-1}\text{K}^{-1}$

Step 2: Identify the given values

• $k_1 = 1.60 \times 10^{-5}\ \text{s}^{-1}$ at $T_1 = 600\ \text{K}$
• $T_2 = 700\ \text{K}$
• $E_a = 209\ \text{kJ/mol} = 209 \times 10^3\ \text{J/mol}$

Step 3: Substitute into the equation

Step 4: Simplify the temperature difference

Step 5: Calculate the right-hand side (carry one extra digit — the final answer is sensitive to rounding here)

$\frac{209 \times 10^3}{8.314} \times \frac{1}{4200}$ …

Common Mistakes with the Arrhenius Equation Problem

Mistake 1: Forgetting to Convert Units of Activation Energy

The error: Students plug $E_a = 209\ \text{kJ/mol}$ directly into the Arrhenius equation without converting to J/mol.

Why it fails: The gas constant $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ is in joules, not kilojoules. Mixing kJ and J gives an answer off by a factor of 1000.

How to avoid: Always convert $E_a$ to J/mol before substituting:

Mistake 2: Using the Wrong Form of the Arrhenius Equation

The error: Using the exponential form $k = A e^{-E_a/RT}$ when $A$ is unknown, leading to a dead end.

Why it fails: You don't have the pre-exponential factor $A$, so you cannot directly compute $k$ at 700 K.

How to avoid: Use the two-point form that eliminates $A$:

Mistake 3: Swapping $T_1$ and $T_2$ in the Subtraction

The error: Writing $\frac{1}{T_2} - \frac{1}{T_1}$ instead of $\frac{1}{T_1} - \frac{1}{T_2}$.

Why it fails: Since $T_2 > T_1$, the correct expression $\frac{1}{T_1} - \frac{1}{T_2}$ is positive. Reversing gives a negative value, leading to $k_2 < k_1$ — which is physically wrong (rate constants increase with temperature).

How to avoid: Remember: higher temperature → larger rate constant. So $\ln(k_2/k_1) > 0$. Always write:

Mistake 4: Arithmetic Errors with Scientific Notation

The error: Mishandling $k_1 = 1.60 \times 10^{-5}$ during multiplication or division.

Why it fails: When solving for $k_2$, you compute $k_2 = k_1 \times e^{\text{stuff}}$. Errors in exponent arithmetic (e.g., $10^{-5} \times 10^3 = 10^{-2}$ vs $10^{-8}$) are common. …