Q.The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is $4\times10^{10}\ \text{s}^{-1}$. Calculate $k$ at 318 K and $E_a$.

Concept: Arrhenius equation — $k = A e^{-E_a/RT}$.

Step 1 – Relate times to rate constants

For a first order reaction, $t = \frac{1}{k}\ln\frac{[A]_0}{[A]}$.

At 298 K, 10% completion means $[A] = 0.9[A]_0$, so

At 308 K, 25% completion means $[A] = 0.75[A]_0$, so

Given $t_{298} = t_{308}$, we get

Step 2 – Find ratio of rate constants

Compute: $\ln(4/3) \approx 0.28768$, $\ln(10/9) \approx 0.10536$, so

Step 3 – Use Arrhenius equation to find $E_a$

$\frac{1}{298} - \frac{1}{308} = \frac{10}{298 \times 308} = \frac{10}{91784} = 1.0895 \times 10^{-4}\ \text{K}^{-1}$.

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

$\ln(2.7304) = 1.0045 = \frac{E_a}{8.314} \times 1.0895 \times 10^{-4}.$ …

Step 1: Translate the condition. For a first order reaction, $t = \dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}$.

• 10% completion at $298\ \text{K}$ ($[A]/[A]_0 = 0.9$): $\;t = \dfrac{2.303}{k_{298}}\log\dfrac{100}{90}$
• 25% completion at $308\ \text{K}$ ($[A]/[A]_0 = 0.75$): $\;t = \dfrac{2.303}{k_{308}}\log\dfrac{100}{75}$

The two times are equal, so:

Step 2: Activation energy. Using the two-temperature Arrhenius form with $T_1 = 298\ \text{K}$, $T_2 = 308\ \text{K}$:

$E_a = \frac{0.4362 \times 19.147}{1.0895 \times 10^{-4}} = 7.665 \times 10^{4}\ \text{J mol}^{-1} \approx 76.65\ \text{kJ mol}^{-1}$ …

Method: Arrhenius Equation with Two-Point Form

We use the Arrhenius equation in its logarithmic form to relate rate constants at different temperatures, combined with first-order kinetics to connect percentage completion to $k$.

Step 1: Relate time to rate constant for first-order reaction

For a first-order reaction, the integrated rate law is:

Let initial concentration $[A]_0 = 100$ (in arbitrary units).

• At 298 K, 10% completion means $[A] = 90$:

• At 308 K, 25% completion means $[A] = 75$:

Since time $t$ is the same for both:

Step 2: Calculate the ratio

So:

Step 3: Apply two-point Arrhenius equation

The Arrhenius equation in two-point form:

Here $T_1 = 298\ \text{K}$, $T_2 = 308\ \text{K}$, and $\frac{k_{298}}{k_{308}} = 0.3667$, so $\frac{k_{308}}{k_{298}} = \frac{1}{0.3667} \approx 2.727$.

Step 4: Solve for $E_a$

$E_a \approx \frac{8.342}{1.0895 \times 10^{-4}} \approx 7.66 \times 10^4\ \text{J/mol}$ …

🧠 Common Mistake #1: Confusing fraction reacted with fraction remaining

The error:

Students often take “10% completion” to mean $[A] = 0.10\,[A]_0$.

But 10% completion means 10% has reacted — so 90% remains.

• For 10% completion: $[A] = 0.90\,[A]_0$
• For 25% completion: $[A] = 0.75\,[A]_0$

How to avoid:

Always write:

fraction remaining = $1 - \frac{\%\ \text{completed}}{100}$

🧠 Common Mistake #2: Using the wrong integrated rate equation

The error:

Using $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$ is correct — but students sometimes plug in $[A]_0/[A]$ backwards.

Correct form for first order:

For 10% completion:

For 25% completion:

How to avoid:

Always write the ratio as $\frac{\text{initial}}{\text{remaining}}$.

🧠 Common Mistake #3: Forgetting that times are equal

The error:

Students solve each $t$ separately and then don’t equate them.

Key given:

So:

Cancel $2.303$:

How to avoid:

Write the equality explicitly before substituting numbers.

🧠 Common Mistake #4: Using $\log$ instead of $\ln$ in the Arrhenius equation

The error:

The Arrhenius equation in log form is:

Students sometimes use $\ln$ but forget the $2.303$ conversion.

How to avoid:

• If using $\log_{10}$, keep $2.303$ in denominator.
• If using $\ln$, the equation is $\ln k = \ln A - \frac{E_a}{RT}$.

🧠 Common Mistake #5: Mixing up $k$ values when finding $E_a$

The error:

After finding the ratio $\frac{k_{308}}{k_{298}}$, students plug into:

But they sometimes swap $T_1$ and $T_2$ or use wrong $k$ ratio.

Correct:

From the time equality:

Then use $T_1 = 298\ \text{K}$, $T_2 = 308\ \text{K}$.

How to avoid: …