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 $E_a$.

The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25%

Exercises · 3.29

4\times10^{10}\ \text{s}^{-1}

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Concept understanding — Arrhenius Equation

The Arrhenius Equation Plot: Why Temperature Changes Reaction Speed

You already know that heating things up makes reactions go faster. A cold chai takes forever to dissolve sugar; hot chai does it in seconds. But how much faster? And is there a pattern that holds for every reaction?

That pattern is the Arrhenius equation, and plotting it in a clever way reveals something fundamental about how molecules need to collide to react.

The core idea: an energy barrier

Imagine a ball sitting in a valley. To get to the next valley, it must first be pushed up over a hill. That hill is the activation energy ( $E_a$ ) — the minimum energy two molecules need to have when they collide, for the reaction to happen.

At a low temperature, most molecules move slowly. Only a tiny fraction have enough energy to climb that hill. Raise the temperature, and suddenly many more molecules have the required energy. The fraction of molecules with energy $\ge E_a$ is given by the Boltzmann distribution:

\text{fraction} = e^{-E_a / RT}

where $R$ is the gas constant and $T$ is the absolute temperature (in Kelvin). This exponential is the heart of the story.

The Arrhenius equation (precise statement)

The rate constant $k$ of a reaction depends on temperature as:

k = A \, e^{-E_a / RT}

$k$ = rate constant (how fast the reaction proceeds)
$A$ = pre-exponential factor (frequency of collisions, times a steric factor — how often molecules hit in the right orientation)
$E_a$ = activation energy (J/mol or kJ/mol)
$R$ = 8.314 J/(mol·K)
$T$ = temperature in Kelvin

Important

$k$ is not the reaction rate itself — it's the proportionality constant in the rate law. But for a fixed concentration, a larger $k$ means a faster reaction.

Why plot it? The linear trick

The equation $k = A e^{-E_a / RT}$ is exponential in $1/T$ . That's hard to eyeball. But take the natural logarithm of both sides:

\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}

This is the equation of a straight line:

y = c + m x

where:

$y = \ln k$
$x = 1/T$
slope $m = -E_a / R$
intercept $c = \ln A$

So if you measure $k$ at several temperatures and plot $\ln k$ versus $1/T$ , you get a straight line — provided the reaction follows Arrhenius behaviour (most do, over moderate temperature ranges).

Tip

Always use Kelvin for $T$ . Celsius will give you a curved mess because $1/T$ is not linear in Celsius.

What the plot tells you

From the slope, you get $E_a$ :

E_a = - (\text{slope}) \times R

A steep negative slope means a large $E_a$ — the reaction is very sensitive to temperature. A shallow slope means a small $E_a$ — temperature doesn't affect it much.

From the intercept, you get $A$ :

A = e^{\text{intercept}}

This tells you about the collision frequency and orientation factor. A high $A$ means molecules are colliding often and in the right geometry.

A typical Arrhenius plot looks like this

$T$ (K)	$k$ (s⁻¹)	$1/T$ (K⁻¹)	$\ln k$
300	0.0012	0.00333	-6.72
310	0.0028	0.00323	-5.88
320	0.0061	0.00313	-5.10
330	0.0125	0.00303	-4.38

Plot $\ln k$ (y-axis) vs $1/T$ (x-axis). The points fall on a straight line. Draw the best-fit line, measure its slope, and compute $E_a$ . …

Why this formula?

Arrhenius Equation: Why It Holds

The Arrhenius equation is not a guess — it emerges from a deep physical picture of how molecules react. Let's build that picture step by step.

1. The Core Question

Why does reaction rate increase dramatically with temperature?

For many reactions, a 10 °C rise can double or triple the rate. This cannot be explained by simple kinetic energy arguments alone — the relationship is exponential.

2. The Key Insight: An Energy Barrier

Before molecules can react, they must collide — but not all collisions lead to products.

There is a minimum energy threshold called activation energy ( $E_a$ ). Only collisions with energy ≥  $E_a$ can break old bonds and form new ones.

Think of it like pushing a boulder over a hill:

The hilltop is the transition state (activated complex).
$E_a$ is the height of that hill from the reactant valley.

3. The Boltzmann Factor — The "Why" of Exponential Dependence

At temperature $T$ , the fraction of molecules with energy ≥  $E_a$ is given by the Boltzmann distribution:

\text{Fraction} = e^{-E_a / (RT)}

where:

$R$ = universal gas constant (8.314 J mol⁻¹ K⁻¹)
$T$ = absolute temperature (Kelvin)

Why this form?

The Boltzmann distribution tells us that the probability of a molecule having energy $E$ is proportional to $e^{-E/(k_B T)}$ .
For 1 mole of molecules, $k_B$ (Boltzmann constant) becomes $R$ via $R = N_A k_B$ .
So the fraction with energy ≥ $E_a$ is the integral of that distribution from $E_a$ to ∞, which yields $e^{-E_a/(RT)}$ .

Key takeaway: This exponential factor is not arbitrary — it comes directly from statistical mechanics.

4. The Pre-Exponential Factor ( $A$ )

Even if a collision has enough energy, it must also:

Have the correct orientation (steric factor)
Occur with sufficient collision frequency

These are bundled into the pre-exponential factor $A$ (also called the frequency factor):

A = \text{(collision frequency)} \times \text{(orientation factor)}

For simple gas-phase reactions, collision frequency can be calculated from kinetic molecular theory — it's on the order of $10^{10}$  L mol⁻¹ s⁻¹.

5. Putting It Together: The Arrhenius Equation

The rate constant $k$ is proportional to:

The number of effective collisions per second (given by $A$ )
The fraction of collisions with sufficient energy (given by $e^{-E_a/(RT)}$ )

Thus:

\boxed{k = A \, e^{-E_a / (RT)}}

This is the Arrhenius equation.

6. Why It Works — The Physical Logic

Component	Physical meaning	Why it's there
$A$	Maximum possible rate if every collision worked	Accounts for collision frequency & geometry

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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

t_{298} = \frac{1}{k_{298}}\ln\frac{1}{0.9} = \frac{1}{k_{298}}\ln\frac{10}{9}.

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

t_{308} = \frac{1}{k_{308}}\ln\frac{1}{0.75} = \frac{1}{k_{308}}\ln\frac{4}{3}.

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

\frac{1}{k_{298}}\ln\frac{10}{9} = \frac{1}{k_{308}}\ln\frac{4}{3}.

Step 2 – Find ratio of rate constants

\frac{k_{308}}{k_{298}} = \frac{\ln(4/3)}{\ln(10/9)}.

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

\frac{k_{308}}{k_{298}} \approx 2.7304.

Step 3 – Use Arrhenius equation to find $E_a$

\ln\frac{k_{308}}{k_{298}} = \frac{E_a}{R}\left(\frac{1}{298} - \frac{1}{308}\right).

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

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

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

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The equal-time condition gives $k_{308}/k_{298} = 2.73$ ; the Arrhenius equation then yields $E_a \approx 76.65\ \text{kJ mol}^{-1}$ , and with $A = 4 \times 10^{10}\ \text{s}^{-1}$ the rate constant at $318\ \text{K}$ is $k_{318} \approx 1.03 \times 10^{-2}\ \text{s}^{-1}$ .

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:

\frac{1}{k_{298}}\log\frac{10}{9} = \frac{1}{k_{308}}\log\frac{4}{3} \implies \frac{k_{308}}{k_{298}} = \frac{\log(4/3)}{\log(10/9)} = \frac{0.1249}{0.0458} = 2.73

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

\log\frac{k_{308}}{k_{298}} = \frac{E_a}{2.303\,R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)

\log 2.73 = \frac{E_a}{2.303 \times 8.314}\left(\frac{10}{298 \times 308}\right)

0.4362 = \frac{E_a}{19.147}\times 1.0895 \times 10^{-4}

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

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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:

k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}

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

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

k_{298} = \frac{2.303}{t} \log \frac{100}{90} = \frac{2.303}{t} \log(1.111)

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

k_{308} = \frac{2.303}{t} \log \frac{100}{75} = \frac{2.303}{t} \log(1.333)

Since time $t$ is the same for both:

\frac{k_{298}}{k_{308}} = \frac{\log(1.111)}{\log(1.333)}

Step 2: Calculate the ratio

\log(1.111) \approx 0.0458, \quad \log(1.333) \approx 0.1249

\frac{k_{298}}{k_{308}} = \frac{0.0458}{0.1249} \approx 0.3667

So:

k_{298} = 0.3667 \, k_{308}

Step 3: Apply two-point Arrhenius equation

The Arrhenius equation in two-point form:

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

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

\log(2.727) = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{298} - \frac{1}{308} \right)

Step 4: Solve for $E_a$

\log(2.727) \approx 0.4357

\frac{1}{298} - \frac{1}{308} = \frac{308 - 298}{298 \times 308} = \frac{10}{91784} = 1.0895 \times 10^{-4}

0.4357 = \frac{E_a}{2.303 \times 8.314} \times 1.0895 \times 10^{-4}

E_a = \frac{0.4357 \times 2.303 \times 8.314}{1.0895 \times 10^{-4}}

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

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🧠 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:

t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}

For 10% completion:

t_{10\%} = \frac{2.303}{k_{298}} \log \frac{1}{0.90}

For 25% completion:

t_{25\%} = \frac{2.303}{k_{308}} \log \frac{1}{0.75}

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:

t_{10\%\ \text{at 298 K}} = t_{25\%\ \text{at 308 K}}

So:

\frac{2.303}{k_{298}} \log\frac{1}{0.90} = \frac{2.303}{k_{308}} \log\frac{1}{0.75}

Cancel $2.303$ :

\frac{\log(1/0.90)}{k_{298}} = \frac{\log(1/0.75)}{k_{308}}

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:

\log k = \log A - \frac{E_a}{2.303\,R\,T}

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:

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

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

Correct:

From the time equality:

\frac{k_{308}}{k_{298}} = \frac{\log(1/0.75)}{\log(1/0.90)}

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

How to avoid: …

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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 4×1010 s−14\times10^{10}\ \text{s}^{-1}4×1010 s−1. Calculate kkk at 318 K and EaE_aEa​.

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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 $E_a$ .

Unlock everything free for 14 days