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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

JKBOSE Class 12 Annual Regular Examination 2024Set SZ2 marks
$Q. Activation energy of a reaction is zero. Will the rate constant of the reaction depends on temperature ? Give reason.$
›Reveal solutionSolution

With Ea = 0, the exponential (Boltzmann) factor in the Arrhenius equation becomes 1 at every temperature, so k = A is temperature-independent.

The Arrhenius equation relates the rate constant k to temperature T and activation energy Ea:

k = A·e^(−Ea/RT)

where A is the pre-exponential (frequency) factor, R is the gas constant, and T is the absolute temperature.

If activation energy Ea = 0, then the exponent −Ea/RT = 0 for any value of T, so:

k = A·e⁰ = A × 1 = A

…

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JKBOSE Class 12 Annual Regular Examination 2018Set SZ2 marks
$Q. Define Activation Energy.$
›Reveal solutionSolution

Not every molecular collision leads to a reaction — only 'sufficiently energetic' collisions do, and activation energy is exactly the extra energy needed to make a collision effective.

According to collision theory, reactant molecules must collide with each other for a reaction to occur, but not every collision is successful. Only those collisions in which the colliding molecules possess energy equal to or greater than a certain threshold (and the correct orientation) result in the formation of products. The activation energy, Ea, is defined as the minimum extra energy (above the average energy of the reactants) that the reacting molecules must acquire so that their collision can lead to the formation of an unstable, high-energy intermediate called the activated complex (or transition state), which then breaks down into the products.

…

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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 $E_a$ .

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