The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of ac

Exercises · 3.30

Q. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Jammu Kashmir JkboseTextbookSubjective· 3mImportance★★★★★

42% · 49/117 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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

Start your 14-day free trial to unlock the full solution →

Using the Arrhenius equation in its two-point form, the activation energy $E_a$ is found to be $52.86\ \text{kJ mol}^{-1}$ when the rate quadruples ( $k_2/k_1 = 4$ ) over a 20 K rise from 293 K to 313 K.

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

k = A e^{-E_a / RT}

Here $A$ is the pre-exponential factor (frequency factor), $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature. The key idea: when temperature increases, more molecules have energy above the activation barrier, so $k$ increases exponentially.

For two different temperatures, we can eliminate $A$ (which is assumed constant) by taking a ratio:

\frac{k_2}{k_1} = \frac{A e^{-E_a / RT_2}}{A e^{-E_a / RT_1}} = e^{-\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)}

Taking natural logs gives the two-point Arrhenius equation — the workhorse for problems like this:

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

Notice the sign: $\frac{1}{T_1} - \frac{1}{T_2}$ is positive when $T_2 > T_1$ , so $\ln(k_2/k_1)$ is positive — the rate increases with temperature, as expected.

Now let’s apply it step by step.

Identify the given data.

$T_1 = 293\ \text{K}$ , $T_2 = 313\ \text{K}$ .

The rate quadruples: $k_2/k_1 = 4$ .

$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ (we’ll get $E_a$ in J/mol, then convert to kJ/mol).
Write the two-point equation with numbers.

\ln 4 = \frac{E_a}{8.314}\left(\frac{1}{293} - \frac{1}{313}\right)

Compute the temperature reciprocal difference.

\frac{1}{293} - \frac{1}{313} = \frac{313 - 293}{293 \times 313} = \frac{20}{293 \times 313}

$293 \times 313 = 293 \times (300 + 13) = 87900 + 3809 = 91709$ .

So the difference is $\frac{20}{91709} \approx 2.1808 \times 10^{-4}\ \text{K}^{-1}$ . …

Start your 14-day free trial to unlock the full solution →

Here are the most common mistakes students make when solving this Arrhenius equation problem, along with how to avoid each one.

1. Misinterpreting "Quadruples" as a Ratio

The Mistake:

Students often write $k_2 = 4$ or $k_1 = 1$ , forgetting that the rate constant ratio is what matters. They might set $\frac{k_2}{k_1} = 4$ but then plug in $4$ incorrectly into the log term.

How to Avoid:

Always define clearly:

Let $k_1$ = rate constant at $T_1 = 293 \text{ K}$
Let $k_2$ = rate constant at $T_2 = 313 \text{ K}$

Since the rate quadruples, the ratio is:

\frac{k_2}{k_1} = 4

Then, directly substitute into the Arrhenius equation:

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

So $\log_{10}(4)$ is the left-hand side — not $4$ itself.

2. Using the Wrong Logarithm Base

The Mistake:

Using natural log ( $\ln$ ) when the formula uses $\log_{10}$ , or vice versa, without adjusting the constant $2.303$ .

How to Avoid:

Memorise both forms and stick to one per problem:

If using $\log_{10}$ :

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

If using $\ln$ :

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

For this problem, $\log_{10}(4) \approx 0.6021$ is the correct value to use.

3. Forgetting to Convert Temperature to Kelvin

The Mistake:

Using temperatures in Celsius (e.g., $20^\circ C$ and $40^\circ C$ ) directly in the $\frac{1}{T}$ term.

How to Avoid:

The Arrhenius equation requires absolute temperature in Kelvin.

Here, the temperatures are already given as 293 K and 313 K — so no conversion is needed. But always double-check: if the problem says " $20^\circ C$ ", convert to $293 \text{ K}$ .

4. Incorrectly Subtracting the Reciprocal Temperatures

The Mistake:

Writing $\frac{1}{T_2} - \frac{1}{T_1}$ instead of $\frac{1}{T_1} - \frac{1}{T_2}$ , leading to a negative value for $E_a$ .

How to Avoid:

The correct formula is:

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

Since $T_2 > T_1$ , $\frac{1}{T_1} - \frac{1}{T_2}$ is positive.

If you get a negative $E_a$ , you have the subtraction reversed.

5. Using the Wrong Value for $R$

The Mistake:

Using $R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1}$ (gas constant for PV = nRT) instead of $R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}$ .

How to Avoid:

For activation energy in J/mol, always use:

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

If the answer is required in kJ/mol, divide the final result by 1000.

6. Arithmetic Errors in the Reciprocal Difference

The Mistake:

Rushing the calculation of $\frac{1}{293} - \frac{1}{313}$ and getting a wrong decimal.

How to Avoid:

Do it step-by-step: …

Start your 14-day free trial to unlock the full solution →

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

…

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →

←Previous question3.29 — The time required for 10% completion of a first order reaction at 298 …→Next section · answersAnswers to Intext Questions

Q.The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Concept understanding — Arrhenius Equation

The Arrhenius Equation Plot: Why Temperature Changes Reaction Speed

The core idea: an energy barrier

The Arrhenius equation (precise statement)

Why plot it? The linear trick

What the plot tells you

A typical Arrhenius plot looks like this

Why this formula?

Arrhenius Equation: Why It Holds

1. The Core Question

2. The Key Insight: An Energy Barrier

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

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

5. Putting It Together: The Arrhenius Equation

6. Why It Works — The Physical Logic

Method: Using the Arrhenius Equation in Logarithmic Form

Step 1: Recall the Arrhenius equation

Step 2: Write the two-point logarithmic form

Step 3: Identify the given data

Step 4: Substitute and solve for $E_a$

1. Misinterpreting "Quadruples" as a Ratio

2. Using the Wrong Logarithm Base

3. Forgetting to Convert Temperature to Kelvin

4. Incorrectly Subtracting the Reciprocal Temperatures

5. Using the Wrong Value for $R$

6. Arithmetic Errors in the Reciprocal Difference

Q.The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days