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.

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

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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: …

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

TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$Q. At what temperature will the RMS velocity of sulphur dioxide molecules at 400 K be the same as the most probable velocity of oxygen molecules? (A) 600 K (B) 200 K (C) 400 K (D) 300 K$
›Reveal solutionSolution

Equating $v_{rms}(\text{SO}_2, 400\,\text{K})$ with $v_{mp}(\text{O}_2, T)$ gives $T = 300$ K.

$v_{rms} = \sqrt{\frac{3RT_1}{M_1}}, \qquad v_{mp} = \sqrt{\frac{2RT_2}{M_2}}.$
With $M(\text{SO}_2) = 64$ , $T_1 = 400$ K, and $M(\text{O}_2) = 32$ , set the two speeds equal:

$\frac{3RT_1}{M_1} = \frac{2RT_2}{M_2}$ …
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TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$^\circ$
›Reveal solutionSolution
The temperature coefficient (Q₁₀ = 2) tells us the rate constant halves for every 10 °C drop. From 25 °C to 15 °C is a 10 °C decrease, so the rate constant at 15 °C is half of $1 \times 10^{-3}$ min⁻¹, i.e., $5 \times 10^{-4}$ min⁻¹.

Concept & Intuition

The temperature coefficient (often denoted $Q_{10}$ ) is defined as the factor by which the rate constant increases when the temperature is raised by 10 °C.

Here $Q_{10} = 2$ means:

Increase temperature by 10 °C → rate constant doubles.

Decrease temperature by 10 °C → rate constant halves.

We are moving down from 25 °C to 15 °C, a drop of exactly 10 °C. So the rate constant at the lower temperature is simply $k_{15} = k_{25} / 2$ .

Step-by-step reasoning

Identify the given data

$k_{25} = 1 \times 10^{-3}$ min⁻¹

Temperature coefficient $Q_{10} = 2$

Temperature change: from 25 °C to 15 °C = −10 °C

Apply the definition of temperature coefficient

The relationship is:

$\frac{k_{T+10}}{k_T} = Q_{10}$
So for a 10 °C decrease:
$\frac{k_{15}}{k_{25}} = \frac{1}{Q_{10}} = \frac{1}{2}$

Calculate the unknown rate constant $k_{15} = k_{25} \times \frac{1}{2} = (1 \times 10^{-3}) \times \frac{1}{2} = 0.5 \times 10^{-3} = 5 \times 10^{-4} \text{ min}^{-1}$ …
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TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
$25^\circ \mathrm{C}$
›Reveal solutionSolution
The temperature coefficient (Q₁₀ = 2) tells us the rate constant halves for every 10°C drop. Going from 25°C to 15°C is a 10°C decrease, so the rate constant at 15°C is half of the given value: $5 \times 10^{-4}~\mathrm{min}^{-1}$ .

Concept & Intuition

The temperature coefficient (often denoted $Q_{10}$ ) is the factor by which the rate constant changes when the temperature is raised by 10°C. Here $Q_{10} = 2$ means that for every 10°C increase, the rate constant doubles; conversely, for every 10°C decrease, it halves. Since we are moving from 25°C down to 15°C (a drop of exactly 10°C), we simply divide the given rate constant by 2.

Step-by-step reasoning

Identify the temperature change

The given rate constant $k_{25} = 1 \times 10^{-3}~\mathrm{min}^{-1}$ is at $25^\circ\mathrm{C}$ . We need $k_{15}$ at $15^\circ\mathrm{C}$ .

The difference: $25^\circ\mathrm{C} - 15^\circ\mathrm{C} = 10^\circ\mathrm{C}$ .

Apply the definition of the temperature coefficient

The temperature coefficient $Q_{10}$ is defined as

$Q_{10} = \frac{k_{T+10}}{k_T}$
Here $Q_{10} = 2$ , so for a 10°C increase the rate constant multiplies by 2. For a 10°C decrease, we use the reciprocal:
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TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
$Q. When salt is added to water, which of the following statement is true? (A) Boiling point decreases (B) Boiling point increases (C) Boiling point remain constant (D) Freezing point increases$
›Reveal solutionSolution
Adding salt to water raises its boiling point and lowers its freezing point — a colligative effect. The correct statement is that the boiling point increases.

The question tests your understanding of colligative properties — properties that depend only on the number of solute particles, not their chemical identity. When salt (sodium chloride, NaCl) dissolves in water, it dissociates into Na⁺ and Cl⁻ ions, increasing the total number of particles in the solution. This changes two key physical properties of water: its boiling point and its freezing point.

Why does this happen? At the boiling point, the vapour pressure of the liquid equals the atmospheric pressure. Adding a non-volatile solute like salt lowers the vapour pressure of the solvent (water). To make the vapour pressure reach atmospheric pressure again, you need to supply more heat — hence the boiling point rises. For freezing, the solute particles disrupt the orderly arrangement of water molecules into ice, so a lower temperature is needed to freeze the solution — hence the freezing point drops.

Let’s examine each option step by step.

Boiling point behaviour

The boiling point elevation is given by $\Delta T_b = i \cdot K_b \cdot m$ , where $i$ is the van’t Hoff factor (for NaCl, $i \approx 2$ ), $K_b$ is the ebullioscopic constant of water, and $m$ is the molality. Since $\Delta T_b > 0$ , the boiling point increases. This eliminates option (A) and (C).

Freezing point behaviour …
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←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.

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