The decomposition of A into product has value of $k$ as $4.5\times10^{3}\ \text{s}^{-1}$ at 10°C and energy of activation $60\ \text{kJ mol}^{-1}$. At what temperature would $k$ be $1.5\times10^{4}\ \text{s}^{-1}$?

The decomposition of A into product has value of k as 4.5×10³s⁻¹ at 10°C and energy of activation 60kJ mol⁻¹.

Exercises · 3.28

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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 temperature at which the rate constant rises to $1.5 \times 10^{4}\ \text{s}^{-1}$ is approximately $297\ \text{K}$ (or $24^\circ\text{C}$ ).

The key to this problem is the Arrhenius equation, which tells us how the rate constant $k$ depends on temperature and activation energy. The equation is:

k = A e^{-E_a / RT}

where $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature. When we have two different temperatures and their corresponding rate constants, we can eliminate $A$ by taking a ratio. This gives the very useful two-point form:

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

This is the formula we'll use. Notice the order: $k_2$ is the rate constant at the unknown temperature $T_2$ , and $k_1$ is the known rate constant at $T_1$ . The activation energy must be in the same energy units as $R$ , so we'll use $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ and convert $E_a$ from kJ to J.

Let's work through it step by step.

Convert all data to consistent units.

The activation energy is $60\ \text{kJ mol}^{-1} = 60\,000\ \text{J mol}^{-1}$ .

The first temperature is $10^\circ\text{C} = 283\ \text{K}$ (since $T(\text{K}) = T(^\circ\text{C}) + 273$ ).

The first rate constant is $k_1 = 4.5 \times 10^{3}\ \text{s}^{-1}$ .

The second rate constant is $k_2 = 1.5 \times 10^{4}\ \text{s}^{-1}$ .

We need to find $T_2$ .
Write the two-point Arrhenius equation with the known values.

\ln\left(\frac{1.5 \times 10^{4}}{4.5 \times 10^{3}}\right) = \frac{60\,000}{8.314}\left(\frac{1}{283} - \frac{1}{T_2}\right)

Simplify the left-hand side.

\frac{1.5 \times 10^{4}}{4.5 \times 10^{3}} = \frac{15}{4.5} = \frac{10}{3} \approx 3.333

So $\ln(3.333) \approx 1.204$ (using a calculator or knowing $\ln(10/3) = \ln 10 - \ln 3 \approx 2.303 - 1.099 = 1.204$ ).

Simplify the right-hand side constant.

\frac{E_a}{R} = \frac{60\,000}{8.314} \approx 7217\ \text{K}

So the equation becomes:

1.204 = 7217 \left(\frac{1}{283} - \frac{1}{T_2}\right)

Solve for the bracket. …

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

1. Forgetting to Convert Celsius to Kelvin

The Mistake:

Plugging $T_1 = 10^\circ\text{C}$ directly into the Arrhenius equation.

The equation uses absolute temperature (Kelvin), not Celsius.

How to Avoid:

Always convert:

T_1 = 10 + 273 = 283\ \text{K}

Write this step explicitly before substituting.

2. Using the Wrong Form of the Arrhenius Equation

The Mistake:

Using $k = A e^{-E_a/RT}$ directly without realising you need the two-point form to compare two temperatures.

How to Avoid:

When you have $k$ at two different $T$ , always use:

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

This avoids needing the pre-exponential factor $A$ .

3. Mixing Up $k_1$ and $k_2$ (or $T_1$ and $T_2$ )

The Mistake:

Assigning $k_1 = 1.5 \times 10^4$ and $k_2 = 4.5 \times 10^3$ , then getting a negative or illogical temperature.

How to Avoid:

Label clearly:

$k_1 = 4.5 \times 10^3\ \text{s}^{-1}$ at $T_1 = 283\ \text{K}$
$k_2 = 1.5 \times 10^4\ \text{s}^{-1}$ at $T_2 = ?$

Since $k_2 > k_1$ , $T_2$ must be greater than $T_1$ . Check your final answer for sense.

4. Unit Errors in $E_a$ and $R$

The Mistake:

Using $E_a = 60\ \text{kJ mol}^{-1}$ but $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ without converting.

How to Avoid:

Convert $E_a$ to J/mol:

E_a = 60 \times 10^3 = 60\,000\ \text{J mol}^{-1}

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

5. Arithmetic Errors in the Fraction $\frac{1}{T_1} - \frac{1}{T_2}$

The Mistake:

Computing $\frac{1}{283} - \frac{1}{T_2}$ incorrectly, or rounding too early.

How to Avoid:

Keep at least 4–5 decimal places during calculation.

Example: …

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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 question3.27 — The rate constant for the first order decomposition of H₂O₂ is given b…→Next question3.29 — The time required for 10% completion of a first order reaction at 298 …

Q.The decomposition of A into product has value of $k$ as $4.5\times10^{3}\ \text{s}^{-1}$ at 10°C and energy of activation $60\ \text{kJ mol}^{-1}$ . At what temperature would $k$ be $1.5\times10^{4}\ \text{s}^{-1}$ ?

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: Two-Point Form of the Arrhenius Equation

Steps

1. Forgetting to Convert Celsius to Kelvin

2. Using the Wrong Form of the Arrhenius Equation

3. Mixing Up $k_1$ and $k_2$ (or $T_1$ and $T_2$ )

4. Unit Errors in $E_a$ and $R$

5. Arithmetic Errors in the Fraction $\frac{1}{T_1} - \frac{1}{T_2}$

Q.The decomposition of A into product has value of kkk as 4.5×103 s−14.5\times10^{3}\ \text{s}^{-1}4.5×103 s−1 at 10°C and energy of activation 60 kJ mol−160\ \text{kJ mol}^{-1}60 kJ mol−1. At what temperature would kkk be 1.5×104 s−11.5\times10^{4}\ \text{s}^{-1}1.5×104 s−1?

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Q.The decomposition of A into product has value of $k$ as $4.5\times10^{3}\ \text{s}^{-1}$ at 10°C and energy of activation $60\ \text{kJ mol}^{-1}$ . At what temperature would $k$ be $1.5\times10^{4}\ \text{s}^{-1}$ ?

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