The rate constant for the first order decomposition of $H_2O_2$ is given by the following equation: $\log k = 14.34 - 1.25\times10^4\ K/T$ Calculate $E_a$ for this reaction and at what temperature will its half-period be 256 minutes?

The rate constant for the first order decomposition of H₂O₂ is given by the following equation: log k = 14.34

Exercises · 3.27

H_2O_2

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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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The Arrhenius equation in logarithmic form gives $E_a$ directly from the slope, and the half-life for a first-order reaction links $k$ to $t_{1/2}$ . Here $E_a = 239.3\ \text{kJ mol}^{-1}$ and the required temperature is $T = 669\ \text{K}$ .

The problem gives a linear equation for $\log k$ versus $1/T$ . That’s the Arrhenius equation in disguise. The Arrhenius equation is:

k = A e^{-E_a / RT}

Taking common logarithms (base 10) on both sides:

\log k = \log A - \frac{E_a}{2.303 R} \cdot \frac{1}{T}

This is of the form $\log k = C - m \cdot \frac{1}{T}$ , where the slope $m = \frac{E_a}{2.303 R}$ . The given equation is:

\log k = 14.34 - 1.25 \times 10^4 \ \frac{1}{T}

So the slope is $1.25 \times 10^4\ \text{K}$ . That’s the key to finding $E_a$ .

E_a = 2.303 \times R \times \text{slope}

Now, the second part: half-life for a first-order reaction is $t_{1/2} = \frac{0.693}{k}$ . Given $t_{1/2} = 256$ minutes, we can find $k$ , then use the equation to find $T$ .

Let’s work it through.

Calculate $E_a$ Slope $= 1.25 \times 10^4\ \text{K}$ , $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ .

E_a = 2.303 \times 8.314 \times 1.25 \times 10^4

First, $2.303 \times 8.314 = 19.147$ (approximately).

Then $19.147 \times 1.25 \times 10^4 = 23.93375 \times 10^4 = 2.393375 \times 10^5\ \text{J mol}^{-1}$ .

Convert to kJ: $E_a = 239.3\ \text{kJ mol}^{-1}$ .

Tip

A quick check: $2.303 \times 8.314 \approx 19.15$ , times $1.25 \times 10^4$ gives $2.394 \times 10^5$ J — always round sensibly for exam answers.

Find $k$ from half-life

For first order: $t_{1/2} = \frac{0.693}{k}$ .

Given $t_{1/2} = 256$ minutes. Convert to seconds? The rate constant $k$ has units of time $^{-1}$ . The Arrhenius equation uses consistent units — here $k$ is in min $^{-1}$ if we keep time in minutes. But the given $\log k$ equation uses $k$ in what units? The constant 14.34 suggests $k$ is in s $^{-1}$ (since typical pre-exponential factors for such reactions are around $10^{14}\ \text{s}^{-1}$ ). Let’s check: if $k$ were in min $^{-1}$ , $\log A$ would be different. The value 14.34 is typical for $A$ in s $^{-1}$ . So we must convert half-life to seconds.

$256\ \text{min} = 256 \times 60 = 15360\ \text{s}$ .

Then $k = \frac{0.693}{15360} = 4.511 \times 10^{-5}\ \text{s}^{-1}$ . …

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Method: Arrhenius Equation Analysis (Two-Part Problem)

This problem uses the Arrhenius equation in logarithmic form to extract activation energy and then solve for temperature given half-life.

Part 1 — Finding Activation Energy ( $E_a$ )

Step 1: Recall the logarithmic Arrhenius form

The standard equation is:

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

where:

$k$ = rate constant
$A$ = pre-exponential factor
$E_a$ = activation energy (J/mol)
$R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$
$T$ = temperature (K)

Step 2: Compare with the given equation

Given:

\log k = 14.34 - 1.25 \times 10^4 \ \frac{\text{K}}{T}

By comparing coefficients:

\frac{E_a}{2.303 \, R} = 1.25 \times 10^4 \ \text{K}

Step 3: Solve for $E_a$

E_a = 1.25 \times 10^4 \times 2.303 \times 8.314

E_a = 1.25 \times 10^4 \times 19.147

\boxed{E_a = 2.39 \times 10^5 \ \text{J/mol} = 239 \ \text{kJ/mol}}

Part 2 — Finding Temperature for Given Half-Life

Step 1: Relate half-life to rate constant

For a first-order reaction:

t_{1/2} = \frac{0.693}{k}

Given $t_{1/2} = 256$ minutes, convert to seconds (since $E_a$ is in J/mol using SI units):

t_{1/2} = 256 \times 60 = 15360 \ \text{s}

Step 2: Calculate $k$

k = \frac{0.693}{15360} = 4.51 \times 10^{-5} \ \text{s}^{-1}

Step 3: Use the given equation to find $T$

\log k = 14.34 - \frac{1.25 \times 10^4}{T}

Substitute $\log k$ :

\log(4.51 \times 10^{-5}) = 14.34 - \frac{1.25 \times 10^4}{T}

Step 4: Compute $\log k$ …

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Common Mistakes with the Arrhenius Equation (H₂O₂ Decomposition Problem)

Students often slip on this problem because it blends logarithmic manipulation, unit conversion, and conceptual clarity. Here are the most frequent errors and how to avoid each.

1. ✗ Misidentifying the Form of the Arrhenius Equation

Mistake:

Students try to match the given equation directly to $\log k = \log A - \frac{E_a}{2.303RT}$ but forget the sign convention.

How to avoid:

Always write the standard form first:

\log k = \log A - \frac{E_a}{2.303R} \cdot \frac{1}{T}

Compare term-by-term with the given:

\log k = 14.34 - 1.25 \times 10^4 \cdot \frac{1}{T}

Here, $1.25 \times 10^4$ is $\frac{E_a}{2.303R}$ , not $E_a$ itself.

2. ✗ Forgetting the Gas Constant Units

Mistake:

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

How to avoid:

Always check:

If $R$ is in J, $E_a$ comes out in J/mol → divide by 1000 for kJ/mol.
If $R$ is in kJ (0.008314), $E_a$ is directly in kJ/mol.

Correct calculation:

\frac{E_a}{2.303 \times 8.314} = 1.25 \times 10^4

E_a = 1.25 \times 10^4 \times 2.303 \times 8.314

E_a = 2.393 \times 10^5\ \text{J/mol} = \mathbf{239.3\ kJ/mol}

3. ✗ Leaving $k$ in min⁻¹ When the Given Equation Implies s⁻¹

Mistake:

The half-life is given as 256 minutes, so students compute $k = \frac{0.693}{256} = 2.707 \times 10^{-3}\ \text{min}^{-1}$ , take $\log k = -2.567$ , and solve

-2.567 = 14.34 - \frac{1.25 \times 10^4}{T} \implies T \approx 739\ \text{K} \quad \text{(wrong)}

Why it's wrong: The given equation $\log k = 14.34 - 1.25\times10^4/T$ fixes the unit of $k$ : an intercept of $\log A = 14.34$ ( $A \approx 10^{14}$ ) is characteristic of $A$ in s⁻¹ — the standard SI treatment for this reaction. Plugging a min⁻¹ value of $k$ into an equation calibrated for s⁻¹ mixes units and lands 70 K too high.

Correct approach: Convert the half-life to seconds first:

t_{1/2} = 256 \times 60 = 15360\ \text{s}, \qquad k = \frac{0.693}{15360} = 4.511 \times 10^{-5}\ \text{s}^{-1}

4. ✗ Sign Errors When Solving for Temperature

Mistake:

Plugging $\log k$ into $\log k = 14.34 - 1.25\times10^4/T$ without rearranging correctly.

How to avoid:

Write the equation clearly (using the correct $\log k = \log(4.511 \times 10^{-5}) = -4.3457$ ):

-4.3457 = 14.34 - \frac{1.25 \times 10^4}{T}

Then isolate $T$ step-by-step:

\frac{1.25 \times 10^4}{T} = 14.34 + 4.3457 = 18.686

T = \frac{1.25 \times 10^4}{18.686} \approx \mathbf{669\ K}

5. ✗ Forgetting to Convert Temperature to Kelvin

Mistake: …

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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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