Q.The activation energy for the reaction
2HI(g)→H2+I2(g)
is 209.5 kJ mol−1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
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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 (Ea) — 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 ≥Ea is given by the Boltzmann distribution:
fraction=e−Ea/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=Ae−Ea/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)
- Ea = activation energy (J/mol or kJ/mol)
- R = 8.314 J/(mol·K)
- T = temperature in Kelvin
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=Ae−Ea/RT is exponential in 1/T. That's hard to eyeball. But take the natural logarithm of both sides:
lnk=lnA−REa⋅T1
This is the equation of a straight line:
y=c+mx
where:
- y=lnk
- x=1/T
- slope m=−Ea/R
- intercept c=lnA
So if you measure k at several temperatures and plot lnk versus 1/T, you get a straight line — provided the reaction follows Arrhenius behaviour (most do, over moderate temperature ranges).
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 Ea:
Ea=−(slope)×R
A steep negative slope means a large Ea — the reaction is very sensitive to temperature. A shallow slope means a small Ea — temperature doesn't affect it much.
From the intercept, you get A:
A=eintercept
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) | lnk |
|---|---|---|---|
| 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 lnk (y-axis) vs 1/T (x-axis). The points fall on a straight line. Draw the best-fit line, measure its slope, and compute Ea. …
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 (Ea). Only collisions with energy ≥ Ea 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).
- Ea 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 ≥ Ea is given by the Boltzmann distribution:
Fraction=e−Ea/(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/(kBT).
- For 1 mole of molecules, kB (Boltzmann constant) becomes R via R=NAkB.
- So the fraction with energy ≥ Ea is the integral of that distribution from Ea to ∞, which yields e−Ea/(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=(collision frequency)×(orientation factor)
For simple gas-phase reactions, collision frequency can be calculated from kinetic molecular theory — it's on the order of 1010 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−Ea/(RT))
Thus:
k=Ae−Ea/(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 |
The key idea is the Arrhenius equation, where the fraction of molecules with energy ≥Ea is given by the exponential factor e−Ea/RT.
Step 1: Write the expression for the fraction f:
f=e−Ea/RT
Step 2: Insert the given values. Ea=209.5×103 J mol−1, R=8.314 J mol−1K−1, T=581 K.
RTEa=8.314×581209.5×103
Step 3: Compute the exponent.
RTEa=4830.434209500=43.371
Step 4: Evaluate the fraction (via base-10 logs, as NCERT does). …
The fraction of molecules with energy ≥ activation energy is given by the Boltzmann factor e−Ea/RT. Using Ea=209.5 kJ mol−1, R=8.314 J mol−1K−1, and T=581 K, the fraction is 1.471×10−19.
The Arrhenius equation tells us that not all collisions lead to a reaction — only those molecules that possess energy equal to or greater than the activation energy Ea can overcome the energy barrier. The fraction of such molecules in a gas at temperature T follows the Boltzmann distribution: it is proportional to e−Ea/RT. This exponential factor is the heart of the temperature dependence of reaction rates.
Here, we are given Ea=209.5 kJ mol−1 and T=581 K. The fraction f is simply:
f=e−Ea/RT
But careful: Ea is in kJ, while R is usually in J. So we must convert units first.
-
Convert activation energy to J/mol
Ea=209.5 kJ mol−1=209.5×103 J mol−1
-
Write the gas constant
R=8.314 J mol−1K−1
-
Compute the exponent
−RTEa=−8.314×581209.5×103
First, denominator: 8.314×581=4830.434
Then, numerator: 209500
So exponent = −4830.434209500=−43.371
- Find the fraction
f=e−43.371
Evaluating via base-10 logarithms (NCERT's route):
logf=−2.30343.371=−18.832 …
Method: Direct Application of the Arrhenius Exponential Factor
This method uses the exponential term from the Arrhenius equation to find the fraction of molecules with energy ≥ activation energy.
Concept (Why this works)
The Arrhenius equation is:
k=Ae−Ea/RT
The factor e−Ea/RT represents the fraction of molecules that have energy equal to or greater than the activation energy Ea. This comes from the Maxwell–Boltzmann distribution of molecular energies.
Steps
Step 1: Identify the given data
- Activation energy, Ea=209.5 kJ mol−1
- Temperature, T=581 K
- Gas constant, R=8.314 J mol−1K−1
Step 2: Convert units to match R
Since R is in Joules, convert Ea from kJ to J:
Ea=209.5×1000=209500 J mol−1
Step 3: Compute the exponent
RTEa=8.314×581209500
First calculate denominator:
8.314×581=4830.434
Then:
RTEa=4830.434209500=43.371
Step 4: Calculate the fraction …
Here are the common mistakes students make when solving this Arrhenius equation problem, along with how to avoid each.
Mistake 1: Forgetting to convert activation energy to the same units as R
The error:
Students plug Ea=209.5 kJ mol−1 directly into the formula but use R=8.314 J mol−1K−1. This gives a completely wrong exponent.
How to avoid:
Always check units before substituting.
- Ea must be in J mol−1 when R is in J mol−1 K−1.
- Convert:
Ea=209.5×103 J mol−1
Mistake 2: Using the wrong formula for "fraction of molecules"
The error:
Some students try to use the full Arrhenius equation k=Ae−Ea/RT and then solve for k, but the question asks only for the fraction of molecules with energy ≥Ea, which is not k.
How to avoid:
The fraction of molecules with energy ≥Ea is given directly by the Boltzmann factor:
Fraction=e−Ea/RT
This is not the rate constant — it is the exponential part alone.
Mistake 3: Sign error in the exponent
The error:
Writing e+Ea/RT instead of e−Ea/RT, or forgetting the negative sign entirely.
How to avoid:
The Arrhenius equation has a negative exponent because higher activation energy means fewer molecules have enough energy. Remember:
Fraction=e−Ea/RT
A quick sanity check: if Ea is large, the fraction should be very small (much less than 1).
Mistake 4: Incorrect temperature unit
The error:
Using temperature in Celsius instead of Kelvin, or forgetting to convert.
How to avoid:
The problem gives T=581 K — no conversion needed here. But always verify: if temperature is in °C, add 273.15.
Mistake 5: Calculation errors in the exponent value
The error:
Mishandling the large numbers when computing RTEa.
How to avoid:
Do the calculation step-by-step:
- Ea=209.5×103=209500 J mol−1
- R=8.314 J mol−1K−1
- T=581 K
- Compute denominator: RT=8.314×581
- 8.314×500=4157
- 8.314×81=673.434
- Total = 4157+673.434=4830.434
- Compute exponent:
RTEa=4830.434209500=43.371
- Fraction = e−43.371
Use a calculator carefully — this is a very small number. Evaluated via logs: logf=−43.371/2.303=−18.832, so f=1.471×10−19.
Mistake 6: Reporting the fraction as a percentage without stating it
The error: …
- TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQQ.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 vrms(SO2,400K) with vmp(O2,T) gives T=300 K.
vrms=M13RT1,vmp=M22RT2.
With M(SO2)=64, T1=400 K, and M(O2)=32, set the two speeds equal:
M13RT1=M22RT2 …
- TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQQ.The rate constant of a reaction at 25 ∘C is 1×10−3 min−1. If the temperature coefficient of the reaction is 2, the rate constant (min−1) at 15 ∘C is (A) 2×10−3 (B) 2×10−4 (C) 5×10−4 (D) 4×10−3
›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 1×10−3 min⁻¹, i.e., 5×10−4 min⁻¹.
Concept & Intuition
The temperature coefficient (often denoted Q10) is defined as the factor by which the rate constant increases when the temperature is raised by 10 °C.
Here Q10=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 k15=k25/2.
Step-by-step reasoning
-
Identify the given data
- k25=1×10−3 min⁻¹
- Temperature coefficient Q10=2
- Temperature change: from 25 °C to 15 °C = −10 °C
-
Apply the definition of temperature coefficient
The relationship is:
kTkT+10=Q10
So for a 10 °C decrease:
k25k15=Q101=21
- Calculate the unknown rate constant k15=k25×21=(1×10−3)×21=0.5×10−3=5×10−4 min−1 …
- TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQQ.The rate constant of a reaction at 25∘C is 1×10−3 min−1. If the temperature coefficient of the reaction is 2, the rate constant (min−1) at 15∘C is (A) 2×10−3 (B) 2×10−4 (C) 5×10−4 (D) 4×10−3
›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: 5×10−4 min−1.
Concept & Intuition
The temperature coefficient (often denoted Q10) is the factor by which the rate constant changes when the temperature is raised by 10°C. Here Q10=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 k25=1×10−3 min−1 is at 25∘C. We need k15 at 15∘C.
The difference: 25∘C−15∘C=10∘C.
-
Apply the definition of the temperature coefficient
The temperature coefficient Q10 is defined as
Q10=kTkT+10
Here Q10=2, so for a 10°C increase the rate constant multiplies by 2. For a 10°C decrease, we use the reciprocal:
-
- TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQQ.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 ΔTb=i⋅Kb⋅m, where i is the van’t Hoff factor (for NaCl, i≈2), Kb is the ebullioscopic constant of water, and m is the molality. Since ΔTb>0, the boiling point increases. This eliminates option (A) and (C).
-
Freezing point behaviour …
-
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