The reaction between $H_2(g)$ and $O_2(g)$ is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of water. Explain.

The reaction between H₂(g) and O₂(g) is highly feasible yet allowing the gases to stand at room temperature in

NCERT Exemplar · Q31

H_2(g)

CBSEShort· 2mImportance★★★★★

68% · 80/117 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 →

The reaction is thermodynamically feasible ( $\Delta G < 0$ ) but kinetically hindered — the activation energy is so high at room temperature that the reaction rate is essentially zero without a spark or catalyst.

The key here is to separate thermodynamics from kinetics. A reaction can be energetically downhill (spontaneous in the Gibbs free energy sense) but still not happen at an observable rate if the molecules lack the energy to overcome the activation barrier.

1. The Arrhenius Equation — why rate depends on temperature and activation energy

The rate constant $k$ for any elementary step is given by:

k = A e^{-E_a / RT}

where $E_a$ is the activation energy, $R$ is the gas constant, $T$ is the absolute temperature, and $A$ is the frequency factor (how often molecules collide with the correct orientation).

Even if collisions occur, only those with kinetic energy $\geq E_a$ lead to reaction. The fraction of such collisions is $e^{-E_a / RT}$ .

\text{Fraction of effective collisions} = e^{-E_a / RT}

At room temperature ( $T \approx 298\ \text{K}$ ), $RT \approx 2.5\ \text{kJ mol}^{-1}$ . For a reaction with $E_a \approx 200\ \text{kJ mol}^{-1}$ (typical for breaking strong $\text{O=O}$ and $\text{H–H}$ bonds), the fraction is:

e^{-200 / 2.5} = e^{-80} \approx 1.4 \times 10^{-35}

That is astronomically small — effectively zero.

2. The specific case of $\mathbf{H_2 + O_2}$

The overall reaction:

2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \quad \Delta G^\circ \approx -237\ \text{kJ mol}^{-1}

is highly exergonic. But the mechanism is a chain reaction that requires first breaking the strong $\text{H–H}$ bond ( $436\ \text{kJ mol}^{-1}$ ) and the $\text{O=O}$ double bond ( $498\ \text{kJ mol}^{-1}$ ). At room temperature, thermal energy is far too low to break these bonds.

Watch out

A common mistake is to think "feasible" means "fast". Feasibility refers to $\Delta G < 0$ (spontaneity), not rate. Many spontaneous reactions are immeasurably slow at room temperature — diamond turning to graphite is another classic example.

3. What is needed to make it happen? …

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

Common Mistakes & How to Avoid Them

This question tests your understanding of thermodynamics vs. kinetics — a classic point of confusion.

Mistake 1: Saying the reaction is “not feasible” or “not spontaneous”

Why it’s wrong: The question explicitly states the reaction is highly feasible. $\Delta G^\circ$ for $2H_2 + O_2 \rightarrow 2H_2O$ is negative (≈ –237 kJ/mol at 25°C), so it is thermodynamically spontaneous.
How to avoid: Always check the sign of $\Delta G$ . If $\Delta G < 0$ , the reaction is feasible. The question is asking why it doesn’t happen fast, not whether it can happen.

Mistake 2: Confusing “feasibility” with “rate”

Why it’s wrong: Feasibility (thermodynamics) tells you the direction and extent of a reaction. The rate (kinetics) tells you how fast it proceeds. They are independent.
How to avoid: Memorise this distinction:
- Thermodynamics: $\Delta G$ — can it happen?
- Kinetics: Activation energy $E_a$ — how fast does it happen?

Mistake 3: Ignoring the role of activation energy

Why it’s wrong: Students often say “the reaction needs a spark” without explaining why.
How to avoid: State clearly: The $H_2$ – $O_2$ mixture has a high activation energy barrier ( $E_a$ ). At room temperature, very few molecules have enough kinetic energy to overcome it. The reaction requires an external energy source (spark, flame, catalyst) to break the strong $H–H$ and $O=O$ bonds.

Mistake 4: Saying “no reaction occurs at all”

Why it’s wrong: A tiny fraction of molecules do react, but the rate is negligible. The reaction is extremely slow, not zero.
How to avoid: Use precise language: “The reaction is kinetically hindered at room temperature.” Mention that the rate is immeasurably slow under these conditions.

Mistake 5: Mixing up equilibrium and kinetics

Why it’s wrong: Some students say “the equilibrium lies far to the right, so it should happen.” That’s true thermodynamically, but equilibrium tells you nothing about the time required. …

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.

CBSE 2024Set 56/3/11 markMCQ
$Q. When a catalyst increases the rate of a chemical reaction, then the rate constant (k) : (A) remains constant (B) decreases (C) increases (D) may increase or decrease depending on the order of the reaction$
›Reveal solutionSolution
A catalyst lowers the activation energy, which directly increases the rate constant $k$ through the Arrhenius equation. The answer is (C).

The rate constant $k$ is not just a number we measure—it encodes how the molecular-scale energy barrier controls reaction speed. To see why a catalyst must increase $k$ , we need the Arrhenius equation.
$k = A e^{-E_a/RT}$
This equation tells us that $k$ depends exponentially on the activation energy. A catalyst works by providing an alternative reaction pathway with a lower $E_a$ —it doesn't change the thermodynamics (reactants and products stay the same), but it reduces the energy hill molecules must climb to react.

Step-by-step reasoning

What a catalyst does at the molecular level

A catalyst participates in the reaction mechanism but is regenerated at the end. It creates intermediate steps with lower energy barriers than the uncatalyzed path. The net effect: $E_a$ (catalyst) $< E_a$ (no catalyst).

Impact on the exponential term

When $E_a$ decreases, the exponent $-E_a/RT$ becomes less negative (closer to zero). Since $e^x$ is an increasing function, $e^{-E_a/RT}$ becomes larger.

The pre-exponential factor $A$

This factor relates to collision frequency and orientation. A catalyst typically doesn't change $A$ significantly—the main effect is on $E_a$ .

Independence from reaction order

The rate constant $k$ appears in the rate law (e.g., $\text{rate} = k[A]^n$ ), but its value is determined by the Arrhenius equation, not by the order $n$ . The order tells us how concentration affects rate; the activation energy tells us the intrinsic speed at given concentrations. A catalyst lowers $E_a$ regardless of whether the reaction is zeroth, first, second, or any other order.

Quantitative example

Suppose $E_a = 100 \, \text{kJ/mol}$ without catalyst and $E_a = 50 \, \text{kJ/mol}$ with catalyst at $T = 300 \, \text{K}$ (with $R = 8.314 \, \text{J/(mol·K)}$ ): …
Start your 14-day free trial to see the full solution →
CBSE 2020Set 56/1/11 mark
$E_{act}$
›Reveal solutionSolution
When activation energy is zero, the Arrhenius equation reduces to $k = A$ , making the rate constant independent of temperature.

Why activation energy matters

The Arrhenius equation connects temperature to the rate constant through the activation energy—the minimum energy barrier reactants must overcome to transform into products. The equation captures a fundamental idea: higher temperatures give molecules more kinetic energy, increasing the fraction that can surmount the barrier.
$k = A e^{-E_a/RT}$
The exponential term $e^{-E_a/RT}$ embodies the temperature dependence. When $E_a$ is large, even small temperature changes dramatically alter $k$ . But what happens when there's no barrier at all?

The special case: $E_a = 0$

Substitute zero activation energy into the Arrhenius equation:

$k = A e^{-0/RT} = A e^0 = A \cdot 1 = A$

Interpret the result:

The rate constant collapses to just the pre-exponential factor $A$ . This factor represents the frequency of collisions with proper orientation—it depends on molecular properties and collision geometry, but crucially, it has no temperature dependence built into the exponential term.

Physical meaning:

A zero activation energy means every collision between properly oriented molecules leads to reaction, regardless of their kinetic energy. There's no energy threshold to cross. Temperature might still affect collision frequency slightly through changes in molecular speed, but the dominant exponential temperature dependence vanishes. …
Start your 14-day free trial to see the full solution →
CBSE 2019Set ANNUAL1 markMCQ
$Q. Arrhenius equation is$
(a) k = -Ae^(-Ea/RT)
(b) k = Ae^(Ea/RT)
(c) k = Ae^(-Ea/RT)
(d) k = -Ae^(Ea/RT)
›Reveal solutionSolution
The Arrhenius equation relating the rate constant to temperature is $k = A e^{-E_a/RT}$ .

The Arrhenius equation expresses how the rate constant $k$ of a reaction varies with absolute temperature $T$ :

$k = A\,e^{-E_a/RT}$

where:

$A$ = the Arrhenius (pre-exponential/frequency) factor, related to the frequency of collisions with proper orientation

$E_a$ = activation energy of the reaction

$R$ = universal gas constant

$T$ = absolute temperature …
Start your 14-day free trial to see the full solution →
CBSE 2017Set ANNUAL1 mark
$Q. Explain Arrhenius equation.$
›Reveal solutionSolution
The Arrhenius equation shows that a rate constant increases exponentially with temperature because more molecules acquire energy equal to or greater than the activation energy.

The Arrhenius equation is:

$k = A\,e^{-E_a/RT}$

where:

$k$ = rate constant of the reaction

$A$ = pre-exponential (frequency) factor, related to the frequency of collisions with proper orientation

$E_a$ = activation energy of the reaction

$R$ = universal gas constant

$T$ = absolute temperature (K)

The equation shows that as temperature $T$ increases, the exponential term $e^{-E_a/RT}$ increases (since $-E_a/RT$ becomes less negative), so a larger fraction of reactant molecules possess energy equal to or greater than $E_a$ , and the rate constant $k$ increases — explaining why reaction rates generally rise sharply with temperature.

Taking the natural log of both sides gives the linear form: …
Start your 14-day free trial to see the full solution →

←Previous questionQ30 — For a general reaction A arrow B, the concentration of A is plotted ag…→Next questionQ32 — Why does the rate of a reaction increase with rise in temperature?

Q.The reaction between $H_2(g)$ and $O_2(g)$ is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of water. Explain.

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

1. The Arrhenius Equation — why rate depends on temperature and activation energy

2. The specific case of $\mathbf{H_2 + O_2}$

3. What is needed to make it happen? …

Concept: Thermodynamic Feasibility vs. Kinetic Feasibility

Method: Thermodynamic vs. Kinetic Analysis

Common Mistakes & How to Avoid Them

Mistake 1: Saying the reaction is “not feasible” or “not spontaneous”

Mistake 2: Confusing “feasibility” with “rate”

Mistake 3: Ignoring the role of activation energy

Mistake 4: Saying “no reaction occurs at all”

Mistake 5: Mixing up equilibrium and kinetics

Step-by-step reasoning

Why activation energy matters

The special case: $E_a = 0$

Q.The reaction between H2(g)H_2(g)H2​(g) and O2(g)O_2(g)O2​(g) is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of water. Explain.

1. The Arrhenius Equation — why rate depends on temperature and activation energy

2. The specific case of H2+O2\mathbf{H_2 + O_2}H2​+O2​

3. What is needed to make it happen? …

🎓Unlock everything free for 14 days

Q.The reaction between $H_2(g)$ and $O_2(g)$ is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of water. Explain.

2. The specific case of $\mathbf{H_2 + O_2}$

Unlock everything free for 14 days