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)

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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 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? …

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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. …

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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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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

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? …

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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 $\mathbf{H_2 + O_2}$

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