Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help

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Q. Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.

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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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Thermodynamic feasibility tells us whether a reaction can happen (spontaneity), but the rate tells us how fast it happens — these are independent. A classic example is the combustion of petrol: thermodynamically highly feasible ( $\Delta G \ll 0$ ), yet it does not proceed at room temperature without a spark because the activation energy barrier is high.

The Core Idea: Feasibility vs. Rate

Many students mix up two separate questions: "Will this reaction occur?" and "How quickly will it occur?" The first is answered by thermodynamics (Gibbs free energy change, $\Delta G$ ). The second is answered by kinetics (activation energy, $E_a$ , and the Arrhenius equation).

A reaction is thermodynamically feasible when $\Delta G < 0$ — the products are more stable than the reactants. But that says nothing about the path from reactants to products. Even if the products are much more stable, there may be a high energy hill (activation barrier) to climb first. If that hill is too steep at a given temperature, the reaction will be imperceptibly slow — it is feasible but not kinetically favoured.

The Arrhenius equation captures this:

k = A e^{-E_a / RT}

Here $k$ is the rate constant, $A$ is the frequency factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature. A large $E_a$ makes $k$ tiny, even if $\Delta G$ is very negative.

Step-by-Step Explanation with an Example

1. Choose a reaction that is clearly feasible but slow.

The combustion of petrol (a hydrocarbon like octane, $\text{C}_8\text{H}_{18}$ ) is a perfect example. The reaction is:

\text{C}_8\text{H}_{18}(l) + \frac{25}{2}\,\text{O}_2(g) \rightarrow 8\,\text{CO}_2(g) + 9\,\text{H}_2\text{O}(g)

This reaction has a large negative $\Delta G$ — it is highly exergonic. Petrol is thermodynamically unstable in air; it wants to burn.

2. Check the thermodynamic feasibility.

At room temperature, $\Delta G^\circ$ for this combustion is roughly $-5.3 \times 10^3\ \text{kJ mol}^{-1}$ . That is an enormous driving force. By thermodynamics alone, the reaction should proceed spontaneously.

3. Observe the actual rate.

Yet, a pool of petrol sitting open in air at $25^\circ\text{C}$ does not burst into flames. It can sit for hours with no visible reaction. The rate is essentially zero. Why?

4. Identify the kinetic barrier.

The reaction has a very high activation energy. The C–H and C–C bonds in the hydrocarbon, and the O=O double bond in oxygen, are strong. To break them, the molecules must collide with enough energy to overcome a large $E_a$ (on the order of several hundred kJ/mol). At room temperature, the fraction of molecules with that much energy is vanishingly small — given by the Boltzmann factor $e^{-E_a/RT}$ .

5. Supply the activation energy externally. …

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Here is a breakdown of the common mistakes students make on this concept, along with how to avoid them.

The Core Concept: Thermodynamics vs. Kinetics

The question asks you to distinguish between thermodynamic feasibility (spontaneity) and kinetic rate (speed).

Thermodynamic Feasibility: Determined by $\Delta G$ (Gibbs free energy). If $\Delta G < 0$ , the reaction is spontaneous (feasible) in the forward direction. It tells you if the reaction can happen.
Rate of Reaction: Determined by activation energy ( $E_a$ ) and the pathway (mechanism). It tells you how fast the reaction happens.

The key mistake is assuming that if a reaction is thermodynamically feasible, it must happen quickly. This is false. A reaction can be highly spontaneous (large negative $\Delta G$ ) but have a very high activation energy, making it extremely slow at room temperature.

Common Mistake #1: Confusing Spontaneity with Speed

The Error: Students often write that a reaction with a large negative $\Delta G$ will occur rapidly. They treat $\Delta G$ as a measure of rate.

Why it happens: The word "feasible" is misleading. In everyday language, "feasible" means "possible and easy." In chemistry, it only means "possible" (spontaneous), not "easy" or "fast."

How to Avoid:

Memorize the distinction: $\Delta G$ = direction (which way the reaction will go). $E_a$ = speed (how fast it gets there).
Use the analogy: A ball at the top of a hill (high potential energy) is thermodynamically "feasible" to roll down (spontaneous). But if it's sitting in a small dip (a high activation energy barrier), it won't roll down until you give it a push (energy input). The ball can roll down, but it won't roll down quickly on its own.

Common Mistake #2: Choosing a Wrong Example

The Error: Students pick an example where the reaction is not thermodynamically feasible (e.g., $\Delta G > 0$ ) and then say it's slow. This misses the point entirely. The question specifically asks for a reaction that is thermodynamically feasible but is slow.

Why it happens: Students confuse "non-spontaneous" with "slow." A non-spontaneous reaction ( $\Delta G > 0$ ) is not feasible at all, so it's not a valid example for this concept.

How to Avoid:

Check the sign of $\Delta G$ : Your example must have $\Delta G < 0$ (spontaneous) but a high activation energy (slow).
Use the classic textbook example: The reaction between hydrogen gas ( $H_2$ ) and oxygen gas ( $O_2$ ) to form water ( $H_2O$ ).

2H_2(g) + O_2(g) \rightarrow 2H_2O(l)

- **Thermodynamics:** This reaction is highly spontaneous. $\Delta G^\circ = -237.2 \text{ kJ/mol}$ (very negative). It is thermodynamically feasible.
- **Kinetics:** At room temperature, the reaction is **extremely slow** (essentially zero rate). A mixture of $H_2$ and $O_2$ can sit for years without reacting.
- **Why slow?** The activation energy is very high because the $H-H$ and $O=O$ bonds are very strong. The reaction needs a spark or a catalyst (like platinum) to overcome this barrier.

Common Mistake #3: Not Explaining the "Why" Behind the Slowness …

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Q.Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.

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

The Core Idea: Feasibility vs. Rate

Step-by-Step Explanation with an Example

Concept: Thermodynamics vs. Kinetics

Method: Compare $\Delta G$ and Activation Energy

Steps:

Example: Combustion of Diamond (or Graphite) in Air

The Core Concept: Thermodynamics vs. Kinetics

Common Mistake #1: Confusing Spontaneity with Speed

Common Mistake #2: Choosing a Wrong Example

Common Mistake #3: Not Explaining the "Why" Behind the Slowness …

Q.Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.

The Core Idea: Feasibility vs. Rate

Step-by-Step Explanation with an Example

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