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

NCERT Exemplar · Q36

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

Andhra Pradesh BieapShort· 2mImportance★★★★★

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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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$\exp\left(-\dfrac{E_a}{RT}\right)$
›Reveal solutionSolution
This tests the meaning of each term in the Arrhenius equation; the exponential term represents the fraction of molecules with energy exceeding the activation energy.

Concept and Intuition

The Arrhenius equation, $k = A e^{-E_a/RT}$ , splits the rate constant into two physically distinct factors: $A$ (the frequency/pre-exponential factor, related to collision frequency and orientation) and $e^{-E_a/RT}$ (the fraction of molecular collisions that have enough energy to cross the activation energy barrier, derived from the Maxwell-Boltzmann energy distribution). Only molecules with energy $\geq E_a$ can react upon collision; the exponential term quantifies what fraction of the population meets this threshold at a given temperature.

Step-by-Step Solution

Write the Arrhenius equation: $k = A\,e^{-E_a/RT}$ .

Recognize $A$ is the frequency factor — related to the rate of collisions and their proper orientation, independent of energy considerations.

Recognize $e^{-E_a/RT}$ arises from integrating the Maxwell-Boltzmann distribution above the energy threshold $E_a$ — it represents the fraction of molecules whose energy exceeds $E_a$ at temperature $T$ . …
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AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$2.5 \times 10^{-4} s^{-1}$
›Reveal solutionSolution
This tests the Arrhenius equation relating rate constants at two temperatures; solving gives $T_2 \approx 625\,K$ .

Concept and Intuition

A faster half-life means a larger rate constant. The Arrhenius equation connects the ratio of two rate constants at two temperatures to the activation energy — a bigger $E_a$ needs a bigger temperature jump to boost $k$ by the same factor.

Step-by-Step Solution

$T_1 = 300°C = 573\,K$ , $k_1 = 2.5\times10^{-4}\,s^{-1}$ .

Desired half-life $t_{1/2}=138.6\,s$ for a first-order reaction: $k_2 = \dfrac{0.693}{138.6} = 5\times10^{-3}\,s^{-1}$ .

Ratio: $k_2/k_1 = \dfrac{5\times10^{-3}}{2.5\times10^{-4}} = 20$ .

Arrhenius (two-temperature form): $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ .

$\dfrac{E_a}{2.303R} = \dfrac{42000}{2.303\times2} = 9118.5$ . …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$1.6\times10^{-5}s^{-1}$
›Reveal solutionSolution
The Arrhenius equation converts rate constants at two temperatures via activation energy; the given antilog hint (antilog(0.4771)=3.0) is a shortcut for the final power-of-ten step, giving $k_2=4.8\times10^{-3}\ s^{-1}$ .

Concept and Intuition

The two-point form of the Arrhenius equation lets us find how much a rate constant changes when temperature changes, purely from the activation energy — a higher $E_a$ makes the rate far more sensitive to temperature. The problem hands us the exact antilog we'll need at the end, which is a strong signal the intended computation lands exactly on $\log(k_2/k_1)=2.4771$ .

Step-by-Step Solution

Write the two-temperature Arrhenius relation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ .

Compute $\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{600}-\dfrac{1}{700} = \dfrac{700-600}{600\times700} = \dfrac{100}{420000} = \dfrac{1}{4200}$ .

Compute $\dfrac{E_a}{2.303R} = \dfrac{198870}{2.303\times8.3} = \dfrac{198870}{19.1149} \approx 10403.9$ .

Multiply: $10403.9 \times \dfrac{1}{4200} \approx 2.4771$ . …
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AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$\to$
›Reveal solutionSolution
This tests the two-point form of the Arrhenius equation to find an unknown temperature given a rate constant ratio. The answer works out to T ≈ 311 K.

Concept and Intuition

The Arrhenius equation $k = A e^{-E_a/RT}$ can be converted, for two temperatures, into a linear relation between $\log(k_2/k_1)$ and $(1/T_1 - 1/T_2)$ , letting us solve for an unknown temperature when the rate constants and activation energy are known.

Step-by-Step Solution

Two-point Arrhenius equation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303 R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ .

Given $k_1 = 5\times10^3\ s^{-1}$ at $T_1 = 300\ K$ , and $k_2 = 1.0\times10^4\ s^{-1}$ at $T_2 = T$ (to find). Ratio $k_2/k_1 = \dfrac{1.0\times10^4}{5\times10^3} = 2$ .

Given $\log 2 = 0.3$ , so LHS $= 0.3$ .

Compute $\dfrac{E_a}{2.303R} = \dfrac{50000}{2.303 \times 8.3} = \dfrac{50000}{19.115} \approx 2615.7$ .

So $0.3 = 2615.7\left(\dfrac{1}{300} - \dfrac{1}{T}\right)$ , giving $\dfrac{1}{300} - \dfrac{1}{T} = \dfrac{0.3}{2615.7} \approx 1.1468\times10^{-4}$ . …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$\left(\dfrac{k_{610}}{k_{600}}\right)$
›Reveal solutionSolution
Same reaction time for 10% completion at 600K and 20% completion at 610K lets the time cancel, leaving $k_{610}/k_{600}=\log(1.25)/\log(1.111)\approx2.12$ .

Concept and Intuition

For a first-order reaction, $k=\dfrac{1}{t}\ln\dfrac{1}{1-x}$ where $x$ is the fraction reacted. If the same time $t$ produces different completions at two temperatures, that common $t$ cancels out when taking the ratio of the two rate constants, leaving a ratio of two logarithmic (completion) factors only.

Step-by-Step Solution

At 600 K, 10% complete in time $t$ : $k_{600} = \dfrac{1}{t}\ln\dfrac{1}{0.90} = \dfrac{2.303}{t}\log(1.111)$ .

At 610 K, 20% complete in the same time $t$ : $k_{610} = \dfrac{1}{t}\ln\dfrac{1}{0.80} = \dfrac{2.303}{t}\log(1.25)$ .

Ratio: $\dfrac{k_{610}}{k_{600}} = \dfrac{\log(1.25)}{\log(1.111)} = \dfrac{0.0969}{0.0457} \approx 2.12$ . …
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AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$E_a$
›Reveal solutionSolution
Statement I is correct because reaction orders are experimentally determined and can be zero, positive integers, or fractions. Statement II is incorrect because the frequency factor (pre-exponential factor) is not a fraction of molecules with energy above $E_a$ ; that fraction is given by the exponential term $e^{-E_a/RT}$ . Therefore, the correct option is (C).

The Arrhenius equation is the backbone of chemical kinetics when it comes to temperature dependence. It is written as:
$k = A e^{-E_a/RT}$
Here, $k$ is the rate constant, $A$ is the frequency factor (also called the pre-exponential factor), $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature. The key intuition:

$A$ represents how often molecules collide in the correct orientation (a frequency, not a fraction).

The exponential term $e^{-E_a/RT}$ is the fraction of molecules that have energy equal to or greater than $E_a$ . Mixing these two up is a classic mistake.

Now, let’s evaluate each statement carefully.

Statement I: "Order of a reaction can be obtained from experiment and can have zero or positive integer or positive fraction values"

The order of a reaction is defined as the sum of the exponents of concentration terms in the experimentally determined rate law.

It is not derived from the stoichiometric coefficients (that would be molecularity, which applies only to elementary steps).

Orders can indeed be zero (e.g., decomposition of ammonia on a platinum surface), positive integers (most common), or positive fractions (e.g., the reaction $H_2 + Br_2 \rightarrow 2HBr$ has order 1.5).

Negative orders are also possible, but the statement only claims zero, positive integers, or positive fractions — all of which are valid.

Conclusion: Statement I is correct.

Statement II: "In the Arrhenius equation, the frequency factor is the fraction of molecules that can have energy higher than $E_a$ "

Let’s revisit the Arrhenius equation: $k = A e^{-E_a/RT}$ . …
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$2\times10^{-5}\ \mathrm{s^{-1}}$
›Reveal solutionSolution
Plugging the two rate constants and temperatures into the two-point Arrhenius equation gives an activation energy of about 88 kJ/mol.

Concept and Intuition

The Arrhenius equation's two-temperature form lets you find activation energy directly from two rate constants at two known temperatures, without knowing the pre-exponential factor.

Step-by-Step Solution

Two-point Arrhenius equation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ .

$\dfrac{k_2}{k_1} = \dfrac{4\times10^{-3}}{2\times10^{-5}} = 200$ ; $\log 200 = \log 2 + \log 100 = 0.3 + 2 = 2.3$ .

$\dfrac{1}{T_1} - \dfrac{1}{T_2} = \dfrac{1}{400} - \dfrac{1}{500} = \dfrac{500-400}{400\times500} = \dfrac{100}{200000} = 5\times10^{-4}\ \mathrm{K^{-1}}$ .

$2.3 = \dfrac{E_a}{2.303 \times 8.3} \times 5\times10^{-4}$ .

$2.303 \times 8.3 = 19.115$ . …
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AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$3.46 \times 10^{-2}\ s^{-1}$
›Reveal solutionSolution
This applies the two-temperature Arrhenius equation to find the rate constant at a higher temperature given the activation energy; the result is closest to $0.692\ s^{-1}$ .

Concept and Intuition

The Arrhenius equation, $k = Ae^{-E_a/RT}$ , shows that rate constants increase sharply with temperature for reactions with a sizeable activation energy. Taking the equation at two temperatures and dividing eliminates the pre-exponential factor $A$ , giving a convenient logarithmic two-point formula for finding a rate constant at a new temperature.

Step-by-Step Solution

Write the two-point Arrhenius equation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ .

Compute $\dfrac{1}{T_1} - \dfrac{1}{T_2} = \dfrac{1}{298} - \dfrac{1}{350} = \dfrac{350-298}{298\times350} = \dfrac{52}{104300} \approx 4.986\times10^{-4}\ \text{K}^{-1}$ .

Compute $\dfrac{E_a}{2.303R} = \dfrac{50100}{2.303\times8.314} = \dfrac{50100}{19.147} \approx 2616.6$ .

Multiply: $\log\dfrac{k_2}{k_1} \approx 2616.6 \times 4.986\times10^{-4} \approx 1.305$ .

So $\dfrac{k_2}{k_1} = 10^{1.305} = 10 \times 10^{0.305}$ . Using $\log 2 = 0.3010$ , $10^{0.305}$ is just slightly above 2, i.e. $\approx 2.02$ , giving $k_2/k_1 \approx 20.2$ . …
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AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
$\mathrm{kJ\,mol^{-1}}$
›Reveal solutionSolution
Plugging the rate-doubling data into the two-point Arrhenius equation gives $E_a \approx 53.33\ \mathrm{kJ\,mol^{-1}}$ — option (D).

Concept and Intuition

The Arrhenius equation, $k = Ae^{-E_a/RT}$ , links a reaction's rate constant to temperature through its activation energy. When we know the rate constant at two temperatures, we can eliminate the pre-exponential factor $A$ and solve directly for $E_a$ using the two-temperature form of the equation — this is the standard way "activation energy from a rate-doubling" problems are solved.

Step-by-Step Solution

Two-point Arrhenius form: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ .

Here $k_2/k_1 = 2$ (rate doubled), so $\log 2 = 0.3$ (given).

$\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{300}-\dfrac{1}{310} = \dfrac{310-300}{300\times310} = \dfrac{10}{93000} = 1.0753\times10^{-4}\ \mathrm{K^{-1}}$ .

$0.3 = \dfrac{E_a}{2.303\times 8.3}\times 1.0753\times10^{-4}$ .

$2.303\times 8.3 = 19.115$ . …
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AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$k_1$
›Reveal solutionSolution
Plugging $E_a$ , $R$ , $T_1=300$ K and $T_2=400$ K into the two-point Arrhenius equation gives $\ln(k_2/k_1)\approx 4.166$ .

Concept and Intuition

The Arrhenius equation $k = A\,e^{-E_a/RT}$ shows that rate constants at two temperatures are related by $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac1{T_1}-\dfrac1{T_2}\right)$ , since the pre-exponential factor $A$ cancels out. A higher activation energy or a larger temperature gap gives a bigger ratio $k_2/k_1$ .

Step-by-Step Solution

Convert $E_a$ to joules: $41.5\ \mathrm{kJ/mol} = 41500\ \mathrm{J/mol}$ .

Compute $\dfrac{E_a}{R} = \dfrac{41500}{8.3} = 5000$ .

Compute $\dfrac1{T_1}-\dfrac1{T_2} = \dfrac1{300}-\dfrac1{400} = \dfrac{4-3}{1200} = \dfrac1{1200}$ . …
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AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
$\mathrm{C_2H_5I(g) \rightarrow C_2H_4(g) + HI(g)}$
›Reveal solutionSolution
Plugging the given rate-constant ratio and temperatures into the two-point Arrhenius equation gives an activation energy of 48.16 kJ/mol.

Concept and Intuition

The Arrhenius equation relates the rate constant at two different temperatures to the activation energy: $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ . Given rate constants at two temperatures, we can solve directly for $E_a$ .

Step-by-Step Solution

$k_1 = x$ at $T_1=600$ K, $k_2=4x$ at $T_2=700$ K, so $k_2/k_1 = 4$ .

$\ln4 = 2.303\log4 = 2.303\times0.6 = 1.3818$ .

$\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac1{600}-\dfrac1{700} = \dfrac{700-600}{600\times700}=\dfrac{100}{420000}=2.381\times10^{-4}\ \text{K}^{-1}$ . …
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AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
$^{-1}$
›Reveal solutionSolution
Applying the two-point Arrhenius equation gives activation energy $\approx 33.45$ kJ/mol.

Concept and Intuition

The Arrhenius equation $k = A e^{-E_a/RT}$ links rate constant to temperature through the activation energy. Taking the ratio of rate constants at two temperatures eliminates the pre-exponential factor $A$ , letting us solve directly for $E_a$ .

Step-by-Step Solution

Write $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ .

$\dfrac{k_2}{k_1} = \dfrac{0.2}{0.02}=10$ , so $\ln 10 = 2.303$ .

$\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{500}-\dfrac{1}{700} = \dfrac{700-500}{500\times700} = \dfrac{200}{350000} = 5.714\times10^{-4}\ \text{K}^{-1}$ . …
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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.

The Core Idea: Feasibility vs. Rate

Step-by-Step Explanation with an Example

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