A potential-energy profile (energy on the vertical axis, reaction coordinate on the horizontal axis) is drawn for a reaction. Reading from left to right: the reactants sit at the lowest energy level; the curve rises to a maximum (the activated complex) and then falls to the products, which lie at a HIGHER energy level than the reactants. The drop from the peak down to the product level is marked $E_1$, and the further gap from the product level down to the reactant level is marked $E_2$. Thus the peak lies $E_1 + E_2$ above the reactants and $E_1$ above the products. On this basis, mark the correct option. (i) Activation energy of the forward reaction is $E_1 + E_2$ and the product is less stable than the reactant. (ii) Activation energy of the forward reaction is $E_1 + E_2$ and the product is more stable than the reactant. (iii) Activation energy of both the forward and the backward reaction is $E_1 + E_2$ and the reactant is more stable than the product. (iv) Activation energy of the backward reaction is $E_1$ and the product is more stable than the reactant.

A potential-energy profile (energy on the vertical axis, reaction coordinate on the horizontal axis) is drawn

NCERT Exemplar · Q4

E_1

(i) Activation energy of the forward reaction is

E_1 + E_2

and the product is less stable than the reactant.

(ii) Activation energy of the forward reaction is

E_1 + E_2

and the product is more stable than the reactant.

(iii) Activation energy of both the forward and the backward reaction is

E_1 + E_2

and the reactant is more stable than the product.

(iv) Activation energy of the backward reaction is

E_1

and the product is more stable than the reactant.

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Concept understanding — Arrhenius Equation

Why reactions speed up when you heat them

You already know that most reactions go faster when you raise the temperature. The intuitive reason is simple: molecules move faster, collide more often, and collide harder. But that alone doesn't explain the dramatic jump in rate — a 10 °C rise can double or triple the rate, even though the collision frequency only increases by a few percent. Something else is at work.

The missing piece is that not every collision leads to a reaction. Only collisions with enough energy — above a certain threshold — actually break bonds and form products. That threshold is the activation energy $E_a$ . Think of it as a hill the reactants must climb before they can roll down into products. At room temperature, only a tiny fraction of molecules have enough energy to get over that hill. Raise the temperature, and that fraction grows exponentially.

The Arrhenius equation

The Swedish chemist Svante Arrhenius captured this relationship in a single compact formula:

k = A e^{-E_a / (RT)}

Here:

$k$ is the rate constant (how fast the reaction proceeds at a given temperature)
$A$ is the pre-exponential factor (roughly, how often collisions happen with the right orientation)
$E_a$ is the activation energy (the energy barrier, in J/mol or kJ/mol)
$R$ is the gas constant (8.314 J/mol·K)
$T$ is the absolute temperature (in Kelvin)

Important

The exponential term $e^{-E_a/(RT)}$ is the fraction of molecules that have energy at least $E_a$ . This fraction is tiny when $E_a$ is large or $T$ is low, and it grows sharply as $T$ increases.

Why the rate jumps so sharply with temperature

The exponential is the key. Suppose $E_a = 50$ kJ/mol. At 300 K, the fraction is $e^{-50000/(8.314 \times 300)} \approx e^{-20.0} \approx 2 \times 10^{-9}$ . At 310 K, it becomes $e^{-50000/(8.314 \times 310)} \approx e^{-19.4} \approx 3.8 \times 10^{-9}$ . That's nearly double — even though the temperature rose only 3%. The collision frequency $A$ barely changed, but the exponential term nearly doubled. That's why a small temperature rise can cause a large rate increase.

Tip

A useful rule of thumb: for many reactions near room temperature, a 10 °C rise roughly doubles the rate constant. This is a consequence of the exponential, not a law — it depends on $E_a$ .

What $A$ and $E_a$ really mean

$A$ (the pre-exponential factor) accounts for how often molecules collide and whether they're oriented correctly. It depends on the size and shape of the molecules. $E_a$ is the minimum energy needed for a successful collision. A reaction with a high $E_a$ is very sensitive to temperature; one with a low $E_a$ is less sensitive.

Watch out

Do not confuse $E_a$ with the overall energy change of the reaction ( $\Delta H$ ). $E_a$ is the barrier height; $\Delta H$ is the net energy difference between reactants and products. A reaction can be highly exothermic ( $\Delta H$ large and negative) but still have a high $E_a$ — that's why some exothermic reactions (like burning wood) need a spark to start.

The logarithmic form …

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The reactants are the lowest point on the profile, the activated complex (peak) is $E_1+E_2$ above them, and the products sit $E_2$ above the reactants (i.e. $E_1$ below the peak). Hence the forward activation energy is $E_1+E_2$ , and because the products are higher in energy they are less stable than the reactants.

Concept

Activation energy is the energy gap between a state (reactants or products) and the top of the barrier (activated complex). Stability is judged by energy: the lower the energy, the more stable the species.

Reading the profile

Reactant level = the bottom of the diagram.
Barrier top (activated complex) lies $E_1+E_2$ above the reactants.
Product level lies $E_2$ above the reactants, i.e. $E_1$ below the barrier top.

Applying it

Forward activation energy $=$ (barrier top) $-$ (reactants) $= E_1+E_2$ .
Backward activation energy $=$ (barrier top) $-$ (products) $= E_1$ . …

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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 23 on this concept.

CBSE 2026Set A1 markMCQ
$Q. Which of the following represents the effect of temperature on reaction rate ?$
(a) Nernst's equation
(b) Gibbs-Helmholtz equation
(c) Arrhenius equation
(d) Van't Hoff equation
›Reveal solutionSolution

The Arrhenius equation, k = A e^(-Ea/RT), describes how the rate constant (and hence reaction rate) depends on temperature.

The Arrhenius equation k = A e^(-Ea/RT) links the rate constant k to temperature T and activation energy Ea. It quantifies the effect of temperature on reaction rate. The Nernst equation relates electrod …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Activation energy of a chemical reaction can be determined by$
(a) rate constant at standard temperature
(b) rate constants at two different temperatures
(c) orientation of reactant molecules during the collision
(d) using catalyst
›Reveal solutionSolution

Activation energy is found from the Arrhenius equation by measuring the rate constant of a reaction at two (or more) different temperatures.

The Arrhenius equation is: k = A e^(-Ea/RT)

Taking the natural log at two temperatures T1 and T2 with rate constants k1 and k2, and subtracting, gives:

ln(k2/k1) = -(Ea/R)(1/T2 - 1/T1)

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The correct form of Arrhenius equation is:$
(a) k = e^(-Ea/RT)
(b) k = Ea/RT
(c) k = log_e (Ea/R)
(d) k = A e^(-Ea/RT)
›Reveal solutionSolution

The Arrhenius equation is $k = Ae^{-E_a/RT}$ , relating the rate constant to temperature and activation energy.

The Arrhenius equation, proposed by Svante Arrhenius, expresses how the rate constant $k$ of a reaction varies with temperature:

$k = Ae^{-E_a/RT}$ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The influence of temperature on the rate of a reaction is determined by$
(a) Nernst equation
(b) Gibbs-Helmholtz equation
(c) van't Hoff equation
(d) Arrhenius equation
›Reveal solutionSolution
Of the four named equations, only the Arrhenius equation directly relates the rate constant of a reaction to temperature.

Why the others are not it:

(a) Nernst equation: relates the EMF of an electrochemical cell to the concentrations (activities) of the species involved — an electrochemistry relationship, not a kinetics one.

(b) Gibbs–Helmholtz equation: relates the temperature dependence of Gibbs free energy change to enthalpy and entropy changes — a thermodynamics relationship about spontaneity, not reaction rate.

(c) van't Hoff equation: relates the equilibrium constant ( $K_c$ or $K_p$ ) to temperature — about the position of equilibrium, not how fast equilibrium is reached.

Arrhenius equation (d): …
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CBSE 2025Set X11 mark
$Q. Arrhenius factor is also called __________ factor.$
›Reveal solutionSolution

The Arrhenius factor $A$ is also called the frequency (pre-exponential) factor.

The Arrhenius equation is $k = A\,e^{-E_a/RT}$ , where $A$ is the Arrhenius constant. It represents the frequency of collisions with correct orientation and is therefore cal …
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CBSE 2025Set D1 markMCQ
$Q. The specific rate constant of a first order reaction depends upon$
(a) concentration of reactants
(b) concentration of products
(c) time
(d) temperature
›Reveal solutionSolution

The rate constant depends on temperature (via the Arrhenius equation), not on concentration or time.

For a first order reaction rate = k[A]. The rate constant k is a proportionality constant characteristic of the reaction at a given temperature. According to the Arrhenius equation:

k = A e^(-Ea/RT) …

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CBSE 2025Set D1 markMCQ
$Q. The temperature coefficient of most of the reactions lies between$
(a) 1 and 3
(b) 2 and 3
(c) 1 and 4
(d) 2 and 4
›Reveal solutionSolution

Temperature coefficient of most reactions is between 2 and 3.

The temperature coefficient is the ratio of the rate constant at (T + 10) K to that at T K:

temperature coefficient = k(T+10)/k(T)

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following represents Arrhenius equation?$
(a) k = Ae^(Ea/RT)
(b) k = Ae^(-Ea/RT)
(c) k = Ae^(Ea/R)
(d) dk = Ae^(Ea/T)
›Reveal solutionSolution

The Arrhenius equation relates the rate constant to temperature through an exponential term with a NEGATIVE activation-energy exponent.

The Arrhenius equation is:

$k = A\,e^{-E_a/RT}$ …

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CBSE 2025Set ANNUAL1 mark
$Q. Define : Activation energy.$
›Reveal solutionSolution

Activation energy is the energy barrier reactant molecules must cross to be converted into products.

According to collision theory, not every collision between reactant molecules leads to product formation — only those collisions in which molecules possess sufficient energy (and correct orientation) are effective. The activation energy, $E_a$ , is defined as the minimum extra energy that reactant molecules, over and above their average energy, must acquire before they can react to form products. Equivalently, it is the difference between the threshold energy (the minimum energy needed …

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CBSE 2024Set ANNUAL1 mark
$Q.The kinetic energy of maximum fraction of reactant molecules is called ______.$
›Reveal solutionSolution

On a Maxwell-Boltzmann distribution-of-energy curve, the energy value at the peak (where the curve is highest) is possessed by the largest fraction of molecules - this is called the most probable kinetic energy.

In any collection of gas/reactant molecules at a given temperature, molecules do not all have the same kinetic energy; their energies are spread out according to the Maxwell-Boltzmann distribution.

When the fraction of molecules (y-axis) is plotted against kinetic energy (x-axis), the curve rises, reaches a maximum, then falls - the energy value at this peak is the one possessed by the largest fraction of molecules, called the most probable kinetic energy. …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Activation energy of a chemical reaction can be determined by :$
(a) determining the rate constant at standard temperature
(b) determining the rate constant at two temperatures
(c) determining probability of collision
(d) using catalyst
›Reveal solutionSolution

Activation energy comes from the Arrhenius equation, which needs the rate constant measured at (at least) two temperatures to solve for $E_a$ .

The Arrhenius equation is $k = A\,e^{-E_a/RT}$ , or in logarithmic form:

$\ln k = \ln A - \frac{E_a}{RT}$

A single rate constant at one temperature carries two unknowns ( $A$ and $E_a$ ) and cannot be solved alone. Measuring $k$ at two different temperatures $T_1$ and $T_2$ gives two equations, which combine to: …

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CBSE 2024Set ANNUAL1 markMCQ
$\Delta H$
(a) 5 kcal/mol
(b) – 35 kcal/mol
(c) 15 kcal/mol
(d) 25 kcal/mol
›Reveal solutionSolution

Subtracting the backward activation energy from the forward one gives ΔH = −5 kcal/mol, meaning the reaction is exothermic by 5 kcal/mol.

The relationship between the enthalpy change of a reaction and the activation energies of its forward and reverse directions is:

$\Delta H = E_a(\text{forward}) - E_a(\text{backward})$

Substituting the given values:

$\Delta H = 15 - 20 = -5\ kcal/mol$

…

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Concept

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