How will the rate of the reaction be affected when (a) Surface area of the reactant is reduced, (b) Catalyst i

Question

Q. How will the rate of the reaction be affected when

(a) Surface area of the reactant is reduced,

(b) Catalyst is added in a reversible reaction, and

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Concept understanding — Factors Affecting Reaction Rate

Reaction Rate Stoichiometry – From Intuition to Precision

Imagine you are watching a simple reaction:

$2 \text{NO}_2 \rightarrow 2 \text{NO} + \text{O}_2$

As NO₂ disappears, NO appears twice as fast as O₂ appears. Why? Because the balanced equation says: for every 2 molecules of NO₂ that break apart, you get 2 molecules of NO and 1 molecule of O₂. The numbers in front of the species — the stoichiometric coefficients — tell you the relative speeds at which reactants vanish and products appear.

That is the core idea: reaction rate stoichiometry is the relationship between the rates of change of different species in a chemical reaction, dictated by their coefficients in the balanced equation.

The Intuitive Picture

Think of a factory assembly line. The balanced equation is like a recipe:

2 units of raw material A → 2 units of product B + 1 unit of byproduct C

If the line runs steadily, every time 2 units of A are consumed, 2 units of B are produced and 1 unit of C is produced. So the rate at which A disappears must be twice the rate at which C appears. The rate at which B appears equals the rate at which A disappears (both have coefficient 2).

Note

The stoichiometric coefficients are not speeds themselves — they are scaling factors that connect the speeds of different species.

The Precise Statement

For a general reaction:

aA + bB \rightarrow cC + dD

The rate of reaction (often called the rate of the process, $r$ ) is defined as:

r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

Here:

$\frac{d[X]}{dt}$ is the instantaneous rate of change of concentration of species X (in mol L⁻¹ s⁻¹).
The minus sign is used for reactants (their concentration decreases with time).
The plus sign is used for products (their concentration increases with time).
Dividing by the coefficient normalises the rate — so $r$ is the same number for every species.

r = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}

This single value $r$ is the intrinsic rate of the reaction, independent of which species you measure.

How to Use It (Step by Step)

Suppose you are given:

$2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}$

And you measure that $\frac{d[\text{H}_2]}{dt} = -0.040 \text{ M/s}$ (negative because H₂ is being consumed).

Step 1: Write the rate relation:

r = -\frac{1}{2}\frac{d[\text{H}_2]}{dt} = -\frac{1}{1}\frac{d[\text{O}_2]}{dt} = \frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt}

Step 2: Plug in the known value:

r = -\frac{1}{2}(-0.040) = +0.020 \text{ M/s}

Step 3: Find the rate for O₂:

-\frac{d[\text{O}_2]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{O}_2]}{dt} = -0.020 \text{ M/s}

Step 4: Find the rate for H₂O:

\frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{H}_2\text{O}]}{dt} = 2r = 0.040 \text{ M/s}

Tip

A quick check: the coefficients tell you the relative rates. Here, H₂ disappears twice as fast as O₂, and H₂O appears at the same rate as H₂ disappears (both coefficient 2). Always verify your numbers match the coefficient ratios.

Common Pitfall to Avoid …

Why this formula?

Factors Affecting Reaction Rate: Understanding the Why Behind the Formulas

In chemical kinetics, the rate of a reaction depends on several factors. The key formulas are not arbitrary — they arise from collision theory and transition state theory. Let’s break down each factor and derive the reasoning step-by-step.

1. Effect of Concentration: The Rate Law

The Formula

For a reaction $\text{aA} + \text{bB} \rightarrow \text{products}$ , the rate is often:

\text{Rate} = k [\text{A}]^m [\text{B}]^n

where $m$ and $n$ are orders (not necessarily equal to $a$ and $b$ ).

Why This Holds

Collision theory: For a reaction to occur, reactant particles must collide with sufficient energy and correct orientation.
Doubling concentration of A doubles the number of A particles per unit volume. This doubles the collision frequency between A and B (if B is constant). Hence, rate ∝ [A].
But the order $m$ $m$ is determined experimentally because:
- Not all collisions are effective.
- Some reactions involve multiple steps (mechanism). The rate depends on the slowest step (rate-determining step), which may involve only some reactants.

Example: For $\text{NO}_2 + \text{CO} \rightarrow \text{NO} + \text{CO}_2$ , the rate law is $\text{Rate} = k[\text{NO}_2]^2$ — because the slow step involves two NO₂ molecules colliding.

2. Effect of Temperature: The Arrhenius Equation

The Formula

k = A e^{-E_a / RT}

where:

$k$ = rate constant
$A$ = frequency factor (collision frequency × orientation factor)
$E_a$ = activation energy (J/mol)
$R$ = gas constant (8.314 J/mol·K)
$T$ = temperature (K)

Why This Holds

Boltzmann distribution: At a given temperature, molecules have a range of kinetic energies. Only those with energy ≥ $E_a$ can react.
The fraction of molecules with energy ≥ $E_a$ is given by the Maxwell–Boltzmann distribution:

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

This is derived from integrating the energy distribution function.

The frequency factor $A$ accounts for how often collisions occur and the fraction with correct orientation.
Temperature increase shifts the distribution to higher energies, exponentially increasing the fraction of molecules that can overcome $E_a$ .

Key insight: A 10°C rise near room temperature roughly doubles the rate — because $e^{-E_a/RT}$ changes significantly.

3. Effect of a Catalyst

No Simple Formula, But a Key Principle

A catalyst lowers the activation energy $E_a$ by providing an alternative reaction pathway (e.g., forming an intermediate).

Why This Works

In the Arrhenius equation, $k = A e^{-E_a/RT}$ .
Lowering $E_a$ increases the exponential factor $e^{-E_a/RT}$ dramatically.
Example: If $E_a$ drops from 100 kJ/mol to 80 kJ/mol at 300 K, the rate constant increases by a factor of:

\frac{e^{-80,000/(8.314 \times 300)}}{e^{-100,000/(8.314 \times 300)}} = e^{(20,000)/(2494)} \approx e^{8.02} \approx 3000

The catalyst itself is not consumed — it participates in the mechanism but is regenerated.

4. Effect of Surface Area (for Heterogeneous Reactions) …

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Reaction rate depends on collision frequency and activation energy. Reducing surface area slows the rate; adding a catalyst speeds up both forward and reverse reactions equally; increasing temperature speeds up the reaction by raising the fraction of molecules with energy above the activation barrier.

Let’s understand the why behind each effect before we write the answer. Reaction rate is governed by collision theory: molecules must collide with sufficient energy (≥ activation energy, $E_a$ ) and proper orientation. Anything that changes the number of effective collisions per second changes the rate.

(a) Surface area of the reactant is reduced

What surface area does: For solid reactants, only the exposed surface can participate in collisions with other molecules (gas or liquid). The interior of the solid is not accessible until the outer layers react away.
Effect of reducing it: When you reduce surface area — for example, by using a solid block instead of powder — fewer reactant molecules are exposed at any given moment. This lowers the frequency of effective collisions between the solid and the other reactant.
Result: The rate of reaction decreases. The reaction becomes slower because the available contact area for reaction is smaller.

Watch out

A common mistake is to think that reducing surface area increases rate because “less area means more concentrated collisions.” That’s wrong — fewer molecules are exposed, so total collisions per second drop.

(b) Catalyst is added in a reversible reaction

What a catalyst does: A catalyst provides an alternative reaction pathway with a lower activation energy ( $E_a$ ). It does not change the enthalpy change ( $\Delta H$ ) or the equilibrium constant ( $K_{eq}$ ).
Effect on forward and reverse rates: In a reversible reaction, the catalyst lowers the activation barrier for both the forward and the reverse reactions by the same amount. This is because the transition state is stabilised equally in both directions.
Result: Both forward and reverse rates increase by the same factor. The system reaches equilibrium faster, but the equilibrium position itself is unchanged.

Tip

Think of a catalyst as a “speed bump remover” on both sides of a hill — you get to the top (transition state) faster from either direction, but the heights of the two sides relative to each other stay the same.

(c) Temperature of the reaction is increased …

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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 27 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Rate constant depends on :$
(a) Temperature
(b) Time
(c) Initial concentration
(d) None of the above
›Reveal solutionSolution

The rate constant k of a reaction depends only on temperature (and the nature of the reactants/catalyst); it is independent of the initial concentration of reactants and of time.

By definition, the rate law is Rate $= k[A]^m[B]^n$ . Here k is a proportionality constant unique to a given reaction at a given temperature — it does NOT change if you alter the starting concentrations, and it does not vary with elapsed time during the reaction. What it DOES depend on is temperature, as captured by the Arrhenius equation:

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

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. For a chemical reaction with rise in temperature by 10°C the rate constant is nearly:$
(a) No change
(b) Tripled
(c) Doubled
(d) Ten time increases
›Reveal solutionSolution

A 10 C temperature rise roughly doubles the rate constant.

As an empirical rule, for many reactions the rate (and rate constant) becomes about two to three times larger for every 10 C rise in temperature. This ratio, k(T+10)/k(T), is called the temperature coefficient and is nearly 2. The reason (from the Arrhenius equation) …

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CBSE 2025Set 56/5/11 markMCQ
$Q. The role of a catalyst is to change : (A) equilibrium constant (B) enthalpy of reaction (C) Gibbs energy of reaction (D) activation energy of reaction$
›Reveal solutionSolution
A catalyst provides an alternative reaction pathway with lower activation energy, speeding up both forward and reverse reactions equally without altering thermodynamic properties. The answer is (D).

Why catalysts work: the energy barrier concept

Every chemical reaction requires reactant molecules to climb an energy hill before they can transform into products. This hill—the activation energy $E_a$ —represents the minimum energy needed to break old bonds and form new ones. Think of it as a mountain pass: molecules must reach the summit before they can descend into the product valley.

A catalyst doesn't push molecules harder or change where they start and end. Instead, it carves a tunnel through the mountain—a new reaction pathway with a lower summit. More molecules now have enough thermal energy to cross, so the reaction speeds up dramatically.

Crucially, the catalyst lowers the barrier in both directions by the same amount. The thermodynamic landscape—the relative energies of reactants and products—remains untouched.

What a catalyst does NOT change

Before examining what changes, let's eliminate what stays constant:

Equilibrium constant $K$

The equilibrium constant depends only on the standard Gibbs energy change: $\Delta G^\circ = -RT \ln K$ . Since a catalyst doesn't alter the stability difference between reactants and products, $K$ remains unchanged. The system reaches equilibrium faster, but the final composition is identical.

Enthalpy of reaction $\Delta H$

This is the heat absorbed or released, determined by the difference in bond energies between products and reactants. A catalyst doesn't strengthen or weaken these bonds—it merely provides an alternative route between them.

Gibbs energy of reaction $\Delta G$

The spontaneity and equilibrium position depend on $\Delta G = \Delta H - T\Delta S$ , both state functions. The catalyst affects how fast the system reaches its natural destination, not where that destination lies.

Watch out

A common misconception is that catalysts "shift equilibrium." They don't. Both forward and reverse rates increase proportionally, so the ratio $k_f/k_r = K$ stays constant.

What a catalyst DOES change

Activation energy $E_a$ This is the catalyst's sole thermodynamic target. By stabilizing the transition state through temporary bond formation (in homogeneous catalysis) or by adsorbing reactants in favorable orientations (in heterogeneous catalysis), the catalyst lowers the energy barrier. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following does not affect the rate of reaction?$
(a) delta H
(b) Temperature
(c) Concentration
(d) Catalyst
›Reveal solutionSolution
Rate of reaction depends on how fast reactants are converted to products (a kinetic quantity), while ΔH is a thermodynamic quantity describing the energy difference between reactants and products — the two are independent.

Temperature: raises rate strongly (more molecules cross the activation-energy barrier; Arrhenius equation k = A·e^(−Ea/RT)).

Concentration: higher concentration of reactants increases collision frequency, increasing rate (rate law dependence).

Catalyst: provides an alternate pathway with lower activation energy, increasing rate without being consumed. …
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CBSE 2025Set ANNUAL1 mark
$Q. Assertion [A] : A catalyst increases the rate of reaction without itself undergoing any permanent chemical change. Reason [R] : A catalyst does not alter Gibbs energy, ΔG of a reaction.$
›Reveal solutionSolution

Both statements are individually true, but the reason a catalyst speeds up a reaction is that it LOWERS THE ACTIVATION ENERGY by providing an alternative pathway — not because ΔG is unchanged. ΔG being unaffected explains why equilibrium is unshifted, not why the rate increases.

[A]: A catalyst does increase reaction rate without itself being permanently consumed (it is regenerated at the end of the catalytic cycle) — TRUE.

[R]: A catalyst indeed does not alter the standard Gibbs energy change (ΔG) of a reaction, since ΔG is a thermodynamic (state-function) quantity depending only on the initial and final states, not on the pathway — TRUE.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. What happens to the value of ΔG for a reaction when a catalyst is added to it ?$
(a) It increases
(b) It decreases
(c) It remains unchanged
(d) It is undefined on addition of catalyst
›Reveal solutionSolution

A catalyst is a thermodynamic bystander — it changes the rate of a reaction, not its free energy change.

ΔG (Gibbs free energy change) is a state function that depends only on the initial and final states (reactants and products) of a reaction, not on the pathway taken to reach the products. A catalyst works by providing an alternative reaction pathway with a lower activation energy, $E_a$ , which speeds up both the forward and reverse reactions equally, but it does not alter the energies of the reactants or products themselves. Hence the equilibrium position and ΔG (and …

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CBSE 2024Set 56/2/11 markMCQ
$Q. The addition of catalyst during a chemical reaction alters which of the following quantities of the reaction ? (A) Enthalpy (B) Activation energy (C) Entropy (D) Internal energy$
›Reveal solutionSolution
A catalyst provides an alternative reaction pathway with a lower activation energy barrier, speeding up both forward and reverse reactions without changing the thermodynamic properties of reactants or products. The answer is (B) Activation energy.

Why catalysts work: the energy landscape

Chemical reactions occur when reactant molecules collide with enough energy to break existing bonds and form new ones. The minimum energy required for this transformation is the activation energy $E_a$ . Think of it as an energy hill that molecules must climb before they can slide down into the product valley.

A catalyst doesn't push molecules harder or change where they start and end. Instead, it offers a different route over a lower hill—a new mechanism with intermediate steps that require less energy to traverse. The reactants and products remain chemically identical, so their inherent energy content (enthalpy, internal energy) and disorder (entropy) are unchanged.

Examining each quantity

Let's see what a catalyst does and doesn't alter:

Activation energy $E_a$

This is precisely what a catalyst reduces. By stabilizing the transition state or forming intermediate complexes, the catalyst lowers the energy barrier. For example, enzymes in biological systems can reduce $E_a$ by factors of $10^6$ or more, which is why reactions that would take years occur in milliseconds in living cells.

Enthalpy $\Delta H$

Enthalpy change is the difference in heat content between products and reactants:

$\Delta H = H_{\text{products}} - H_{\text{reactants}}$
Since the catalyst doesn't change the chemical identity of reactants or products, their bond energies and hence their enthalpies remain the same. The reaction is still exothermic or endothermic by the same amount.

Entropy $\Delta S$

Entropy measures the disorder or number of accessible microstates. The initial and final states of the system are identical with or without a catalyst, so $\Delta S$ for the overall reaction is unchanged. The catalyst may create transient intermediate states, but these don't affect the entropy difference between reactants and products.

Internal energy $\Delta U$

Internal energy is related to enthalpy by $\Delta H = \Delta U + \Delta(PV)$ . For the same reason enthalpy is unaffected, internal energy change is also constant. The thermodynamic state functions depend only on the initial and final states, not the path taken. …
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CBSE 2023Set ANNUAL1 mark
$Q. Write two factors that affect the rate of reaction.$
›Reveal solutionSolution
The rate of a chemical reaction increases with an increase in the concentration of reactants and with an increase in temperature.

Concentration of reactants: According to the law of mass action, the rate of reaction is directly proportional to the concentration(s) of the reacting species (raised to their respective orders). Increasing concentration increases the frequency of effective collisions between reactant molecules, increasing the rate. …
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CBSE 2023Set ANNUAL1 mark
$Q. When the added substance reduces the rate of reaction, then it is called ________ in place of catalyst.$
›Reveal solutionSolution

A substance that slows down (retards) a reaction, opposite in effect to a catalyst, is called an inhibitor or negative catalyst.

A catalyst speeds up a reaction without itself being consumed, by providing an alternative pathway of lower activation energy. When an added substance instead REDUCES the rate of reaction, it is called an inhibitor (or neg …

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CBSE 2023Set ANNUAL1 mark
$Q. Write the name of enzyme used in decomposition of urea into ammonia and carbondioxide.$
›Reveal solutionSolution

The enzyme urease specifically catalyses the breakdown of urea into ammonia and carbon dioxide, a classic textbook example of enzyme catalysis.

Urease is a biological (enzyme) catalyst that speeds up the hydrolysis reaction: NH2-CO-NH2 + H2O --urease--> 2NH3 + CO2. This is one of the standard examples use …

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CBSE 2022Set ANNUAL1 mark
$Q. State law of Mass action. OR What is active mass?$
›Reveal solutionSolution

The Law of Mass Action states that reaction rate is proportional to the product of the reactants' active masses (molar concentrations), each raised to the power of its stoichiometric coefficient.

Statement (Guldberg and Waage, 1864): At a constant temperature, the rate of a chemical reaction at any given instant is directly proportional to the product of the 'active masses' (i.e. molar concentrations, expressed in mol/L) of the reacting species, each concentration term raised to a power equal to that species' stoichiometric coefficient in the balanced chemical equation.

Mathematical form: For a general reaction

$aA + bB \longrightarrow \text{products}$

the law of mass action gives:

$\text{Rate} \propto [A]^a[B]^b$

This relationship is the historical basis both for writing equilibrium constant expressions (at equilibrium, forward rate = backward rate) and, in its kinetic form, for rate laws of elementary reactions.

…

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CBSE 2020Set ANNUAL1 mark
$Q. Read the given passage and answer the following questions: A substance which alters the rate of chemical reaction without itself undergoing any change in mass and chemical composition at the end of reaction is called catalyst. It may be noted that a catalyst which increase the speed of a reaction are called positive catalyst and catalyst which decrease the speed of reaction are called negative catalyst. The promoters are substances that enhance the activity of catalyst and poisons which decrease the activity of catalyst. What is catalyst?$
›Reveal solutionSolution

A catalyst provides an alternate reaction pathway of lower activation energy, speeding up the reaction, while being chemically unchanged at the end.

A catalyst is a substance that changes the rate of a chemical reaction without itself undergoing any permanent change in mass and chemical composition at the end of the reaction. It works by adsorbing reactant molecules or forming a short-lived intermediate, opening up an alternative reaction pathway that has a lower activation energy than the uncatalysed path. Because more reactant molecules can then cross the (lower) energy barrier at a given temperature, the reaction proceeds faster. Although the catalyst actively participates in intermediate steps of …

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Q.How will the rate of the reaction be affected when

(a) Surface area of the reactant is reduced

(b) Catalyst is added in a reversible reaction

(c) Temperature of the reaction is increased …

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