The role of a catalyst is to change _______. (i) gibbs energy of reaction. (ii) enthalpy of reaction. (iii) ac

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Q. The role of a catalyst is to change ______________.

(i) gibbs energy of reaction.

(ii) enthalpy of reaction.

(iii) activation energy of reaction.

(iv) equilibrium constant.

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Concept understanding — Average Rate Of Reaction

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

Watch out

Do not write $\frac{d[\text{reactant}]}{dt}$ as a positive number and then forget the minus sign. The rate of change of a reactant is negative (concentration falls). The minus sign in the definition flips it to a positive $r$ . If you skip the sign, you will get the wrong magnitude for other species.

Why This Matters

In exams (JEE, NEET, etc.), you will often be given the rate for one species and asked to find the rate for another. The stoichiometric relation is the only tool you need — no extra formulas. It also appears in more advanced topics like the rate law (where the exponents are not the coefficients) — but that is a separate concept. Reaction rate stoichiometry is purely about the definition of the reaction rate itself.

Final takeaway: The coefficients in the balanced equation are the conversion factors between the rates of different species. Always normalise by dividing by the coefficient to get the universal reaction rate $r$ .

Average rate of reaction is one of the first ideas introduced in the NCERT/CBSE Class 12 Chemistry Chemical Kinetics chapter, and ‘average rate of reaction formula’ or ‘average vs instantaneous rate’ are common important-question topics in board exams and JEE Main chemistry. A solid grasp of this basic definition is also assumed in nearly every subsequent kinetics numerical asked in NEET and state CETs.

Why this formula?

Average Rate of Reaction — Why the Formula Holds

Let’s build this from the ground up. The goal is to understand why the average rate formula looks the way it does — not just memorise it.

1. What does "rate of reaction" mean physically?

A chemical reaction changes the concentration of reactants (decreasing) and products (increasing) over time.

Rate = how fast this change happens.
If you measure the change over a finite time interval, you get the average rate.

2. The core idea: change per unit time

For any quantity $X$ that changes from $X_1$ to $X_2$ over time $t_1$ to $t_2$ :

\text{Average rate of change of } X = \frac{\Delta X}{\Delta t} = \frac{X_2 - X_1}{t_2 - t_1}

This is just the slope of the straight line connecting the two points on a concentration vs. time graph.

3. Applying this to a reaction

Consider a simple reaction:

A \rightarrow B

Reactant A is consumed: $[A]$ decreases.
Product B is formed: $[B]$ increases.

For reactant A (disappearing):

\text{Average rate} = -\frac{\Delta [A]}{\Delta t}

Why the minus sign?

Because $\Delta [A] = [A]_2 - [A]_1$ is negative (concentration drops). The rate itself must be positive (speed is never negative). So we multiply by $-1$ .

For product B (appearing):

\text{Average rate} = +\frac{\Delta [B]}{\Delta t}

Here $\Delta [B]$ is positive, so no minus sign needed.

4. The general formula for any reaction

For a balanced reaction:

aA + bB \rightarrow cC + dD

The average rate is defined per mole of reaction — so it’s the same number regardless of which species you track.

We divide each $\frac{\Delta [\text{species}]}{\Delta t}$ by its stoichiometric coefficient:

\text{Average rate} = -\frac{1}{a} \frac{\Delta [A]}{\Delta t} = -\frac{1}{b} \frac{\Delta [B]}{\Delta t} = +\frac{1}{c} \frac{\Delta [C]}{\Delta t} = +\frac{1}{d} \frac{\Delta [D]}{\Delta t}

Why divide by the coefficient?

If 2 moles of A disappear for every 1 mole of C formed, then $\frac{\Delta [A]}{\Delta t}$ is twice as large as $\frac{\Delta [C]}{\Delta t}$ . Dividing by the coefficient normalises them to the same "per mole of reaction" rate.

5. Key exam point: the formula in one line

For any species $X$ with stoichiometric coefficient $\nu_X$ (negative for reactants, positive for products):

\text{Average rate} = \frac{1}{\nu_X} \frac{\Delta [X]}{\Delta t}

$\nu_X$ is negative for reactants → the minus sign is already built in.
$\nu_X$ is positive for products.

6. Why this is the average rate (not instantaneous)

Average rate uses a finite $\Delta t$ — it’s the slope of the chord between two points.
Instantaneous rate uses $\Delta t \rightarrow 0$ — it’s the slope of the tangent at a single point.

The average rate formula is just the discrete version of the derivative:

\text{Instantaneous rate} = \frac{1}{\nu_X} \frac{d[X]}{dt}

Summary — the "why" in one sentence

The average rate formula holds because it measures change in concentration per unit time, uses a minus sign to keep rates positive for reactants, and divides by stoichiometric coefficients to give a single, comparable value for the whole reaction.

Always remember:

$\Delta [\text{reactant}]$ is negative → minus sign makes it positive.
$\Delta [\text{product}]$ is positive → no minus sign.
Divide by coefficient → normalise to "per mole of reaction".

A catalyst provides an alternative reaction pathway with a lower activation energy, so it changes the activation energy of reaction without affecting thermodynamic quantities like Gibbs energy, enthalpy, or the equilibrium constant.

The key idea here is the distinction between thermodynamics and kinetics. A catalyst is a substance that speeds up a reaction without being consumed — but it does not change the inherent energy difference between reactants and products. Let's see why each option stands or falls.

Gibbs energy of reaction ( $\Delta G$ ) is a thermodynamic property. It depends only on the initial and final states (reactants and products), not on the path taken. A catalyst does not alter these states, so $\Delta G$ remains unchanged. In fact, if a catalyst could change $\Delta G$ , it would violate the laws of thermodynamics by making a non-spontaneous reaction spontaneous — which is impossible.
Enthalpy of reaction ( $\Delta H$ ) is also a state function. It measures the heat absorbed or released when reactants convert to products. Again, the catalyst does not change the reactants or products themselves, so $\Delta H$ stays the same.
Activation energy ( $E_a$ ) is the energy barrier that must be overcome for the reaction to occur. This is where the catalyst acts. It provides an alternative pathway — a different mechanism — with a lower activation energy. More molecules now have enough energy to cross the barrier, so the reaction rate increases. This is the only quantity among the options that a catalyst changes.

k

Equilibrium constant ( $K$ ) is determined by the standard Gibbs energy change: $\Delta G^\circ = -RT \ln K$ . Since $\Delta G^\circ$ is unchanged by a catalyst, $K$ remains the same. A catalyst speeds up both the forward and reverse reactions equally, so the equilibrium position is reached faster — but it is the same equilibrium.

Watch out

A common mistake is to think that a catalyst changes the equilibrium constant because it makes the reaction go faster. Speed and equilibrium position are different concepts. The catalyst does not shift the balance; it only helps you get there sooner.

Tip

Think of a catalyst as a tunnel through a mountain instead of a path over it. The tunnel is shorter (lower activation energy), but the height of the mountain itself (the energy difference between the two sides) hasn't changed. That height is $\Delta G$ or $\Delta H$ — untouched.

✓Final answer

The correct option is (iii) activation energy of reaction.

Let's break this down step-by-step — first the concept, then the common mistakes, and finally how to avoid them.

The Concept in One Line

A catalyst provides an alternative reaction pathway with lower activation energy. It does not change the thermodynamic quantities of the reaction.

Correct Answer

(iii) activation energy of reaction.

Common Mistakes & How to Avoid Them

✗ Mistake 1: Choosing "Gibbs energy of reaction" (option (i))

Why students pick it:

They confuse activation energy (kinetic barrier) with Gibbs free energy change (thermodynamic spontaneity).

Why it's wrong:

$\Delta G^\circ$ (standard Gibbs energy change) depends only on the initial and final states of reactants and products — not on the path. A catalyst does not change the initial or final states.

✓ How to avoid:

Remember:

Catalyst → affects rate (kinetics)
$\Delta G$ → tells if reaction is spontaneous (thermodynamics) They are separate concepts.

✗ Mistake 2: Choosing "Enthalpy of reaction" (option (ii))

Why students pick it:

They think "catalyst lowers energy barrier" means it changes the total heat released or absorbed.

Why it's wrong:

$\Delta H$ (enthalpy change) is a state function — same logic as $\Delta G$ . The catalyst does not alter the enthalpy difference between products and reactants.

✓ How to avoid:

Draw an energy profile diagram. The vertical distance between reactants and products (enthalpy change) stays the same. Only the peak height (activation energy) changes.

✗ Mistake 3: Choosing "Equilibrium constant" (option (iv))

Why students pick it:

They think "catalyst speeds up both forward and backward reactions equally" means it changes the equilibrium constant.

Why it's wrong:

A catalyst speeds up both forward and reverse reactions by the same factor. The equilibrium constant $K$ depends only on $\Delta G^\circ$ :

\Delta G^\circ = -RT \ln K

Since $\Delta G^\circ$ is unchanged, $K$ is unchanged.

✓ How to avoid:

Remember:

Catalyst → reaches equilibrium faster
$K$ → tells where equilibrium lies (unchanged by catalyst)

Quick Summary Table

Option	What it changes?	Does catalyst change it?
(i) Gibbs energy	Spontaneity	✗ No
(ii) Enthalpy	Heat change	✗ No
(iii) Activation energy	Energy barrier	✓ Yes
(iv) Equilibrium constant	Position of equilibrium	✗ No

Final tip for exams:

Whenever you see "catalyst" in a multiple-choice question, immediately think kinetics (rate, activation energy) — not thermodynamics or equilibrium.

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

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. In a reaction 2HI -> H2 + I2, the concentration of HI decreases from 0.5 mol L^-1 to 0.4 mol L^-1 in 10 minutes. What is the rate of reaction during this interval?$
(a) 5 x 10^-3 M min^-1
(b) 2.5 x 10^-3 M min^-1
(c) 5 x 10^-2 M min^-1
(d) 2.5 x 10^-2 M min^-1
›Reveal solutionSolution

For 2HI → H2 + I2, the rate of reaction is defined using the stoichiometric coefficient of HI: Rate = −(1/2)(Δ[HI]/Δt).

Given: [HI] falls from 0.5 mol L⁻¹ to 0.4 mol L⁻¹ over 10 minutes.

Δ[HI] = 0.4 − 0.5 = −0.1 mol L⁻¹, over Δt = 10 min

Since 2 mol of HI are consumed for every 1 unit of reaction progress, the rate of reaction is:

Rate = −(1/2) × (Δ[HI]/Δt) = −(1/2) × (−0.1/10) = −(1/2) × (−0.01) = 0.005 mol L⁻¹ min⁻¹ = 5 × 10⁻³ M min⁻¹

✓Final answer
$(a) 5 x 10^-3 M min^-1.$
GUJCET 2023Set 091 markMCQ
$= K[A]^{1/2}\,[B]^{3/2}$
›Reveal solutionSolution

Unit of $k$ for order $n$ is $(\text{mol L}^{-1})^{1-n}\,s^{-1}$ .

Concept: Overall order $=\tfrac12+\tfrac32=2$ . For a second-order reaction,
$[k]=(\text{mol L}^{-1})^{1-2}\,s^{-1}=\text{mol}^{-1}\,\text{L}\,s^{-1}.$
✓Final answer
$Mol^{-1}.lit.Sec^{-1}$

ANSWER: (C)
GUJCET 2021Set 151 markMCQ
$K = 4.5 \times 10^{-4}$
›Reveal solutionSolution

Units $\text{L mol}^{-1}\text{s}^{-1}$ = $M^{-1}s^{-1}$ ⇒ second order.

Concept: The unit of a rate constant is $\text{mol}^{1-n}\,L^{n-1}\,s^{-1}$ for order $n$ . Solving $L\,mol^{-1}\,s^{-1}$ :
$1-n = -1 \Rightarrow n = 2$
✓Final answer
$(B) Second$

ANSWER: (B)

←Previous section · answersAnswers to Intext Questions →Next questionQ2 — In the presence of a catalyst, the heat evolved or absorbed during the…

Q.The role of a catalyst is to change ______________.

Concept understanding — Average Rate Of Reaction

Reaction Rate Stoichiometry – From Intuition to Precision

The Intuitive Picture

The Precise Statement

How to Use It (Step by Step)

Common Pitfall to Avoid

Why This Matters

Why this formula?

Average Rate of Reaction — Why the Formula Holds

1. What does "rate of reaction" mean physically?

2. The core idea: change per unit time

3. Applying this to a reaction

For reactant A (disappearing):

For product B (appearing):

4. The general formula for any reaction

5. Key exam point: the formula in one line

6. Why this is the average rate (not instantaneous)

Summary — the "why" in one sentence

Concept: Catalysis and Reaction Kinetics

Method: Thermodynamic vs. Kinetic Control

Final Answer

The Concept in One Line

Correct Answer

Common Mistakes & How to Avoid Them

✗ Mistake 1: Choosing "Gibbs energy of reaction" (option (i))

✗ Mistake 2: Choosing "Enthalpy of reaction" (option (ii))

✗ Mistake 3: Choosing "Equilibrium constant" (option (iv))

Quick Summary Table

Q.The role of a catalyst is to change ______________.

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