Explain, using the idea of molecular collisions, why increasing the concentration of a reactant generally increases the rate of a reaction.

Question

Accepted Answer

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

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 $r$ .

Factors affecting reaction rate, including concentration, temperature, catalysts and surface area, are covered in the NCERT/CBSE Class 12 Chemistry Chemical Kinetics chapter, and ‘factors affecting rate of reaction class 12’ is a very commonly searched important-question topic for board exams, JEE Main and NEET. This concept is also frequently tested through application-based scenario questions in competitive chemistry papers.

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)

The Reasoning

For reactions involving solids (e.g., $\text{Zn} + \text{HCl}$ ), reaction occurs only at the surface.
Rate ∝ surface area because more reactant molecules are exposed for collision.
Derivation: For a fixed mass, breaking a solid into smaller pieces increases the total surface area. Example: A 1 cm cube has area 6 cm². If cut into 1 mm cubes (1000 cubes), total area = $1000 \times 6 \times (0.1)^2 = 60$ cm² — 10 times larger.

Thus: More surface → more collisions per unit time → higher rate.

Summary Table

Factor	Formula/Principle	Why It Works
Concentration	$\text{Rate} = k[\text{A}]^m$	More particles → more collisions
Temperature	$k = A e^{-E_a/RT}$	More molecules have energy ≥ $E_a$
Catalyst	Lowers $E_a$	Exponential increase in $k$
Surface area	Rate ∝ area	More exposed reactant sites

Final Takeaway for Exams

Never memorize blindly — always connect the formula to the physical picture (collisions, energy barriers, orientation).
For Arrhenius, remember the log form for calculations:

\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}

This is a straight line ( $y = mx + c$ ) — slope gives $E_a$ .

For rate laws, the order is experimental — do not assume it matches stoichiometric coefficients.

Concentration is a measure of how many molecules of a substance are packed into a given volume. Increasing the concentration of a reactant increases the number density of its molecules, so on average each molecule of that reactant, and each molecule of any other reactant it must meet, encounters the other far more often per unit time -- exactly analogous to how people cross paths more often in a crowded room than in a nearly empty one. Since collision theory holds that reaction can occur only when reacting particles physically meet, this higher collision frequency directly increases the number of collisions occurring per second, and, since a roughly constant fraction of these collisions are energetic and correctly oriented enough to be effective (collision theory, developed later in this chapter), the overall rate of effective, reaction-producing collisions rises correspondingly. [!ANSWER] Raising concentration increases the number of reactant molecules per unit volume, which increases collision frequency; since only colliding molecules can react, a higher collision frequency generally produces a higher observed rate.

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:

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

As temperature T increases, k increases (more molecules cross the activation energy barrier). This is why rate constants are always quoted at a specific temperature.

✓Final answer
$(a) Temperature — the rate constant changes only with temperature (via the Arrhenius equation), not with time or initial concentration.$
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) is that a higher temperature sharply increases the fraction of molecules with energy exceeding the activation energy.

✓Final answer
$(c) Doubled.$
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.

The Arrhenius equation shows why this matters:
$k = A e^{-E_a/RT}$
Even a modest reduction in $E_a$ produces an exponential increase in the rate constant $k$ . A reaction that would take hours might complete in minutes.

Tip

Remember the mnemonic: catalysts are kinetic helpers, not thermodynamic changers. They affect the path (activation energy), not the destination (equilibrium, enthalpy, Gibbs energy).

The mechanism in action

Consider the decomposition of hydrogen peroxide, catalyzed by iodide ions:

Uncatalyzed: $2\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_2$ (high $E_a$ , slow)

Catalyzed: Two-step pathway via $\text{H}_2\text{O}_2 + \text{I}^- \to \text{H}_2\text{O} + \text{IO}^-$ followed by $\text{H}_2\text{O}_2 + \text{IO}^- \to \text{H}_2\text{O} + \text{O}_2 + \text{I}^-$

Each step has a lower activation energy than the direct route. The iodide ion is regenerated, unchanged. The overall $\Delta H$ and $\Delta G$ are identical to the uncatalyzed reaction.

✓Final answer
$The correct option is (D) — a catalyst changes the activation energy of reaction.$
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.

ΔH (enthalpy change): tells us whether the reaction is exothermic or endothermic overall and by how much energy, but says nothing about how fast the reaction proceeds — a reaction can be highly exothermic yet extremely slow (e.g., diamond to graphite), or only mildly exothermic yet very fast. Rate depends on the activation energy (the barrier height), not on ΔH (the net energy difference).

✓Final answer
$(a) ΔH does not affect the rate of reaction.$
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.

However, [R] does not explain [A]: a catalyst speeds up a reaction by providing an alternate reaction pathway with LOWER ACTIVATION ENERGY (a kinetic effect), which is a completely different reason from the unchanged ΔG (a thermodynamic fact, which is why the catalyst does not shift the position of equilibrium).

✓Final answer
$(b) Both [A] and [R] are true, but [R] is not the correct explanation of [A].$
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 equilibrium constant K, since $\Delta G^\circ = -RT\ln K$ ) stay exactly the same with or without a catalyst.

✓Final answer
$Adding a catalyst leaves ΔG unchanged (option c) — it only lowers the activation energy and speeds up attainment of equilibrium.$

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 solution

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.

Important

Thermodynamic vs. kinetic quantities: Catalysts affect kinetics (how fast equilibrium is reached) but not thermodynamics (where equilibrium lies). Enthalpy, entropy, internal energy, and Gibbs free energy are all state functions—path-independent—so they're unaffected by the mechanism.

Tip

A quick way to remember: catalysts change the speed of reaching equilibrium (by lowering $E_a$ ) but not the position of equilibrium (which depends on $\Delta G = \Delta H - T\Delta S$ ).

Watch out

Students sometimes think a catalyst "provides energy" to the reaction. This is incorrect. A catalyst lowers the energy requirement but doesn't supply or remove energy from the system. The heat released or absorbed ( $\Delta H$ ) remains exactly the same.

The energy profile

Consider the energy diagram for a reaction with and without a catalyst:

Feature	Without catalyst	With catalyst
Reactant energy	$E_R$	$E_R$ (same)
Product energy	$E_P$	$E_P$ (same)
Activation energy	$E_a$ (high)	$E_a'$ (lower)
$\Delta H$	$E_P - E_R$	$E_P - E_R$ (same)

The peaks are at different heights, but the starting and ending points are identical.

✓Final answer

The correct option is (B) Activation energy .

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.

Temperature: Raising the temperature increases the average kinetic energy of the reacting molecules, so a larger fraction of molecules possess energy equal to or greater than the activation energy (as described by the Arrhenius equation, k = A e^(-Ea/RT)). This increases the frequency of effective collisions and hence the rate; typically the rate roughly doubles for every 10 degC rise in temperature.

(Other factors affecting rate include the nature of reactants, surface area of solid/heterogeneous reactants, presence of a catalyst, and pressure for gas-phase reactions.)

✓Final answer
$Concentration of reactants and temperature are two factors that affect the rate of reaction.$
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 negative catalyst) - e.g., glycerol is added as a negative catalyst/inhibitor to slow the decomposition of H2O2.

✓Final answer
$Inhibitor (negative catalyst).$
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 used to illustrate enzyme catalysis - highly specific catalytic action of a biocatalyst on one particular substrate.

✓Final answer
$Urease.$
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.

'Active mass' here specifically means the molar concentration of a reacting species (moles per litre); for pure solids and liquids taking part in a reaction, the active mass is taken as constant (essentially 1) and so is normally omitted from the expression.

✓Final answer
$aA+bB\to\text{products}$
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 the mechanism, it is regenerated in a later step, so its total amount and chemical identity are unchanged at the end.

✓Final answer
$A catalyst is a substance that alters the rate of a chemical reaction without itself undergoing any permanent change in mass and chemical composition at the end of the reaction.$

Q.Explain, using the idea of molecular collisions, why increasing the concentration of a reactant generally increases the rate of a reaction.

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