Factors Affecting The Rate of a Chemical Reaction

Q: What is the temperature coefficient of a reaction?

The temperature coefficient mu is the ratio of rate constants at temperatures differing by 10 K: mu = k(T+10)/k(T). For most reactions mu lies between 2 and 3, meaning the rate roughly doubles or triples for every 10 K rise in temperature.

ChemistryChemical KineticsFor NEET aspirants

The rate of a chemical reaction is not fixed. It can be changed by adjusting five main factors: concentration of reactants, temperature, nature of the reacting substances, surface area (for heterogeneous reactions), and the presence of a catalyst. Radiation can also drive photochemical reactions. Each factor works through a physical mechanism explained by collision theory: reactions happen when molecules collide with enough energy and correct orientation, so anything that increases the number, energy, or effectiveness of collisions speeds up the reaction.

Key Formulas - Quick Reference

Rate law: $rate = k [A]^{m} [B]^{n}$ ; rate rises with concentration.
Temperature coefficient: $μ = \frac{k _{T + 10}}{k _{T}} \approx 2 to 3$ for most reactions.
Arrhenius equation: $k = A e^{- E_{a} / R T}$
Van't Hoff (two temperatures): $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$
Catalyst effect: $\frac{k _{c a t}}{k} = e^{(E_{a} - E_{a}^{'}) / R T}$ where $E_{a}^{'} < E_{a}$

1. The Five Main Factors

The rate of a chemical reaction depends on:

Concentration of the reactants (or partial pressure for gases).
Temperature of the system.
Nature of the reacting substances (bonds broken, ionic vs covalent).
Surface area of solids in heterogeneous reactions.
Presence of a catalyst.

A sixth factor, exposure to radiation, is important for photochemical reactions. We examine each of the five main factors below.

2. Concentration of Reactants

Increasing the concentration of a reactant increases the number of reactant particles per unit volume, which in turn increases the frequency of collisions. Since more collisions per second means more effective collisions per second, the rate rises.

Mathematically, this is expressed in the rate law:

$rate = k [A]^{m} [B]^{n}$

For gaseous reactants, concentration is replaced by partial pressure ( $p = c R T$ ). Increasing pressure of a gaseous system compresses it and raises concentration, so the rate rises.

3. Effect of Temperature

The rate of almost every reaction rises sharply with temperature. As a rough rule, the rate of a reaction roughly doubles for every $1 0^{\circ}$ rise in temperature. This is quantified by the temperature coefficient:

Temperature coefficient

μ = \frac{k _{T + 10}}{k _{T}} = \frac{rate constant at ( T + 10 ) K}{rate constant at T K} \approx 2 to 3

Usually $μ$ is measured between $298 K$ and $308 K$ .

3.1 Why does a small temperature rise cause such a large rate increase?

At first sight it seems surprising: a $10 K$ rise increases the average speed of molecules by only about $308/298 \approx 1.016$ , that is, a 1.6% increase in collision frequency. Yet the rate can double. Where does the factor of two come from?

The answer is that only effective collisions - those with kinetic energy at least equal to the activation energy $E_{a}$ - lead to reaction. From the Maxwell-Boltzmann distribution, the fraction of molecules with energy $\geq E_{a}$ is $e^{- E_{a} / R T}$ . This fraction is very sensitive to temperature: a small rise in $T$ produces a large rise in the number of molecules that can react.

Maxwell–Boltzmann distribution. Raising temperature from

T_{1}

T_{2}

broadens the distribution and shifts more molecules above the activation energy

E_{a}

(shaded), sharply increasing the fraction of effective collisions.

The rate can be written as

$rate = f \times z$

where $z$ is the collision frequency and $f$ is the fraction of effective (sufficiently energetic and correctly oriented) collisions. The rise in $z$ is small (about $1.6%$ for $10 K$ ); the rise in $f$ is dramatic (often $\times 2$ or more). Hence the observed sharp temperature sensitivity of the rate.

Van't Hoff equation: The temperature dependence of the equilibrium constant

K

is described by

lo g \frac{K _{2}}{K _{1}} = \frac{Δ H}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

. The analogous relation for the rate constant is the two-point form of the Arrhenius equation, covered in the next concept.

4. Nature of the Reacting Substances

Reactions between ions in solution are essentially instantaneous because ionic bonds are already broken and only rearrangement is needed. Reactions involving the breaking of stronger covalent bonds are slower.

A classic contrast: both $F e^{2 +}$ and $C_{2} O_{4}^{2 -}$ are oxidised by $M n O_{4}^{-}$ in acidic medium, but at very different rates:

Solved Example 1

Compare the rates of the following two reactions with

K M n O_{4}

in acid solution:
(i)

M n O_{4}^{-} + 8 H^{+} + 5 F e^{2 +} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

(fast, instantaneous)
(ii)

2 M n O_{4}^{-} + 16 H^{+} + 5 C_{2} O_{4}^{2 -} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

(slow, needs warming)

Solution:

Reaction (i) involves only electron transfer between hydrated ions. No covalent bonds need breaking, so it is essentially instantaneous.

Reaction (ii) requires breaking $C - C$ and $C - O$ covalent bonds inside the oxalate ion, so it needs additional energy (thermal activation) and is much slower.

5. Surface Area

For heterogeneous reactions (e.g. a solid reacting with a gas or liquid), the reaction takes place at the surface of the solid. Increasing the surface area exposes more reactant to the other phase, so the rate rises.

Examples:

Powdered $C a C O_{3}$ reacts with dilute $H C l$ much faster than a single lump of the same mass. The powder has vastly more surface area.
Coal dust in a mine burns explosively even though a coal block is difficult to ignite.
A catalyst is often used in finely divided form (e.g. platinum black, Raney nickel) to maximise its active surface.

6. Catalyst

A catalyst is a substance that increases the rate of a reaction without itself being consumed at the end of the reaction. The phenomenon is called catalysis. A substance that decreases the rate of reaction is called a negative catalyst or an inhibitor.

6.1 Auto-catalysis

Sometimes one of the products of a reaction acts as a catalyst for that reaction. This is called auto-catalysis and the substance is called an auto-catalyst. The classic example is the titration of oxalic acid with acidified $K M n O_{4}$ :

$2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} C_{2} O_{4} \to K_{2} S O_{4} + 8 H_{2} O + 10 C O_{2} + 2 M n S O_{4}$

The reaction is slow at first, then speeds up as $M n^{2 +}$ (the auto-catalyst) is produced.

6.2 General characteristics of a catalyst

A catalyst does not initiate the reaction; it only accelerates a reaction that is already thermodynamically feasible.
A small amount of catalyst is usually sufficient to catalyse a large amount of reactant.
A catalyst does not change the position of equilibrium or the equilibrium constant $K$ . It only reduces the time needed to reach equilibrium by speeding up both the forward and backward reactions equally.
A catalyst does not change $Δ G^{\circ}$ , $Δ H$ , or the identity of reactants and products.
A catalyst is specific: a substance that catalyses one reaction may have no effect on another.
A catalyst provides an alternative pathway with lower activation energy $E_{a}$ .

6.3 Types of catalysts

Type	Description	Example
Homogeneous	Catalyst and reactants in the same phase.	$N O (g)$ catalyses $2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)$ (lead chamber process).
Heterogeneous	Catalyst in a different phase, usually solid catalysing gas or solution.	Fe catalyses $N_{2} + 3 H_{2} \to 2 N H_{3}$ (Haber process); $V_{2} O_{5}$ catalyses $S O_{2} \to S O_{3}$ (contact process).
Enzymes (biocatalysts)	Protein catalysts of high specificity for biological reactions.	Amylase catalyses hydrolysis of starch; urease catalyses hydrolysis of urea.
Auto-catalyst	A product of the reaction that catalyses the same reaction.	$M n^{2 +}$ in $K M n O_{4}$ -oxalate titration.

6.4 Promoters and poisons

A promoter increases the efficiency of a catalyst (e.g. $M o$ acts as a promoter for iron in the Haber process). A catalyst poison deactivates a catalyst (e.g. traces of arsenic poison $V_{2} O_{5}$ in the contact process; that is why gas is purified before entering the catalyst chamber).

6.5 How a catalyst reduces the activation energy

A catalyst drives the reaction through a different route in which the energy barrier is lower. Because $E_{a}$ is smaller, a much larger fraction of collisions have enough energy to react.

Effect of catalyst on activation energy. A catalyst provides an alternative reaction path with lower activation energy ($E_a' < E_a$). The overall enthalpy change $\Delta H$ between reactants and products is unchanged.

Quantitatively, if $E_{a}$ and $E_{a}^{'}$ are the activation energies without and with catalyst, and $A$ is the same pre-exponential factor for both paths, the ratio of rate constants is

$\frac{k _{c a t}}{k} = \frac{A e ^{- E_{a}^{'} / R T}}{A e ^{- E_{a} / R T}} = e^{(E_{a} - E_{a}^{'}) / R T}$

Since $E_{a}^{'} < E_{a}$ , the exponent is positive and $k_{c a t} > k$ : the catalyst has multiplied the rate constant. A modest reduction of $E_{a}$ can accelerate the reaction by many orders of magnitude.

Solved Example 2

A hydrogenation reaction is carried out at

500 K

. In the presence of a catalyst, the same rate is obtained at

400 K

. The catalyst lowers the activation energy by

20 kJ/mol

. Find the activation energy in the absence of the catalyst.

Solution:

Let the activation energy without catalyst be $x kJ/mol$ ; then with catalyst it is $(x - 20) kJ/mol$ . The rates are equal, so $A e^{- x / (R \cdot 500)} = A e^{- (x - 20) / (R \cdot 400)}$ .

Taking log and cancelling $A$ : $\frac{x}{500} = \frac{x - 20}{400}$ .

$400 x = 500 (x - 20) \Rightarrow 100 x = 10000 \Rightarrow x = 100 kJ/mol$ .

Activation energy without catalyst = $100 kJ/mol$ .

Solved Example 3

The rate of a reaction doubles when the temperature rises from

2 7^{\circ}

C to

3 7^{\circ}

C. Calculate the activation energy.

Solution:

$T_{1} = 300 K$ , $T_{2} = 310 K$ , $k_{2} / k_{1} = 2$ .

Using the Arrhenius two-point form: $lo g (k_{2} / k_{1}) = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$ .

$lo g 2 = 0.301 = \frac{E _{a}}{2.303 \times 8.314} \cdot \frac{10}{300 \times 310}$ .

$E_{a} = \frac{0.301 \times 2.303 \times 8.314 \times 93000}{10} = 53.6 kJ/mol$ .

Common Mistakes to Avoid

Watch out

Assuming a catalyst changes the equilibrium composition. It does not: it changes only the time to reach equilibrium, not the position of equilibrium.
Confusing "increases the rate" with "makes the reaction spontaneous". A catalyst cannot make a thermodynamically infeasible ( $Δ G > 0$ ) reaction occur.
Attributing the doubling of rate on a $10 K$ rise to increased collision frequency. Collision frequency rises only about $1.6%$ ; the dominant factor is the sharp rise in the fraction of molecules with energy $\geq E_{a}$ .
Forgetting that surface area matters only for heterogeneous reactions. For a homogeneous liquid-phase reaction, "increasing surface area" is meaningless.
Assuming the temperature coefficient $μ$ is exactly 2. It varies with reaction (typically 2-3) and is only a rough guide.
Ignoring the catalyst poison problem in industrial processes. In the contact process, un-purified feedstock quickly deactivates $V_{2} O_{5}$ ; this is a common exam theme.

Frequently Asked Questions

Q1. What are the main factors that affect the rate of a chemical reaction?

The five main factors are: concentration of reactants, temperature, nature of the reacting substances, surface area (for heterogeneous reactions), and presence of a catalyst. Exposure to radiation is a sixth factor for photochemical reactions.

Q2. Why does the rate of a reaction increase with temperature?

Raising the temperature sharply increases the fraction of molecules with kinetic energy at least equal to the activation energy $E_{a}$ . From the Maxwell-Boltzmann distribution this fraction is $exp (- E_{a} / R T)$ , which is very sensitive to $T$ . Collision frequency also rises slightly, but the effective-fraction increase is the dominant factor.

Q3. What is the temperature coefficient of a reaction?

The temperature coefficient $μ$ is the ratio of rate constants at temperatures differing by $10 K$ : $μ = k_{T + 10} / k_{T}$ . For most reactions $μ$ lies between 2 and 3, meaning the rate roughly doubles or triples for every $10 K$ rise in temperature.

Q4. How does a catalyst speed up a reaction?

A catalyst offers an alternative reaction path with a lower activation energy $E_{a}$ . Because $E_{a}$ is smaller, a much larger fraction of molecules have sufficient energy to react, so the rate increases dramatically. The catalyst itself is regenerated at the end and does not appear in the overall equation.

Q5. Does a catalyst change the equilibrium constant?

No. A catalyst equally accelerates both the forward and backward reactions, so the position of equilibrium and the value of $K$ are unchanged. Only the time needed to reach equilibrium is reduced. Thermodynamic quantities like $Δ G^{\circ}$ and $Δ H$ also remain unchanged.

Q6. What is auto-catalysis?

Auto-catalysis is the phenomenon in which one of the products of a reaction catalyses the same reaction. The reaction starts slowly and speeds up as the auto-catalyst accumulates. A standard example is the titration of oxalic acid with acidic $K M n O_{4}$ , where the $M n^{2 +}$ produced catalyses further oxidation.

Q7. What is the difference between a homogeneous and a heterogeneous catalyst?

A homogeneous catalyst is in the same phase as the reactants (e.g. $N O$ gas catalyses gas-phase $S O_{2}$ oxidation). A heterogeneous catalyst is in a different phase, usually a solid catalysing a gaseous or liquid reaction (e.g. $F e$ catalyses the Haber $N H_{3}$ synthesis; $V_{2} O_{5}$ catalyses the contact process for $H_{2} S O_{4}$ ).

Q8. Why is a catalyst usually used in finely divided form?

A finely divided catalyst has a much greater surface area than a lump of the same mass. Since heterogeneous catalysis happens on the surface, more surface exposes more active sites and gives a much higher rate. This is why catalysts like Pt, Ni, and Pd are used as black powders or supported on high-surface-area substrates.

Previous year questions on Factors Affecting The Rate of a Chemical Reaction

13 questions from past papers, each with a step-by-step solution.

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