Does a catalyst get used up in a reaction?

No. A catalyst may temporarily bond to reactants as part of the alternate pathway, but it is regenerated by the end, so the same amount is present when the reaction finishes. That is why catalysts are written above the reaction arrow rather than as reactants, and why a small quantity can process a large amount of material.

What is the difference between activation energy and enthalpy change?

Activation energy is the height of the barrier from the reactants up to the transition state, and it determines how fast the reaction goes. Enthalpy change, H, is the energy difference between products and reactants, and it determines whether the reaction is exothermic or endothermic. A reaction can be very exothermic and still very slow if E_a is high.

Why does raising temperature speed up a reaction so much more than raising concentration?

Doubling concentration roughly doubles collision frequency. Raising temperature increases collision frequency a little, but it also shifts the distribution of molecular energies so that a much larger fraction of molecules exceeds E_a. That fraction grows exponentially with temperature, which is why a modest 10 degree Celsius rise can roughly double many reaction rates.

Why is the rate of a reactant written with a negative sign?

Reactant concentration decreases over time, so [A] is negative. Multiplying by -1 makes the reported rate a positive number, since a rate describes how fast something happens and is conventionally positive. Product rates already come out positive, so no negative sign is needed for them.

CHEM-9.3

Reaction Rates & Collision Theory

Learn collision theory, activation energy, and the five factors that speed up reactions — plus how to calculate an average reaction rate from concentration data.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Reaction Rates & Collision Theory, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Some reactions finish before you can blink; others, like iron rusting, take years. Chemistry explains that difference with one simple picture: particles have to crash into each other, hard enough and lined up correctly, before old bonds break and new ones form. That picture is called collision theory, and once you have it, every rate factor your teacher lists — concentration, temperature, surface area, catalysts, and the nature of the reactants — stops being a memorized list and becomes something you can reason out.

In this lesson you will connect collision theory to activation energy, explain each rate factor in terms of collision frequency or collision energy, and calculate an average rate of reaction from concentration-versus-time data. These ideas set up the next topic in the unit, where forward and reverse rates become equal at equilibrium.

Collision Theory: Three Conditions for a Reaction

Collision theory says that for particles to react, three things must all be true.

First, the particles must collide. Molecules that never meet cannot exchange atoms. Second, the collision must have at least a minimum amount of energy, the activation energy (

E_a

). Bonds in the reactants have to stretch and break before product bonds form, and that costs energy up front. Third, the particles must collide with the correct orientation. If the reactive part of one molecule hits the wrong end of the other, the particles simply bounce apart.

Collisions that satisfy all three conditions are called effective collisions. The reaction rate is proportional to the number of effective collisions per second, not the total number of collisions. This distinction matters: in a beaker of solution, trillions of collisions happen every second, and only a tiny fraction of them produce product.

A very common misconception is that adding energy or concentration changes how much product a reaction can make. It does not. Rate factors change how fast the reaction gets there, not how much heat is released or how much product forms overall. Another frequent error is saying "heating gives the molecules more activation energy." Activation energy is a fixed property of a particular reaction pathway. Heating does not lower

E_a

; it raises the fraction of molecules that already have enough energy to clear it. Only a catalyst changes

E_a

, by providing a different pathway entirely.

Activation Energy and the Energy Profile

An energy profile (reaction coordinate diagram) plots potential energy on the vertical axis against reaction progress on the horizontal axis. Reactants start at one level, climb a hill, and settle at the product level.

The height of the hill measured from the reactants is the activation energy

E_a

. The peak itself is the activated complex, or transition state — a short-lived arrangement in which old bonds are partly broken and new bonds are partly formed. The difference between product energy and reactant energy is

\Delta H

. If products sit lower, the reaction is exothermic; if higher, endothermic.

Here is the key separation students often blur:

E_a

controls rate, while

\Delta H

controls energy released or absorbed. A strongly exothermic reaction can still be extremely slow if

E_a

is large. A gasoline–air mixture sits in a container all day at room temperature even though burning releases enormous energy, because almost no molecules have the energy to reach the transition state until a spark supplies it.

Temperature fits here through the distribution of molecular speeds. At any temperature, molecules have a spread of kinetic energies; only those above

E_a

can react. Raising the temperature shifts that distribution to higher energies, and the fraction above

E_a

grows sharply — much faster than the modest increase in how often particles collide. That is why a rise of only 10 degrees Celsius often roughly doubles the rate of a reaction: the effect comes mostly from harder collisions, not merely more frequent ones.

The Five Factors, Explained by Collisions

Every rate factor works by changing either the frequency of collisions or the fraction of collisions that are energetic enough.

Factor	Change	Effect on rate	Collision-theory reason
Concentration	Increase	Faster	More particles per unit volume, so more collisions per second
Pressure (gases)	Increase	Faster	Squeezing gas into less volume raises concentration
Temperature	Increase	Much faster	More collisions and, more importantly, a larger fraction exceeding $E_a$
Surface area	Grind solid finer	Faster	More particles exposed at the surface where collisions can occur
Catalyst	Add	Faster	Provides an alternate pathway with lower $E_a$ , so more collisions succeed
Nature of reactants	—	Varies	Ionic solutions in water react almost instantly; covalent molecules must break strong bonds first

A catalyst is not consumed and does not appear in the overall balanced equation. It lowers

E_a

for both the forward and reverse directions equally, so it speeds up the approach to equilibrium without shifting where equilibrium lies. Enzymes are biological catalysts that do this with remarkable selectivity.

Surface area only applies when a solid (or another separate phase) is involved. A cube of zinc reacts slowly with acid; the same mass as powder can react violently, because only surface atoms can be struck by acid particles. Grain-elevator and flour-mill dust explosions are the industrial version of this idea.

Where students slip: writing "more collisions" as the reason for a catalyst. A catalyst does not change how often particles meet. Name the specific mechanism — frequency, energy, or pathway — for each factor.

Calculating an Average Reaction Rate

Rate is a change in concentration per unit time, usually in molarity per second,

\text{M/s}

. For a reactant A,

\text{rate} = -\frac{\Delta[\text{A}]}{\Delta t} = -\frac{[\text{A}]_f - [\text{A}]_i}{t_f - t_i}

The negative sign appears because reactant concentration decreases, and rates are reported as positive numbers. For a product, drop the negative sign, since the concentration is increasing.

Stoichiometry links the rates of different species. For

aA + bB \rightarrow cC

, the single rate of reaction is

\text{rate} = -\frac{1}{a}\frac{\Delta[\text{A}]}{\Delta t} = \frac{1}{c}\frac{\Delta[\text{C}]}{\Delta t}

So in

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

, hydrogen disappears twice as fast as oxygen does.

Two cautions. First, this is an average rate over an interval, not the rate at a single instant. Reaction rates usually slow down as reactants are used up, so the average over the first 60 seconds is larger than the average over the second 60 seconds. The instantaneous rate is the slope of the tangent to a concentration-versus-time curve at one point.

Second, watch units. If the problem gives moles and a volume, convert to molarity first. If it gives a time in minutes but asks for

\text{M/s}

, convert. Also, rate can be tracked with any measurable property tied to concentration: gas volume produced, mass lost, or how long a solution takes to turn cloudy.

Key terms

Collision theory.: The model stating that reactions occur only when particles collide with sufficient energy and correct orientation.
Effective collision.: A collision that meets both the energy and orientation requirements and therefore produces product.
Activation energy ( $E_a$ ).: The minimum energy colliding particles must have for a reaction to occur; the height of the barrier from reactants to the transition state.
Activated complex (transition state).: The unstable, highest-energy arrangement of atoms at the peak of the energy profile, with bonds partly broken and partly formed.
Catalyst.: A substance that speeds a reaction by providing a pathway with lower activation energy; it is not consumed and does not change $\Delta H$ .
Reaction rate.: The change in concentration of a reactant or product per unit time, typically in $\text{M/s}$ .
Average rate.: The total concentration change divided by the total time interval, equal to the slope of a straight line between two points on a concentration-time graph.
Instantaneous rate.: The rate at one specific moment, found from the slope of the tangent line to a concentration-versus-time curve.

Worked example

Dinitrogen pentoxide decomposes according to

2\text{N}_2\text{O}_5 \rightarrow 4\text{NO}_2 + \text{O}_2

. In a sealed flask,

[\text{N}_2\text{O}_5]

falls from

0.0800\ \text{M}

0.0560\ \text{M}

in 120. seconds. (a) Find the average rate of disappearance of

\text{N}_2\text{O}_5

. (b) Find the average rate of appearance of

\text{NO}_2

. (c) Predict and explain what happens to the rate if the flask is warmed.

(a) Start with the definition.

\Delta[\text{N}_2\text{O}_5] = 0.0560\ \text{M} - 0.0800\ \text{M} = -0.0240\ \text{M}

-\frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} = -\frac{-0.0240\ \text{M}}{120.\ \text{s}} = 2.00 \times 10^{-4}\ \text{M/s}

The answer is positive because of the negative sign in the definition. Reporting a negative rate here is the most common slip.

(b) Use the coefficients. Four moles of

\text{NO}_2

form for every two moles of

\text{N}_2\text{O}_5

consumed, a ratio of

4/2 = 2

\frac{\Delta[\text{NO}_2]}{\Delta t} = 2 \times 2.00 \times 10^{-4}\ \text{M/s} = 4.00 \times 10^{-4}\ \text{M/s}

As a check, the single rate of reaction is

\frac{1}{2}(2.00 \times 10^{-4}) = 1.00 \times 10^{-4}\ \text{M/s}

, and multiplying by the coefficient 4 gives the same

4.00 \times 10^{-4}\ \text{M/s}

.

(c) Warming the flask increases the rate for two reasons, and the second dominates. Faster-moving molecules collide more often, so collision frequency rises slightly. More importantly, a much larger fraction of molecules now has kinetic energy at or above

E_a

, so a greater percentage of collisions are effective. Note that

E_a

itself does not change — only the fraction of molecules able to clear it.

Practice questions

A student adds a catalyst to a reaction mixture. Which statement best explains why the reaction speeds up?

The catalyst increases the number of collisions per second between reactant particles.
The catalyst provides an alternate pathway with a lower activation energy, so a larger fraction of collisions is effective.
The catalyst raises the average kinetic energy of the reactant particles.
The catalyst increases the amount of product the reaction can eventually form.

Answer: The catalyst provides an alternate pathway with a lower activation energy, so a larger fraction of collisions is effective.

A catalyst does not change how often particles meet, and it does not heat the mixture, so the first and third statements describe the wrong mechanism. It also cannot change the maximum yield or the value of

\Delta H

— it only changes how quickly the reaction gets there. Lowering

E_a

means more of the collisions that were already happening now carry enough energy to succeed.

Magnesium ribbon and magnesium powder of equal mass are each dropped into identical samples of hydrochloric acid. The powder fizzes far more vigorously. Explain this observation using collision theory, and state whether the total volume of hydrogen gas produced will differ.

Answer: The powder has far more surface area, exposing many more magnesium atoms to the acid, which increases the frequency of collisions between acid particles and magnesium atoms; the total hydrogen produced is the same for both because the mass of magnesium (and therefore the moles available to react) is identical.

Only atoms at the surface of a solid can be struck by particles in solution. Breaking the same mass into powder multiplies the exposed surface, so collisions per second rise sharply and the rate increases. The distinction that catches students out is rate versus amount: a faster reaction finishes sooner, but stoichiometry — not speed — sets how much product forms. Both samples contain the same moles of magnesium, so both yield the same moles of hydrogen (assuming acid is in excess).

In an experiment,

[\text{H}_2\text{O}_2]

drops from

0.750\ \text{M}

0.510\ \text{M}

over 2.00 minutes. Calculate the average rate of disappearance of hydrogen peroxide in

\text{M/s}

Answer: $2.00 \times 10^{-3}\ \text{M/s}$

First convert the time:

2.00\ \text{min} = 120.\ \text{s}

. Then

\Delta[\text{H}_2\text{O}_2] = 0.510 - 0.750 = -0.240\ \text{M}

, so the rate is

-(-0.240)/120. = 2.00 \times 10^{-3}\ \text{M/s}

. Forgetting the minute-to-second conversion gives

0.120\ \text{M/min}

, which is the same physical rate but not in the requested units — always match units to what the question asks for.

FAQ

Does a catalyst get used up in a reaction?: No. A catalyst may temporarily bond to reactants as part of the alternate pathway, but it is regenerated by the end, so the same amount is present when the reaction finishes. That is why catalysts are written above the reaction arrow rather than as reactants, and why a small quantity can process a large amount of material.
What is the difference between activation energy and enthalpy change?: Activation energy is the height of the barrier from the reactants up to the transition state, and it determines how fast the reaction goes. Enthalpy change, $\Delta H$ , is the energy difference between products and reactants, and it determines whether the reaction is exothermic or endothermic. A reaction can be very exothermic and still very slow if $E_a$ is high.
Why does raising temperature speed up a reaction so much more than raising concentration?: Doubling concentration roughly doubles collision frequency. Raising temperature increases collision frequency a little, but it also shifts the distribution of molecular energies so that a much larger fraction of molecules exceeds $E_a$ . That fraction grows exponentially with temperature, which is why a modest 10 degree Celsius rise can roughly double many reaction rates.
Why is the rate of a reactant written with a negative sign?: Reactant concentration decreases over time, so $\Delta[\text{A}]$ is negative. Multiplying by $-1$ makes the reported rate a positive number, since a rate describes how fast something happens and is conventionally positive. Product rates already come out positive, so no negative sign is needed for them.

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