Lesson Explainer: Reaction Rate Chemistry • 7th Grade

In this explainer, we will learn how to describe the rate of a chemical reaction and explain the effect of heat on the rate of reaction using the collision theory.

Chemical reactions start with reactants and finish with products: reactantsproducts

Reactions involve changes in the positions of atoms, ions, electrons, and nuclei. The modern world depends on a huge number of chemical reactions actually happening. However, it is equally important how fast or slow these reactions happen.

Sometimes, even if a reaction can happen, we do not necessarily want it to happen. For instance, it is generally better if rusting is very slow; we want iron to retain its strength over time. On the other hand, our bodies use enzymes to break down toxins very quickly. If these toxins are not broken down fast enough, we can get very ill.

When reactants turn into products, lots of things could change. Some are harder to measure than others. The most common ways we express reaction rate are the rate of change in mass or the rate of change in moles.

Definition: Rate of Reaction

The rate of reaction is the rate at which the reactants are converted into the products (usually expressed as mass per unit time or moles per unit time).

If we measure the rate of a reaction using mass, the rate that the reactant mass decreases will be the same as the rate that the product mass increases—remember that for a simple chemical reaction, the total mass stays constant, so any decrease in mass of the reactants must be matched by an increase in the mass of the products: therateatwhichreactantmassdecreasestherateatwhichproductmassincreases=.

The symbol Δ is used to denote a change in something; that is, Δ𝑚 is a change in mass. We can express this concept using rate expressions: rateofreactionbymasstimetime()=Δ𝑚Δ=Δ𝑚Δ.reactantsproducts

You may have noticed that the reaction rate is defined as the negative of the rate of change of the reactant mass. This is because the rate of change of reactant mass will already be negative (the reactant mass is decreasing). By putting a negative sign in front of it, the reaction rate will always be expressed as a positive number: a certain positive amount of mass involved in the reaction per unit time.

We can also use changes in the amount, in moles, to describe the rate of a reaction—this is often preferable for small-scale work because it is easier to calculate reaction rates of individual chemicals from the reaction stoichiometry: rateofreactioninmolestimetime()=Δ𝑛Δ=Δ𝑛Δ.reactantsproducts

However, if we have a reaction like the one below, the amount of A plus the amount of B at the start will not equal the amount of C at the end (because it takes one of A and one of B to make just one of C): A+BC

In this case, the rate of the production of C (in moles) is not equal to the rate of consumption of A plus the rate of consumption of B. Instead, we need to look at the equation: A+BC

This can be read as follows: for each A and each B, we produce one C.

So the rate of consumption of A must equal the rate of consumption of B and the rate of production of C: A+BCΔ𝑛Δ𝑡=Δ𝑛Δ𝑡=Δ𝑛Δ𝑡.ABC

If the stoichiometry is different, we have to adjust the equations. For example, a bigger coefficient for A will mean that A will be consumed faster, so we have to divide the rate of consumption of A by 2 to get the rate of consumption of B: 212×Δ𝑛Δ𝑡=Δ𝑛Δ𝑡=Δ𝑛Δ𝑡.A+BCABC

We can generalize this for all possible coefficients: abcA+BC1𝑎×Δ𝑛Δ𝑡=1𝑏×Δ𝑛Δ𝑡=1𝑐×Δ𝑛Δ𝑡.ABC

When we calculate rates like this, what we are really doing is coming up with an average rate, which is equal to the change in mass or amount divided by the time taken.

Some chemical reactions can be incredibly slow, like the conversion of organic matter to coal and oil. Or they can be fast enough for us to observe at a reasonable rate, like a candle burning. Or they can be so fast that they cause an explosion, like the detonation of a hydrogen balloon. We will now discuss what we mean by the rate of a specific reaction.

Next, we are going to look at an example: the burning of coal. Under ideal circumstances, the reaction of coal and oxygen in the air produces carbon dioxide. Here, we treat coal like it is pure carbon: C()+O()CO()sgg22

We can assess the rate of this reaction by monitoring changes in mass or amount.

If we consume 1 mol of carbon each second, we will be consuming 12 grams of carbon per second. We can calculate this by multiplying the rate (in moles per second) by the molar mass of carbon (12 g/mol). For this exercise, we will use simple rounded values of the molar masses to keep it simple: 1×12=12/.molCsgmolCgs

In this reaction, carbon and oxygen are in a ratio of 11 (CO2), so we must be consuming 1 mole per second of oxygen gas as well, using the molar mass of oxygen atoms of 16 g/mol (so the molar mass of oxygen is 32 g/mol): 1×32=32/.molOsgmolOgs22

At the same time, carbon dioxide would be produced. We will start by calculating the molar mass of carbon dioxide: molarmassofCOgmolgmol2=(12+2×16)/=44/.

Now, in the reaction equation, CO2 is in 11 ratio with carbon, so we will generate 1 mol/s of carbon dioxide: 1×44=44/.molCOsgmolCOgs22

At this point, we should check our working. If the total mass cannot change, then the rate at which the mass of reactants decreases must equal to the rate at which the product mass increases: 12/+32/=44/.gsgsgs

This example is relatively easy because all the reactants and products are in a 11 stoichiometric ratio with each other. That is not typically the case. Next, we will look at a similar example, where carbon monoxide is produced instead: 2C()+O()2CO()sgg2

If carbon monoxide is the only product when burning carbon in oxygen, one molecule of oxygen can react with two carbon atoms—CO=212.

If we burn carbon at the same rate (1 mol/s), we have to adjust the calculations to get the correct rate of consumption of oxygen: rateofconsumptionofOrateofconsumptionofCmolsmols2=2=1÷2=0.5/.

It makes sense that if each carbon atom needs half as much oxygen as before, oxygen will be consumed at half the rate.

In general, if we want to calculate the rate of consumption of a reactant or the rate of production of a product, we just need to use the coefficients from the balanced equation: abcA+BCrateofconsumptionofArateofconsumptionofBrateofproductionofC=𝑎𝑏×=𝑎𝑐×.

Be careful with plus and minus signs when dealing with rates. For a reaction that is moving forward,

  • the reaction rate is positive,
  • the rates of change of the masses or amounts of reactants are negative, as their masses are decreasing (but their rates of consumption are positive),
  • the rates of change of the masses of amounts of products are positive, as their masses are increasing (and their rates of production are also positive).

When doing calculations with reaction rates, it is important to check which, if any, of the reactants is a limiting reactant. A reaction rate can be limited in just the same way that a reaction yield can be limited.

Example 1: Calculating the Rate of Reaction for the Reaction of Hydrochloric Acid and Calcium Carbonate given the Mass of Calcium Carbonate Consumed in a Period of Time

When excess hydrochloric acid reacts with calcium carbonate, carbon dioxide is produced. If 5 grams of calcium carbonate is consumed in 3 minutes and 20 seconds, what is the average rate of reaction?


Hydrochloric acid (clearly an acid) and calcium carbonate (a base) react according to the equation 2HCl()+CaCO()CaCl()+HO()+CO()aqsaqlgAcidMetalcarbonateMetalsaltWaterCarbondioxide3222

We have been given the time over which a certain mass of calcium carbonate is consumed: in 3 minutes and 20 seconds, a total of 5 grams of calcium carbonate reacts with hydrochloric acid and is converted into calcium chloride, water, and carbon dioxide.

The rate of the reaction, in terms of calcium carbonate, is equal to the mass consumed divided by the time taken: reactionratebymassofCaCOgmins()=5÷3203

We have not been told what unit of time to use, minutes or seconds. However, it is generally easier to convert minutes to seconds than to convert seconds to minutes, so we will do that using the identity that 60 seconds equals 1 minute (60=1smin): reactionratebymassofCaCOgminsgssgsgs()=53×+20=5180+20=5200=0.025/.3smin

We could also give the answer in grams per minute; we can convert from grams per second (g/s) and grams per minute (g/min) by multiplying by 60 seconds per minute (60 s/min): 0.025×60=1.5/.gssmingmin

The average rate of the reaction, in terms of the mass of calcium carbonate consumed, is 0.025 g/s (1.5 g/min).

When we have a rate, we can calculate the amounts produced in a period of time (in moles or in terms of mass): rateamountmolesperunittimeorratemassperunittime=()=.

By rearranging the question in terms of amount (moles) or mass, we can work out how long a reaction needs to occur for a given amount of reactant to be consumed or amount of product to be produced: amountmolesratetimeormassratetime()=×=×.

If, for example, the rate of production of a product is 3 g/min and we run the reaction for an hour and we want to know the total mass of product produced, this would be our calculation: massratetimegminh=×=3×1.

We need to convert hours to minutes: 1=601×601=60.hminhminhmin

And then we can finish the calculation: massratetimegminming=×=3×60=180.

We may need to do similar calculations for converting between units of mass: 1000=1,1000=1.mgggkg

Example 2: Calculating the Total Mass of Product given The Rate of Production and Total Time

In the following reaction, the rate of sulfur formation was determined to be 0.002 g/s: NaSO()+2HCl()2NaCl()+HO()+SO()+S()22322aqaqaqlgs How much sulfur was formed after 5 minutes?


This question includes a reaction equation; however, in order to answer the question, we need three things:

  • the rate of production of sulfur (0.002 g/s),
  • the time it is being produced for (5 minutes),
  • the understanding that the rate of production multiplied by the time will give us the total mass produced.

We have the equation massratetime=×.

If we insert the values in the equation, this is what we get: massgsmin=0.002×5.

There are two different time units in this equation: seconds (s) and minutes (min). 1 minute is equal to 60 seconds. We can convert the “per second” (/s) or the minute (min)—it is easier to convert minutes to seconds, so we will do that: massgsminsmingssg=0.002×5×601=0.002×300=0.6.

So, the mass of sulfur produced in 5 minutes (300 seconds) is 0.6 grams.

Nonetheless, all of this is based on what we can measure. We should try to actually understand what is going on and work out how we can use that knowledge to our advantage. The first question we need to answer about reactions and reaction rate is this: why do reactions happen in the first place?

Collision theory suggests that reactions will only happen when particles collide with each other. If particles never touch, they cannot possibly react.

However, not every collision is going to lead to a successful reaction. In fact, the air around us is filled with particles that are colliding and not reacting. The other necessary component is that there is enough energy in the collision in order to break bonds and rearrange particles.

But, even if the collision has the right energy, it may not be successful because the particles are not arranged properly. If they are not in the right orientation, the energy will not be transferred into breaking the right bonds.

This means that only when particles collide with enough energy in the right orientation will a reaction occur.

We call the minimum energy needed for a reaction to occur the activation energy.

Definition: Activation Energy

Activation energy is the minimum amount of energy required for a reaction to occur.

We can think of the activation energy as the energy required to break just the right bonds that allow the new bonds in the products to form (although activation energy is not actually measured in this way). Let’s see how we can put it to use by changing reaction rates.

Collision theory suggests there are a few things that could be changed to make a reaction faster:

  • If we can find a way of making collisions happen more frequently, the reaction should be faster.
  • If we can find a way of increasing the energy of those collisions, the proportion of successful collisions should increase, and the rate of the reaction should go up.
  • If we can lower the activation energy, we will increase the proportion of collisions that will be successful.

We could also increase the reaction rate if we can control the orientation of the reactants and arrange them in the right way. This is a part of how catalysts work, but we will not be looking at this in detail.

Example 3: Relating a Real-World Phenomenon to Collision Theory

When lightning strikes, molecules of oxygen (O2) and nitrogen (N2) in the atmosphere are given enough energy to combine together to form nitric oxide (NO). Under normal conditions, oxygen and nitrogen molecules collide but do not react. What does this suggest about the reaction?

  1. It cannot be performed without a catalyst.
  2. It is irreversible.
  3. It is reversible.
  4. It is catalyzed by lightning.
  5. It has a very large activation energy.


The question tells us that nitrogen and oxygen gases react to form nitrogen monoxide (nitric oxide, NO) during a lightning strike, but not under everyday conditions.

This suggests that a lot of energy is required to make the reaction happen. The temperature inside a lightning strike can be higher than 50000C!

Collision theory aims to describe how and why reactions happen. It suggests that reactions happen when particles collide with enough energy and in the right orientation (otherwise, particles will simply bounce off one another).

Nitrogen and oxygen gases in air (at normal temperatures) do not have enough energy to react, so no matter how many times they collide, no reaction occurs.

The minimum energy required for a reaction to occur is called the activation energy—the total energy of a collision must be greater than or equal to the activation energy for a reaction to occur.

With this information alone, we can deduce that the activation energy of the reaction of nitrogen and oxygen that produces nitrogen monoxide oxide (NO) has a very large activation energy. The extremely high temperature caused by lightning is sufficient to provide this to the particles of nitrogen and oxygen in the air.

Lightning is not a single chemical agent, and all it does it provide the energy necessary to overcome the activation barrier—it is not a catalyst.

There is not enough information given to assess whether the reaction is reversible or not (it could be that NO also degrades when struck by lightning—we just do not know as this is not given in the question).

What we do know is that the reaction definitely can happen, so we know that a catalyst is not fundamentally necessary.

The answer is E: it has a very large activation energy.

Controlling the frequency of collisions, the energy of collisions, and the activation energy can be done in various ways. Here, we will only be looking in detail at the effect of heat and temperature.

We can take the typical example of a gas. If we heat up the gas, the particles will move more quickly.

The faster the particles are moving, the more often they will collide. And since they are moving faster, their collisions will be more energetic: moreheathighertemperaturemorefrequenthigher-energycollisions(),.

If we do the opposite and cool the system down, the particles will move more slowly: lessheatlowertemperaturelessfrequentlower-energycollisions(),.

Although we generally depict this scenario with gases, heating up solids and liquids will also increase the rate at which they react. So broadly speaking, at a higher temperature, the rate of reactions will be higher.

The table below shows some other changes that are commonly applied to reactions to increase their rates.

ChangesSuitable ReactionsHow It Works
Increasing the pressureReactions involving gasesIncreases the collision frequency
Increasing concentrationsReactions between solutes in solutionIncreases the collision frequency
Increasing the available surface areaReactions involving solids (and occasionally those involving liquids)Increases the collision frequency
Using a catalystVarious types of reactionsA catalyst provides an alternative reaction route with lower activation energy

The key points of this explainer are summarized below.

Key Points

  • The rate of a reaction is the rate at which the reactants are converted into the products.
  • The rate of a reaction can be expressed using the rate of consumption of a reactant or the rate of production of a product.
  • Rates are commonly expressed as changes in mass over time (e.g., grams per second) or changes in the amount over time (e.g., moles per second).
  • Collision theory is the theory that describes the necessary requirements for reactions to happen.
  • Collision theory says that for a reaction to happen, there must be
    • collisions between reactant particles,
    • collisions between reactant particles with the total energy being equal to or greater than the activation energy,
    • collisions between reactant particles with the particles having the right orientation to one another.
  • Activation energy is the minimum amount of energy required for a reaction to occur.
  • An increase in the rate of a reaction will occur if
    • collisions are more frequent (more collisions per unit time),
    • collisions are of higher energy,
    • the activation energy of the reaction is low.
  • Heating reactants causes more frequent, higher-energy collisions to occur.
  • Increases in reaction rate can also be caused by
    • increasing pressure (for reactions involving gases),
    • increasing concentration (for reactions involving solutes in solutions),
    • increasing surface areas (for reactions involving solids),
    • using a catalyst.

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