Subject: Chemistry | Level: GCSE | Exam Board: OCR
Master the mathematics and mechanics of chemical reactions with this comprehensive guide to C5. From mole calculations to dynamic equilibrium, you'll learn how to predict, control, and measure chemical changes to secure top marks in your GCSE exams.
Revision Notes & Key Concepts
Revision Podcast Transcript
Welcome to your GCSE Chemistry revision podcast. I'm your tutor, and today we're diving deep into Topic C5: Monitoring and Controlling Chemical Reactions. Whether you're revising for your mocks or your final exams, this episode is going to give you everything you need to feel confident walking into that exam hall. We'll cover the key concepts, work through some calculations, talk exam technique, and finish with a quick-fire quiz to test your recall. So grab a pen, get comfortable, and let's get started. SECTION ONE: CORE CONCEPTS Let's begin with the big picture. Topic C5 is all about the quantitative side of chemistry — that means using numbers and measurements to understand and control what happens in chemical reactions. There are three main areas: first, moles, concentration, and gas volumes; second, reaction rates and how to control them; and third, dynamic equilibrium and Le Chatelier's Principle. Let's take each one in turn. THE MOLE AND CONCENTRATION The mole is chemistry's counting unit. Just like a dozen means twelve, one mole means six point zero two times ten to the power of twenty-three particles — that's Avogadro's number. But in practice, you won't need to use that number directly. What you do need is the formula: moles equals mass divided by relative formula mass. Write that down: moles equals mass over Mr. So if you have 4 grams of sodium hydroxide, NaOH, and the Mr is 40, then moles equals 4 divided by 40, which gives you 0.1 moles. Simple. Now, concentration. Concentration tells us how many moles are dissolved in one cubic decimetre of solution — that's one litre. The formula is: concentration equals moles divided by volume. And here's the critical thing that trips up so many candidates: the volume must be in cubic decimetres, not cubic centimetres. If you're given a volume in centimetres cubed, you must divide by 1000 first. So 25 centimetres cubed becomes 0.025 cubic decimetres. Examiners see this mistake constantly, and it costs marks every single time. TITRATION Titration is the required practical that tests your ability to measure concentration accurately. You use a burette to add a solution of known concentration — called the standard solution — to a measured volume of the unknown solution in a conical flask. An indicator tells you when the reaction is complete — this is called the end point. In the exam, you'll often be given the volume used and the concentration of one solution, and asked to find the concentration of the other. The method is always the same: calculate moles of the known solution, use the mole ratio from the balanced equation to find moles of the unknown, then calculate concentration using concentration equals moles divided by volume. A key exam tip: always use the mean titre from your concordant results — that means results that are within 0.1 cubic centimetres of each other. Examiners specifically look for this in six-mark practical questions. MOLAR GAS VOLUME At room temperature and pressure — which we abbreviate as RTP — one mole of any gas occupies 24 cubic decimetres. This is a fixed value you must memorise. The formula is: volume of gas equals moles multiplied by 24. Or rearranged: moles equals volume divided by 24. So if a reaction produces 0.5 moles of carbon dioxide gas, the volume produced at RTP is 0.5 times 24, which equals 12 cubic decimetres. Remember: this only works at room temperature and pressure. If the question specifies different conditions, the value will change. PERCENTAGE YIELD AND ATOM ECONOMY These two calculations come up frequently and candidates often confuse them. Percentage yield compares what you actually made in the lab to what you theoretically should have made. The formula is: percentage yield equals actual yield divided by theoretical yield, multiplied by 100. Yields are less than 100 percent because of incomplete reactions, side reactions, or losses during collection. Atom economy is different — it's about sustainability. It measures how much of the mass of the reactants ends up in the desired product. The formula is: atom economy equals relative formula mass of desired product divided by sum of relative formula masses of all products, multiplied by 100. A high atom economy means less waste, which is better for the environment and more cost-effective industrially. SECTION TWO: REACTION RATES AND COLLISION THEORY Now let's talk about how fast reactions go and how we can control them. Collision theory tells us that for a reaction to happen, particles must collide with sufficient energy — that energy threshold is called the activation energy. The rate of reaction depends on the frequency of successful collisions. That phrase — frequency of successful collisions — is exactly what examiners want to see in your answers. Use it every time. There are four main factors that affect reaction rate: temperature, concentration, surface area, and pressure for gases. Let's go through each. TEMPERATURE: When you increase temperature, particles move faster. This means they collide more frequently AND more of those collisions have energy greater than or equal to the activation energy. Both effects increase the rate. In your answers, you must mention both — more frequent collisions AND more successful collisions. CONCENTRATION: Increasing concentration means more particles in the same volume. This increases the frequency of collisions, so more successful collisions occur per unit time, and the rate increases. For gases, increasing pressure has the same effect — it's essentially the same as increasing concentration. SURFACE AREA: For solid reactants, breaking them into smaller pieces increases the surface area exposed to the other reactant. This means more collisions can happen at the surface, increasing the rate. Think of dissolving a sugar cube versus granulated sugar — the granules dissolve faster because more surface is exposed. CATALYSTS: A catalyst is a substance that increases the rate of reaction without being used up. This is crucial — catalysts are not consumed. They work by providing an alternative reaction pathway with a lower activation energy. On an energy profile diagram, a catalyst lowers the peak of the curve but doesn't change the energy levels of the reactants or products. Examiners frequently ask candidates to draw or interpret these diagrams. SECTION THREE: DYNAMIC EQUILIBRIUM AND LE CHATELIER'S PRINCIPLE This is often the most challenging part of C5, but once you understand the logic, it becomes quite elegant. A reversible reaction is one that can go in both directions — reactants form products, but products can also reform reactants. We show this with a double-headed arrow. When a reversible reaction takes place in a closed system — meaning nothing can enter or leave — the reaction reaches dynamic equilibrium. At this point, the forward and reverse reactions are happening at the same rate, so the concentrations of reactants and products remain constant. Notice I said constant, not equal. This is a very common misconception — equilibrium does NOT mean the concentrations are equal. It just means they're not changing. Le Chatelier's Principle tells us what happens when we disturb that equilibrium. It states: if a system at equilibrium is subjected to a change, the equilibrium will shift in the direction that opposes that change. Let's apply this to three types of change: CONCENTRATION: If you increase the concentration of reactants, the equilibrium shifts to the right — toward the products — to reduce the concentration of reactants. If you remove products, the equilibrium also shifts right to replace them. TEMPERATURE: This one requires you to know whether the forward reaction is exothermic or endothermic. If the forward reaction is exothermic and you increase temperature, the equilibrium shifts LEFT — toward the endothermic direction — to absorb the extra heat. If you decrease temperature, it shifts right. PRESSURE: This only applies to reactions involving gases. If you increase pressure, the equilibrium shifts toward the side with fewer moles of gas, to reduce the pressure. Count the moles of gas on each side of the equation to determine which way it shifts. A classic example is the Haber Process: nitrogen plus three hydrogen gives two ammonia. The forward reaction is exothermic, and there are four moles of gas on the left and two on the right. High pressure favours ammonia production. Low temperature also favours ammonia, but it slows the rate. So industrially, a compromise temperature of around 450 degrees Celsius is used, along with an iron catalyst to speed things up. SECTION FOUR: EXAM TIPS AND COMMON MISTAKES Now let's talk about how to maximise your marks in the exam. Tip one: Always show your working in calculations. Even if your final answer is wrong, you can still earn method marks. Examiners are instructed to award marks for correct steps, so never skip steps. Tip two: Check your units. The most common error in C5 is using centimetres cubed instead of cubic decimetres in concentration calculations. Before you calculate, convert: divide centimetres cubed by 1000. Tip three: When explaining rate of reaction, always use the phrase "frequency of successful collisions." Vague answers like "particles collide more" will not earn full marks. Tip four: For equilibrium questions, always identify whether the forward reaction is exothermic or endothermic before predicting the effect of temperature. Don't guess — work it out from the energy values given. Tip five: Remember that catalysts do NOT shift the equilibrium position. They speed up both the forward and reverse reactions equally, so the equilibrium is reached faster but the position doesn't change. Tip six: In titration calculations, always state the mole ratio you are using. Examiners give a mark for this step, and many candidates skip it and lose an easy mark. Common mistake one: Confusing moles, mass, and number of particles. These are three different things. Moles is the amount. Mass is in grams. Number of particles uses Avogadro's number. Common mistake two: On rate graphs, candidates often describe the gradient incorrectly. The rate of reaction is the gradient of the graph — steeper means faster. The reaction is complete when the line becomes flat. Don't say "the reaction stopped" — say "the reaction is complete." Common mistake three: Thinking dynamic equilibrium means equal concentrations. It doesn't. It means constant concentrations. SECTION FIVE: QUICK-FIRE RECALL QUIZ Right, let's test your memory. I'll ask a question, pause, then give the answer. Try to answer before I do! Question one: What is the formula for calculating moles from mass and Mr? ... Moles equals mass divided by Mr. Question two: What volume does one mole of gas occupy at room temperature and pressure? ... 24 cubic decimetres. Question three: Name four factors that affect the rate of reaction. ... Temperature, concentration, surface area, and pressure for gases. Question four: What does dynamic equilibrium mean? ... The forward and reverse reactions occur at the same rate in a closed system, so concentrations remain constant. Question five: A catalyst lowers the activation energy — true or false? ... True! And it does this by providing an alternative reaction pathway. Question six: If you increase pressure on a reaction where there are more moles of gas on the left than the right, which way does equilibrium shift? ... To the right — toward fewer moles of gas. SUMMARY AND SIGN-OFF Let's wrap up. Topic C5 covers three interconnected areas. First, quantitative chemistry — using moles, concentration, and gas volumes to calculate what happens in reactions. Second, reaction rates — understanding collision theory and how temperature, concentration, surface area, pressure, and catalysts control how fast reactions go. Third, equilibrium — understanding dynamic equilibrium in closed systems and using Le Chatelier's Principle to predict how equilibrium shifts. The key things to take away: always convert centimetres cubed to cubic decimetres; always show your working; use the phrase "frequency of successful collisions"; remember catalysts are not consumed and don't shift equilibrium; and never confuse "constant concentrations" with "equal concentrations." You've got this. Keep practising those calculations, use the worked examples in your study guide, and don't forget to do past paper questions under timed conditions. Good luck with your revision — I'll see you in the next episode!
Key Terms & Definitions
- Mole
- The unit for amount of substance. One mole contains Avogadro's number ($6.02 \times 10^{23}$) of particles.
- Activation Energy
- The minimum amount of energy that colliding particles must possess for a chemical reaction to occur.
- Dynamic Equilibrium
- The point in a reversible reaction in a closed system where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant.
- Catalyst
- A substance that increases the rate of a chemical reaction by providing an alternative reaction pathway with a lower activation energy, without being consumed in the process.
- Atom Economy
- A measure of the amount of starting materials that end up as useful products, expressed as a percentage.
- Closed System
- A system in which no substances can enter or leave during a reaction.
Worked Examples
Worked Example
Question: Calculate the mass of magnesium oxide produced when 6.0g of magnesium burns completely in oxygen. (The balanced equation is $2Mg + O_2 \rightarrow 2MgO$. Relative atomic masses: Mg = 24, O = 16)
Solution: Step 1: Calculate the moles of magnesium used. Moles = Mass / $A_r = 6.0 / 24 = 0.25$ moles. Step 2: Use the molar ratio from the balanced equation. The ratio of Mg to MgO is 2:2, which simplifies to 1:1. Therefore, moles of MgO produced = 0.25 moles. Step 3: Calculate the relative formula mass ($M_r$) of MgO. $M_r$ of MgO = 24 + 16 = 40. Step 4: Calculate the mass of MgO produced. Mass = Moles $\times M_r = 0.25 \times 40 = 10.0g$. Final answer: 10.0g
Worked Example
Question: In a titration, $25.0 cm^3$ of $0.100 mol/dm^3$ sodium hydroxide ($NaOH$) neutralised $20.0 cm^3$ of sulfuric acid ($H_2SO_4$). Calculate the concentration of the sulfuric acid. (Equation: $2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$)
Solution: Step 1: Convert volumes to $dm^3$. Volume of NaOH = $25.0 / 1000 = 0.025 dm^3$. Volume of $H_2SO_4$ = $20.0 / 1000 = 0.020 dm^3$. Step 2: Calculate moles of NaOH used. Moles = Concentration $\times$ Volume = $0.100 \times 0.025 = 0.0025$ moles. Step 3: Use the molar ratio to find moles of $H_2SO_4$. The ratio of NaOH to $H_2SO_4$ is 2:1. Moles of $H_2SO_4$ = $0.0025 / 2 = 0.00125$ moles. Step 4: Calculate the concentration of $H_2SO_4$. Concentration = Moles / Volume = $0.00125 / 0.020 = 0.0625 mol/dm^3$. Final answer: $0.0625 mol/dm^3$
Worked Example
Question: Explain how increasing the temperature affects the rate of a chemical reaction. Use collision theory in your answer.
Solution: Increasing the temperature increases the kinetic energy of the particles, causing them to move faster. This results in a higher frequency of collisions between reactant particles. Furthermore, a greater proportion of the colliding particles possess energy equal to or greater than the activation energy. Consequently, there is an increased frequency of successful collisions, leading to a faster rate of reaction.
Practice Questions
Question: Calculate the percentage yield if the theoretical yield of a reaction is 50g but only 40g is obtained.
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Question: Explain why a catalyst increases the rate of a reaction.
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Question: The Haber process is used to produce ammonia: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$. The forward reaction is exothermic. Predict and explain the effect of increasing the pressure on the yield of ammonia.
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Question: Calculate the atom economy for making hydrogen by reacting coal with steam: $C(s) + 2H_2O(g) \rightarrow CO_2(g) + 2H_2(g)$. ($A_r$: C=12, H=1, O=16)
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Question: A student investigates the rate of reaction between marble chips and hydrochloric acid. State two ways the student could increase the rate of this reaction without changing the temperature.
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