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GCSE Chemistry: How to do calculations from calorimeter chemistry experiments

CALORIMETER EXPERIMENTS and CALCULATION of ENERGY TRANSFER in CHEMICAL REACTIONS

PART D Exothermic and Endothermic Energy Changes – Chemical Energetics  Methods of determining energy transfers and calculation of energy changes from calorimetric data

6(a) The basic principles of calorimetry - introduction

6(b) A simple calorimeter

Method for measuring the energy change for: 6c(i) Energy change for a neutralisation

6c(iv) Precipitation reactions

7

7.2 The energy change calculation for fuel combustion

7.3 The energy change calculation for neutralisation

7.4 The energy change calculation for a metal displacement reaction

What next? Associated Pages

6. The experimental determination of energy changes using simple calorimeters

(a) The basic principles of calorimetry -introduction

Introduction

It is not difficult to measure the quantity of energy released or absorbed in a chemical reaction if it happens in solution.

e.g. mixing two solutions of soluble substances or adding a solid to a solution, and the mixing causes a reaction to happen. All you do is accurately measured quantities of reactants, mix them in a calorimeter (diagram on right), take the temperature of the reactants at the start and of the products at the end of the reaction.

Make sure everything starts at the same temperature - fair test point.

The quantities of chemicals used and temperature change can be used to measure the energy change.

One of the biggest sources of error is losing heat to the surroundings (if exothermic) or absorbing heat from the surroundings (endothermic).

This is why you use an insulated calorimeter - with careful working, you can get quite accurate results by using an insulated polystyrene cup - diagram below right.

Using this simple procedure you can:

vary the reactants to look at different reactions,

vary the concentration of a soluble reactant e.g. of an acid or an alkali,

vary the quantity of a solid reactant, but you must know which is the limiting reactant to get your calculation of the energy change to be correct Using this apparatus you can observe changes in heat energy accompanying the following changes - (i) salts dissolving in water,  (ii) neutralisation reactions, (iii) displacement reactions, (iv) precipitation reactions and when these reactions take place in solution, the temperature changes can be measured to calculate and compare the relative heat energy changes. 6(b) A simple calorimeter This method 6.1 is can be used for any non–combustion reaction that will happen spontaneously at room temperature involving liquids or solid reacting with a liquid. This method involves using a simple polystyrene cup as the calorimeter container - polystyrene is a poor conductor of heat - good insulator. For extra insulation you can place the polystyrene container in a larger beaker containing a thick layer of cotton wool wrapped around to minimise heat loss - the main source of error in the experiment. Lastly, a sealing lid to stop convection in air. All these measures help reduce the heat loss and quite accurate results can be obtained even with this simple apparatus. Solid reactants are weighed into the calorimeter already containing a known volume of water at a measured temperature if investigating dissolving salts. Known volumes of any liquids involved - aqueous solutions of reactants of known concentration are added using a measuring cylinder or pipette. The mixture could be a salt and water (heat change on dissolving) or an acid or an alkali solution (heat change of neutralisation), metal displacement and precipitation reactions can be investigated too. It doesn't matter whether the change is exothermic (heat released or given out, temperature increases) or endothermic (heat absorbed or taken in, temperature decreases). See the energy change calculations section on this page You measure the initial temperature of the reactants at the start and the final maximum/minimum temperature when the reaction is done. You should gently stir the mixture after the initial temperature reading to ensure all reactants react! Subtracting one from the other gives the temperature change. If the reaction is exothermic the temperature rises and if the temperature falls its an endothermic change. From the temperature rise or temperature fall and the heat capacity of the water you can calculate how much heat was released or how much heat energy was absorbed by a specific quantities of chemicals reacting (or even the energy transfer when a salt dissolves in water, which can be exothermic or endothermic). Energy transferred in J = mass of solution in grams X specific heat capacity of water (4.2J/goC) X temperature change in oC You can then calculate the amount of energy released/absorbed per gram or per mole. See the energy change calculations section on this page Whether the reaction is exothermic or endothermic depends on the reactant chemicals involved and the type of reaction. Some types of reaction are always exothermic or always endothermic, other reactions could be either exothermic or endothermic. Some examples are briefly described below and all can be investigated with the humble polystyrene calorimeter. 6(c)(i) Mixing an acid with an alkali always gives an exothermic reaction - the temperature rises. e.g. neutralising hydrochloric acid with sodium hydroxide to give sodium chloride and water HCl(aq)  +  NaOH(aq)  ===>  NaCl(aq)  +  H2O(l)e.g. measure 25 cm3 of 0.5 mol/dm3 hydrochloric acid into a beaker. Have 25 cm3 of 0.5 mol/dm3 sodium hydroxide ready in a 2nd beaker. Check they are both at the same temperature - you could have them in a water bath or just by each other on the lab bench. Pour one solution into the calorimeter, followed by the other and record the temperature every 30 seconds until you have gone past the highest temperature. You subtract the initial lab temperature from the highest temperature, and the difference is used in the calculations ... ... See the energy change 7.3 calculation for neutralisation You can repeat the experiment with different concentrations of the same acid and alkali or you can investigate different acids and alkalis using constant concentrations (in mol/dm3).   6(c)(ii) Dissolving salts (or anything else that dissolves in water without reacting) Adding the salts ammonium nitrate or ammonium chloride to water gives a temperature fall, an endothermic dissolving change. e.g. dissolving ammonium nitrate in water  NH4NO3(s)  +  aq  ===> NH4NO3(aq) Dissolving anhydrous calcium chloride gives a temperature rise - exothermic See the 7.1 energy change calculation for dissolving a salt in water   6(c)(iii) A metal displacement reaction Adding zinc to copper sulfate produces an exothermic displacement reaction shown by the rise in temperature. zinc  +  copper(II) sulfate  ===>  zinc sulfate  +  copper Zn(s)  +  CuSO4(aq)  ===>  ZnSO4(aq)  +  Cu(s)The more reactive zinc displaces the less reactive copper out of solution.  6(c)(iv) Precipitation reactions You can mix solutions of lead(II) nitrate and potassium iodide solution. The precipitation of yellow lead(II) iodide is an exothermic change. lead(II) nitrate  +  potassium iodide  ===> lead(II) iodide  +  potassium nitrate Pb(NO3)2(aq)  + 2KI(aq)  ===> PbI2(s)  +  2KNO3(aq)You see quite a colour change in this reaction as the precipitate forms, as well as the temperature rise. You can investigate burning liquid hydrocarbons like hexane - but smokey and inaccurate, alcohols like ethanol burn more efficiently with a 'cleaner' blue flame 6(c)(v) The heat energy released on burning a liquid fuel This method 6.2 is specifically for determining the heat energy released (given out) for burning fuels. The burner is weighed before and after combustion to get the mass of liquid fuel burned. The thermometer records the temperature rise of the known mass of water (1g = 1cm3). The heat from the fuel combustion heats up the water. From the heat capacity of the water and the temperature rise you can calculate how much heat was released by a specific mass of fuel. You measure the temperature of the reactants at the start and the final maximum/minimum temperature when the reaction is done. Subtracting one from the other gives the temperature change. You should stir the water gently before recording the final raised temperature to get a true average value of the bulk liquid. Energy transferred in J = mass of water in grams X specific heat capacity of water (4.2J/goC) X temperature change in oC You can then calculate the amount of energy released per gram or per mole. You can use this system to compare the heat output from burning various fuels. The bigger the temperature rise, the more heat energy is released. See below for expressing calorific values. BUT you must conduct the experiments under 'fair test' conditions. apart from repeating experiments (to eliminate anomalous results): you must use the same burner & wick (if possible), try and keep the wick the same length giving the same height of flame same volume (mass) of water, same calorimeter, and kept at the same height above the spirit burner, burn for the same length of time, ideally to burn the same mass of fuel or burn to give the same temperature rise - 'trial and error' - see what works and seem to give consistent results, same insulation set-up e.g. the lid, and using a draught shield, all designed to minimise heat losses. This is a very inaccurate method because of huge losses of heat e.g. radiation from the flame and calorimeter, conduction through the copper calorimeter, convection from the flame gases passing by the calorimeter etc. BUT, at least using the same burner and set–up, you can do a reasonable comparison of the heat output of different fuels. You can burn series of simple hydrocarbons like , or alcohols like (alcohols give the best results) and even and it is possible to do a crude calibration of the calorimeter using a fuel of known energy output on complete combustion. The colorimeter data calculation methods are described in the last . You can investigate the temperature rise produced in a known mass of water by the combustion of the series alcohols e.g. methanol, ethanol, propanol, butanol using this simple calorimeter system to get a pattern for a homologous series of organic compounds. Alcohols burn more cleanly with a pale blue flame compared to the inefficient combustion with yellow smokey flames you see with hydrocarbons and oils See GCSE/IGCSE/O Level notes on chemistry of alcohols BOMB CALORIMETER Advanced Level students need to know about the for determining enthalpies of combustion as well as the methods described above.

6(d) A note on graphical analysis - how to obtain a more accurate temperature change

To get the best value for the temperature change (ΔT) you should take multiple readings before and after mixing the reactants and then plotting a graph versus time.

This is required in advanced A level chemistry courses, but maybe treated more simply at GCSE level.  Above are two graphs from simple calorimeter experiments (picture on right).

On the left are typical results from an exothermic reaction e.g. metal plus acid or metal plus metal salt displacement reaction.

On the right are typical results from an endothermic change e.g. when certain salts dissolve in water.

The initial readings give you a baseline, but the reaction may take a few seconds or a few minutes, and so you cannot get an immediate true ΔT. However by drawing a baseline for the initial temperature and extrapolating back to the start of the reaction (e.g. at 1.5 minutes) you can then estimate the real temperature change.

What happens is quite simple, but it leads to inaccuracy:

For exothermic reactions the system will continuously lose heat once the reaction has started, so the temperature starts to fall once the reaction is complete, so extrapolating back up gives the true temperature rise.

At GCSE level, it might be ok to just subtract the actual difference between the maximum and minimum readings. In this case a rise in temperature of 27.8 - 20. 4 = 7.4oC.

In this case ΔT (corrected) = 28.4 - 20.4 = 8.0oC (and this is more accurate than a ΔT of 7.4)

For endothermic reactions the system will continuously gain heat once the reaction has started, so the temperature starts to rise once the reaction is complete, so extrapolating back down gives the true temperature fall.

Again, at GCSE level, it might be ok to just subtract the actual difference between the maximum and minimum readings. In this case a fall in temperature of 19.5 - 11.9 = 7.6oC.

In this case ΔT (corrected) = 19.5 - 11.2 = 8.3oC (and this is more accurate than a ΔT of 7.6)

Without allowing for these unavoidable experimental circumstances, you will always measure too low a temperature change.

7. Calculations from the experimental calorimeter results

• PLEASE NOTE that section 7. is for higher GCSE students and an introduction for advanced level students of how to do energy change (enthalpy change) calculations from experimental data.

• The calculation method described below applies to both experimental methods 6.1 and 6.2 described above.

• You need to know the following:

• the mass of material reacting in the calorimeter (or their concentrations and volume),

• the mass of water in the calorimeter,

• the temperature change (always a rise for method 6.2 combustion),

• the specific heat capacity of water, (shorthand is SHCwater), and this is 4.2J/goC (for advanced 4.2J g–1 K–1),

• this means it means the addition of 4.2 J of heat energy to raise the temperature of 1g of water by 1oC.

• Example Calculation 7.1 typical of calorimeter method 6.1

• Measuring the energy transfer when a salt dissolves in water

• 5g of ammonium nitrate (NH4NO3) was dissolved in 50cm3 of water (50g) and the temperature fell from 22oC to 14oC.

• Temperature change = 22 – 14 = 8oC (endothermic, temperature fall, heat energy absorbed)

• Heat absorbed by the water = mass of water x SHCwater x temperature

• = 50 x 4.2 x 8 = 1680 J (for 5g)

• heat energy absorbed on dissolving = 1680 / 5 = 336 J/g of NH4NO3

• this energy change can be also expressed on a molar basis.

• Relative atomic masses Ar: N = 14, H = 1, O = 16

• Mr(NH4NO3) = 14 + (1 x 4) + 14 + (3 x 16) = 80, so 1 mole = 80g

• Heat absorbed by dissolving 1 mole of NH4NO3 = 80 x 336 = 26880 J/mole

• At A level this will be expressed as enthalpy of solution = ΔHsolution = +26.88 kJ/mol

• The data book value is +26 kJmol–1

•

• Example Calculation 7.2 typical of calorimeter method 6.2

• Determining the energy change for a typical fuel combustion reaction

• 100 cm3 of water (100g) was measured into the calorimeter.

• The spirit burner contained the fuel ethanol C2H5OH ('alcohol') and weighed 18.62g at the start.

• The initial temperature of the water is taken.

• After burning some time, the flame is extinguished, the water stirred gently and the final water temperature is taken to get the temperature rise.

• The burner and fuel are then reweighed to see how much fuel had been burned.

• After burning it weighed 17.14g and the temperature of the water rose from 18 to 89oC.

• The temperature rise = 89 – 18 = 71oC (exothermic, heat energy given out).

• Mass of fuel burned = 18.62–17.14 = 1.48g.

• Heat absorbed by the water = mass of water x SHCwater x temperature

• = 100 x 4.2 x 71 = 29820 J (for 1.48g)

• heat energy released per g = energy supplied in J / mass of fuel burned in g

• heat energy released on burning = 29820 / 1.48 = 20149 J/g of C2H5OH

• this energy change can be also expressed on a molar basis.

• Relative atomic masses Ar: C = 12, H = 1, O = 16

• Mr(C2H5OH) = (2 x 12) + (1 x 5) + 16 + 16 = 46, so 1 mole = 46g

• Heat released (given out) by 1 mole of C2H5OH = 46 x 20149 = 926854 J/mole or 927 kJ/mol (3 sf)

• At A level this will be expressed as the ...

• Enthalpy of combustion of ethanol = ΔHcombustion (ethanol) = –927 kJmol–1

• This means 926.9 kJ of heat energy is released on burning 46g of ethanol ('alcohol').

• The data book value for the heat of combustion of ethanol is –1367 kJmol–1, showing lots of heat loss in the experiment!

• It is possible to get more accurate values by calibrating the calorimeter with a substance whose energy release on combustion is known.

•

Example Calculation 7.3

Determining the energy change of neutralisation of hydrochloric acid and sodium hydroxide

You can do this experiment by mixing equal volumes of equimolar concentrations of dilute hydrochloric acid and dilute sodium hydroxide. e.g. 25 cm3 of each in the polystyrene calorimeter as previously described.

Suppose after mixing, via accurate pipettes, 25.0 cm3 of 1.0 mol dm–3 hydrochloric acid and 25.0 of 1.0 mol dm–3, sodium hydroxide solutions the temperature rise with an accurate thermometer was 7.1oC.

Calculate the energy of neutralisation for the reaction:

HCl(aq)  +  NaOH(aq)  ===>  NaCl(aq)  +  H2O(l)

Calculation   (SHC shorthand for specific heat capacity, 4.18 is more accurate than 4.2)

Using the SHC for water and the total mass is effectively 50 g (actually ~50 cm3 of NaCl solution).

heat released (J) = mass x  SHCH2O x temperature change (ΔT, oC)

=  50 x 4.18 x 7.1 = 1483.9 J, 1.4839 kJ

From the equation: mol HCl = mol NaOH = 1.0 x 25/1000 = 0.025 mol

Therefore scaling up to 1 mol gives an energy change of 1.4839 x 1/0.025 = 59.4 kJ per mole in equation

Since the temperature rose indicating an exothermic reaction, the energy of this neutralisation is ..

At A level this would be expressed as

Enthalpy of neutralisation for HCl + NaOH = 59.4 kJ mol–1  (only accurate to 3 sf)

Example 7.4 Determining the energy change of the displacement of copper by zinc

25 cm3 of copper(II) sulfate solution was measured into the calorimeter.

The initial temperature of the solution and zinc powder was 22.0oC.

Excess zinc was added and the mixture gently stirred until the maximum temperature was reached and recorded.

This simple observation alone tells you its an exothermic reaction. The final temperature rise was 72.0oC.

Assume the mass of the solution is 25 g and its heat capacity is the same as water  (4.2 J/goC.).

If the concentration of the copper(II) was 1.0 mol/dm3, calculate the energy change per mole of copper.

temperature change = 72.0 - 22.0 = 50.0oC

heat released (J) = mass x  SHCH2O x temperature change (ΔT, oC)

= 25 x 4.18 x 50.0 = 5225 J CuSO4(aq)  + Zn(s)  ===> ZnSO4(aq)  +  Cu(s)

From the balanced equation 1 mol of copper sulfate = 1 mol of copper displaced

mol = molarity x volume(dm3)

mol Cu displaced = 1.0 x 25/1000 = 0.025

Therefore displacing 0.025 mol of Cu releases 5225 J of heat energy

Scaling up to 1 mole

Heat released = 5225 x 1/0.025 = 209000 J per mole of copper displaced

(= 209 kJ/mol, quite exothermic!)

At A level this would be written as: enthalpy of reaction = ΔHθreaction = -209 kJ mol-1

• In some exothermic changes, no heat is not released e.g. in batteries and fuel cells, where the energy is released as electrical energy.

• See the separate pages on

• See other pages related to fuels

What next? Associated Pages

Sub–index for ENERGY CHANGES:

and enthalpy calculations from calorimetry data for Advanced A Level chemistry students GCSE/IGCSE/O level notes on Oil–Fuel burning GCSE/IGCSE/O level Types of Chemical Reaction Notes GCSE/IGCSE/O Level Rates of Reaction Revision Notes

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Experimental methods for obtaining vales for energy transfer changes in chemical reactions are described and how to do the calculations based on calorimeter experiment results. Calculation of energy transferred from experimental data is explained. A simple calorimeter is described and how to obtain energy transfer measurements. Revision notes for GCSE/IGCSE/O Level/basic stuff for GCE Advanced Level AS students. These revision notes on calorimeter experiments, procedures and calculations of energy transfers in chemical reactions should prove useful for the new AQA chemistry, Edexcel chemistry & OCR chemistry GCSE (9–1, 9-5 & 5-1) science courses.

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