CALORIMETER EXPERIMENTS & CALCULATION of ENERGY TRANSFER

Doc Brown's Chemistry KS4 science GCSE/IGCSE/O level/A Level Chemistry Revision Notes

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

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.

Sub–index for ENERGY CHANGES: 1. Heat changes in chemical/physical changes – exothermic and endothermic  *  2. Reversible reactions and energy changes  *  3. Activation energy and reaction profiles  *  4. Catalysts and activation energy  *  5. Introduction to bond energy/enthalpy calculations  *  6. Calorimeter methods of determining energy changes and examples of experiments  *  7. Energy transfer calculations from calorimeter results (this page)

and enthalpy calculations from calorimetry data for Advanced A Levels students

6. The experimental determination of energy changes using simple calorimeters

The basic principles of calorimetry

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 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 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 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

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Sub–index for ENERGY CHANGES:

and enthalpy calculations from calorimetry data for Advanced A Levels students