3.2.1 (e) Q=mcΔT

(e) determination of enthalpy changes directly from appropriate experimental results, including use of the relationship: q = mcΔT

The terms of the equation - q = mcΔT

This is a topic once widely covered at GCSE that you may never have seen

The equation does not appear on the new data-sheet so you will have to learn it.

And you'll need to know what the symbols mean and what units they are in.

Q is Heat energy - who can say why. It's Physics.

All energy is in Joules (J) or kilojoules (kJ)

m is Mass in g - particularly it is the mass of what is being heated.

This will pretty nearly always be water.

Either you will be heating water by burning a fuel - in which case the mass of the water you are heating is the same as its volume in cm³

1cm³ = 1g ----for water

Or perhaps a solid is reacting with a solution in which case you ignore the mass of the solid, and use the volume of the solution in cm³ as its mass even though the solution is not pure water.

Or two solutions react in which case you use the combined volumes of both solutions in cm³ as its mass even though the solution is not pure water.

C is the Specific Heat Capacity of Water.

This is given to you on the data sheet (and often in questions)

Specific heat capacity of water, c = 4.18 J g^–1 K^–1

Physics tells us that it is the amount of energy (in J) needed to raise the temperature of 1g of water by 1 degree. But they won't ask this on a Chemistry paper.

ΔT is temperature change.

Strictly this is in Kelvin.

But ΔT = T_final - T_initial

And it won't make any difference whether you use Kelvin or Celsius to work it out.

ΔT = 283.15K - 0K = 283.15K

ΔT = 10^oC - -273 ^oC = 283.15 ^oC and a difference of is also a difference 283.15 K

At GCSE Q=mcΔT would have been enough, but an A level paper is likely to ask you to calculate ΔH_{neutralisation} or ΔH_reaction which are always defined as being an enthalpy change for 1 mole.

So you would have to work out how many moles had reacted and then remember to divide that answer into your answer for Q - having first converted Q from J to kJ so that your ΔH_{neutralisation} will be in kJ/mol.

eg. In a reaction between 30cm³ of 0.1 mol/dm³ HCl and 25 cm³ of 0.2 mol/dm³ NaOH the temperature rises from 25^oC to 38^oC. Calculate ΔH_{neutralisation} of the reaction. { HCl + NaOH → NaCl + H₂O }

Firstly, we would need to find out which reactant is the limiting reagent (the one which runs out) and which is in excess (which we'll ignore)

i)

Moles (HCl) = Conc x Vol/1000 = 0.1 x 0.03 = 0.003 moles

Moles (NaOH) = Conc x Vol/1000 = 0.2 x 0.025 = 0.005 moles

Looking at the equation, the reacting ration is 1:1 so clearly it is the HCl which will run out.

So this is the important number of moles since not all of the NaOH will react.

ii)

Q = mcΔT

The mass is the mass of "water" being heated - which is the same as the combined volume of water in the two solution, measured in cm³. We ignore the inconvenient fact that neither solution is pure water.

Total volume of "water" = 30cm³ + 25cm³ = 55cm³

Total mass of "water" = = 55g

ΔT = 38 - 25 = 13 K

Q = 55 x 8.314 x 13 = 5944.51 J = 5.945 kJ

iii)

At this point we must think about the definition of the enthalpy change we are measuring.

ΔH_{neutralisation} is defined as the enthalpy change when one mole of water is made.

So, since exactly 0.003 moles of HCl reacted the equation tells us that 0.003 moles of water should be made

So, ΔH_{neutralisation} =Q/moles of water made

ΔH_{neutralisation} =5.945/0.003 = 1981.5 kJ/mol

eg. 30cm³ of water is heated by burning ethanol in a spirit burner. The temperature rises from 25^oC to 38^oC. The mass of the spirit burner and fuel was 48.295g. After burning it had decreased to 48.100g. Calculate ΔH_combustion of the reaction. { C₂H₅OH+ 3O₂ → 2 CO₂ + 3H₂O }

This time we don't really have to worry about calculating a limiting reagent since the ethanol is burning in air which is clearly in excess.

We'll still need to know how much ethanol reacted, however.

And we'll work that out from the mass of ethanol that reacted which is equal to the change of mass of the burner.

i)

M_r (C₂H₅OH) = (12 x 2) + (5 x 1) + 16 + 1 = 46 g/mol

Mass (C₂H₅OH) = 48.295 - 48.100 = 0.195 g

Moles (C₂H₅OH) = Mass/M_r= 0.195/46= 4.24 x 10^-3 moles

ii)

Q = mcΔT

The mass is the mass of "water" being heated - which is the same as the volume of water heated.

Total volume of water = 30cm³

Total mass of water = = 30g

ΔT = 38 - 25 = 13 K

Q = 30 x 8.314 x 13 = 3242.46 J = 3.243 kJ

iii)

At this point we think about the definition of the enthalpy change we are measuring again.

ΔH_combustion is defined as the enthalpy change when one mole of a substance burns.

So, since exactly 4.24 x 10^-3 moles of ethanol reacted this will be the number to muse

So, ΔH_combustion =Q/moles of ethanol burned

ΔH_combustion = 3.243/4.24 x 10^-3 = 746.73 kJ/mol

1(a) An experiment was carried out to determine a value for the enthalpy of combustion of liquid methylbenzene using the apparatus shown in the diagram.

Burning 2.5 g of methylbenzene caused the temperature of 250 g of water to rise by 60°C. Use this information to calculate a value for the enthalpy of combustion of methylbenzene, C7H8

(The specific heat capacity of water is 4.18 J K–1 g–1. Ignore the heat capacity of the container.)

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(b) A 25.0 cm3 sample of 2.00 mol dm–3 hydrochloric acid was mixed with 50.0 cm3 of a 1.00 mol dm–3 solution of sodium hydroxide. Both solutions were initially at 18.0°C.

After mixing, the temperature of the final solution was 26.5°C.

Use this information to calculate a value for the standard enthalpy change for the following reaction.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

In your calculation, assume that the density of the final solution is 1.00 g cm–3 and that its specific heat capacity is the same as that of water. (Ignore the heat capacity of the container.)

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(c) Give one reason why your answer to part (b) has a much smaller experimental error than your answer to part (a).

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2. A 50.0 cm3 sample of a 0.200 mol dm–3 solution of silver nitrate was placed in a polystyrene beaker. An excess of powdered zinc was added to this solution and the mixture stirred. Zinc nitrate, Zn(NO3)2, and silver were formed and a rise in temperature of 3.20 °C was recorded.

(a) Write an equation for the reaction between silver nitrate and zinc.

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(b) Calculate the number of moles of silver nitrate used in the experiment.

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(c) Calculate the heat energy evolved by the reaction in this experiment assuming that all the energy evolved is used to heat only the 50.0 g of water in the mixture.

(Specific heat capacity of water is 4.18 J g–1 K–1)

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(d) Calculate the heat energy change for the reaction per mole of zinc reacted.

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(e) Explain why the experimental value for the heat energy evolved in this experiment is less than the correct value.

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(Total 8 marks)

Answers

1(a) Heat change = m c DT 1

= 250 × 4.18 × 60 = 62700J = 62.7kJ 1

Moles C7H8 = 2.5 /92 = 0.0272 1

DH = 62.7 / 0.0272 = – 2307 kJ mol–1 1

(allow –2300 to –2323)

(b) Mass of water heated = 25 + 50 = 75g

Temp rise = 26.5 – 18 = 8.5 oC 1

both for (1) mark

Heat change = 75 × 4.18 × 8.5 = 2665 J = 2.665 kJ 1

Moles HCl = 0.05 1

DH = – 2.665 / 0.05 = –53.3 kJmol–1 1

(allow –53 to –54)

(c) Less heat loss 1

2. (a) 2AgNO3 + Zn ® Zn(NO3)2 + 2Ag (1)

Accept an ionic equation i.e.2Ag+ +Zn ® 2Ag + Zn2+

1

(b) Moles = mv / 1000 (1) = 0.20 × 50/1000 = 1.00 × 10–2 2

(c) Heat energy change = mCDT (1) = 50 × 418 × 3.2 J

= 669 J (Ignore signs) (1)

Allow 668, 67.0 0.67kJ

Penalise wrong units if given

2

(d) (2 x 669)/1x10^-2 = 134 kJ mol–1

Mark one : 2 × (answer to (c))

Mark two : Dividing by answers to (b)

Allow 133 – 134

Penalise incorrect units

Mark conseq to equation in (a) for full marks, also to that in (c)

If No working is shown and answer is incorrect zero

2

(e) Incomplete reaction or Heat loss (1)

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