Energetics

Section 1 of 17Enthalpy change

Enthalpy

• Enthalpy the total chemical energy (heat content) of a substance

Enthalpy change, ΔH = the heat change measured under conditions of constant pressure

Sign

• ΔH is negative for exothermic and positive for endothermic reactions, shown on enthalpy level diagrams

Exothermic

• Heat is given out to the surroundings, so their temperature increases.
• Products have less energy than reactants, so ΔH is negative.

Endothermic

• Heat is taken in from the surroundings, so their temperature decreases.
• Products have more energy than reactants, so ΔH is positive.

Section 2 of 17Standard conditions and standard enthalpy changes

Standard conditions

• Standard conditions pressure 100 kPa, temperature 298 K (25 °C), and every substance in its normal physical state

Symbol

• Symbol the standard symbol θ is added to ΔH, sometimes written ΔH₂₉₈

Standard enthalpy of combustion, ΔH_c = the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions with everything in standard states

Combustion sign

• Combustion sign it is always negative because combustion is exothermic

Standard enthalpy of formation, ΔH_f = the enthalpy change when one mole of a substance is formed from its constituent elements under standard conditions with everything in standard states

Formation sign

• Formation sign it can be positive or negative, and it is 0 for elements

Section 3 of 17Calorimetry of fuels

• Method
  • a fuel burned in a spirit burner heats water in a copper can, with a lid to prevent heat loss

1

Measure a fixed volume of water into a copper can.

2

Weigh the spirit burner with fuel.

3

Measure the initial water temperature.

4

Burn the fuel and stir the water.

5

Extinguish the flame when the temperature has risen by about 20 °C.

6

Record the final temperature and re-weigh the burner.

Calorimetry of fuels

Specific heat capacity, c

• Specific heat capacity, c the energy to raise 1 g of a substance by 1 °C, 4.18 J/g/°C for water

Energy released

• Energy released q = m × c × ΔT, with q in J, m in g and ΔT in °C

  What the symbols mean: E, Q or q = energy transferred (J) · m = mass · c = specific heat capacity (J kg⁻¹ °C⁻¹; water 4.18 J g⁻¹ °C⁻¹) · ΔT or Δθ = temperature change (°C or K)
  • (kg; g in chemistry calorimetry)

Per gram

• Per gram energy released per gram = q ÷ mass of fuel burned

Per mole

• ΔH = q ÷ n

  What the symbols mean: ΔH = enthalpy change (kJ mol⁻¹) · q = heat energy measured · n = amount of the limiting reactant (mol); negative sign = exothermic
  • (J, divide by 1000 for kJ)

Example

• 0.01 mol of propanol heats 250 g of water by 12 °C, so q = 12 540 J and ΔH = −1254 kJ mol^-1

Section 4 of 17Calorimetry of reactions in solution

Use

• Use neutralisation and displacement reactions, from the temperature change when solutions are mixed

Insulation

• Insulation use an insulated container such as a polystyrene cup to prevent heat loss

Calorimetry of reactions in solution

Add a fixed volume of one reagent to the calorimeter and record its temperature.

Add an excess of the second reagent while stirring continuously.

Record the maximum temperature and calculate the rise.

Assumption

• Assumption the solution has the density of water, 1 g cm^-3, so mass in g equals volume in cm³

Calculation

• Calculation q = m × c × ΔT, then ΔH = q ÷ n in kJ mol^-1

Example

• 50 cm³ of 1.00 mol dm^-3 HCl and 50 cm³ of NaOH raise the temperature by 12 K, so q = 5016 J, n = 0.05 mol and ΔH = −100.32 kJ mol^-1

Temperature units

• Temperature units a change is the same in K and °C

Section 5 of 17Cooling curves

• Problem
  • if a reaction is slow, the mixture loses heat before the maximum temperature is reached, so ΔH is inaccurate

1

Place one reactant in the polystyrene cup.

2

Record its temperature every minute for about 4 minutes.

3

Add the second reactant and keep recording every minute for about 6 minutes.

4

Plot temperature against time.

5

Extrapolate the cooling part of the line back to the time when the second reactant was added.

• Result
  • the temperature change is taken at the time of mixing from the extrapolated line

Section 6 of 17Required Practical 2: aim and apparatus

Aim

• Aim to find the enthalpy change for converting anhydrous copper(II) sulfate to hydrated copper(II) sulfate, ΔH₃

Why indirect

• ΔH₃ cannot be measured directly, so it is calculated from ΔH₁ and ΔH₂ (hydrated to solution)
  • (anhydrous to solution)

Apparatus

• Apparatus hydrated and anhydrous copper(II) sulfate, polystyrene cups with lids, 250 cm³ beaker to hold the cup, stand and clamp, thermometer, two 25 cm³ measuring cylinders, two weighing bottles, stopwatch, graph paper, stirrer, distilled water and a digital balance

Safety

• Safety avoid anhydrous copper(II) sulfate on skin, as it can irritate

Section 7 of 17Required Practical 2: method

Experiment 1: ΔH₁

• Weigh 3.90–4.10 g of anhydrous copper(II) sulfate and record the mass.
• Measure 25 cm³ of deionised water into a polystyrene cup and record the temperature at t = 0.
• Record the temperature every minute for three minutes.
• At minute 4 add the powder rapidly and stir, without recording the temperature.
• Stir and record the temperature every minute from minute 5 to 15.

Experiment 2: ΔH₂

• Weigh 6.20–6.30 g of hydrated copper(II) sulfate and record the mass.
• Use 24 cm³ of deionised water in a polystyrene cup and record the temperature at t = 0.
• Record every minute for three minutes, add the hydrated solid at minute 4 without recording, and record each minute from 5 to 15.

Table

• Table columns for experiment 1 and experiment 2, with minute 4 left blank because the solid is being added

Section 8 of 17Required Practical 2: analysis

1

Plot temperature on the y-axis against time on the x-axis for each experiment.

2

Draw two best-fit lines, one for the points before the solid was added and one after.

3

Extrapolate both lines to minute 4.

4

Read the temperature change at minute 4.

5

Use q = m × c × ΔT and ΔH = q ÷ n for ΔH₁ and ΔH₂.

6

Calculate ΔH₃ = ΔH₁ − ΔH₂.

• Water mass
  • 25 cm³ of water has a mass of 25 g because the density is 1 g cm^-3

Section 9 of 17Required Practical 2: example calculation

Experiment 1

• 4.94 g of anhydrous CuSO₄ (M_r 159.6) is 0.0310 mol and the temperature rises by 8.4 °C, so q = 25 × 4.2 × 8.4 = 882 J and ΔH₁ = −28.45 kJ mol^-1

Experiment 2

• 7.72 g of hydrated CuSO₄ (M_r 249.6) is 0.0309 mol and the temperature falls by 1.7 °C, so q = 178.5 J and ΔH₂ = +5.78 kJ mol^-1

Result

• ΔH₃ = −28.45 − (+5.78) = −34.2 kJ mol^-1

Section 10 of 17Hess’s law

Hess’s law = the total enthalpy change of a reaction is independent of the route taken

Meaning

• Meaning a reaction in one step or several steps has the same total enthalpy change, so the direct route equals the indirect route

Use

• Use it finds enthalpy changes that cannot be measured by calorimetry,
  • such as ΔH_f of propane from 3C(s) + 4H₂(g) → C₃H₈(g),
  • since carbon and hydrogen do not react under standard conditions

Section 11 of 17Finding ΔH_f from combustion data

• Why
  • formation enthalpies are hard to measure, but combustion enthalpies can often be measured

1

Write the equation for the formation of the compound.

2

Write the combustion products beneath.

3

Draw downward arrows from each substance to its combustion products.

4

Write the ΔH_c values on the arrows and multiply by the moles.

5

Go from reactants to products, changing the sign when an arrow is followed against its direction.

• Propanone example
  • 3C + 3H₂ + ½O₂ → CH₃COCH₃ gives ΔH_f = −1182 − 858 + 1821 = −219 kJ mol^-1, reversing the −1821 value

Section 12 of 17Finding ΔH_r from formation data

1

Write the equation for the reaction.

2

Write the elements, with moles and state symbols, beneath.

3

Draw upward arrows to each compound.

4

Write the ΔH_f values on the arrows and multiply by the moles.

5

Go from reactants to products, changing the sign when an arrow is followed against its direction.

Example

• NH₄NO₃(s) + ½C(s) → N₂(g) + 2H₂O(g) + ½CO₂(g) gives ΔH_r = +365 − 484 − 197 = −316 kJ mol^-1

Elements

• Elements no arrows are needed for carbon and nitrogen because ΔH_f of an element is zero

Section 13 of 17Bond enthalpy: bond breaking and making

Bond breaking

• Bond breaking it is endothermic because energy is needed to overcome the attractive forces between atoms

Bond making

• Bond making it is exothermic because energy is released to the surroundings

Overall

• Overall if more energy is needed to break bonds than is released making bonds, the reaction is endothermic, and the reverse is exothermic

Reality

• Reality only some bonds in the reactants are broken before new ones are formed

Section 14 of 17Bond dissociation and mean bond enthalpy

Bond dissociation enthalpy (bond enthalpy), E = the energy needed to break one mole of a specific covalent bond in the gas phase

Mean bond enthalpy

• Mean bond enthalpy an average over the same bond in different environments, because other atoms in the molecule affect it

Determination

• Determination bond enthalpies cannot be measured directly, so enthalpy cycles are used

Equal and opposite

• Equal and opposite the energy released when a bond forms equals the energy absorbed when it breaks,
  • so O=O is +498 kJ mol^-1 to break and −498 kJ mol^-1 to form

Section 15 of 17Calculating ΔH from mean bond enthalpies

• Equation
  • ΔH^θ = enthalpy change for bonds broken − enthalpy change for bonds formed, using positive bond enthalpies

1

Write the balanced equation.

2

Set out a balance sheet of bonds broken and bonds formed.

3

Multiply each bond enthalpy by the number of bonds.

4

Total each side and subtract formed from broken.

1Haber process

• N≡N 945, H–H 436 and N–H 391 give 2253 broken and 2346 formed, so ΔH = −93 kJ mol^-1

2Ethyne combustion

• Ethyne combustion for one mole of ethyne, C≡C 839, 2 × C–H 828 and 2½ × O=O 1245 give 2912 broken, and 2 × O–H 926 plus 4 × C=O 3216 give 4142 formed,
• so ΔH = −1230 kJ mol^-1

3Sign

• Sign a negative result means the reaction is exothermic

4Difference from Hess

• Difference from Hess mean bond enthalpies give different results from Hess cycles, which use data from the actual compounds

Section 16 of 17Key terms

Term	Meaning
Enthalpy change	The heat change at constant pressure, ΔH.
Standard enthalpy of combustion	The enthalpy change when one mole of a substance burns completely in oxygen under standard conditions.
Standard enthalpy of formation	The enthalpy change when one mole of a compound forms from its elements under standard conditions.
Specific heat capacity	The energy needed to raise 1 g of a substance by 1 °C.
Calorimetry	Measuring heat changes using the temperature change of a known mass of water.
Cooling curve	A graph of temperature against time used to correct for heat loss.

Term	Meaning
Hess’s law	The total enthalpy change is independent of the route taken.
Bond enthalpy	The energy to break one mole of a specific covalent bond in the gas phase.
Mean bond enthalpy	An average bond enthalpy over different environments.
Enthalpy cycle	A diagram linking reactants and products by alternative routes.
Balance sheet	A layout listing bonds broken and bonds formed for a calculation.

Section 17 of 17Exam technique

1

Using the final temperature instead of the extrapolated value — heat loss makes ΔH too small.

2

Leaving out the lid or insulation — heat loss gives inaccurate values.

3

Giving ΔH for an exothermic reaction as positive — ΔH is negative when temperature rises.

4

Using the mass of the solid instead of the mass of water in q = mcΔT — use the water (or solution).

1

Using ΔH_f = products minus reactants for combustion data — use reactants minus products for ΔH_c.

2

Not reversing the sign when following an arrow backwards in a cycle — change the sign.

3

Using unmultiplied bond enthalpies — multiply by the number of each bond.

4

Counting the ethyne equation for two moles for ΔH_c — ΔH_c is per mole of fuel.