DF defg: Energetics

Question 1

What is thermochemistry?

Answer 1

The study of the energy and heat associated with chemical reactions.

Question 2

Draw an enthalpy level diagram for an exothermic reaction.

Answer 2

Consider enthalpy in terms of total energy stored in system (bonds of reactants + products). Reactants store more energy than products, so energy is lost, so ΔH is -ve.

Question 3

Enthalpy change is measured in what units?

Answer 3

kJ mol^-1

Question 4

How is “enthalpy change of a reaction at standard conditions” expressed symbolically?

Answer 4

Δ_rH^⦵₂₉₈

Question 5

What are the standard conditions of temperature, pressure and concentration used to compare enthalpy changes?

Answer 5

• 298 K (25^oC)
• 100 kPa (1 atm)
• 1 mol dm^-3 (molar)

Question 6

How are enthalpy changes often measured in the laboratory?

Answer 6

• Energy involved in reaction transferred to/from water surrounding reaction vessel
• ΔT of water measured
• ΔH calculated using ΔE = mcΔT

Question 7

Explain how a bomb calorimeter accurately measures enthalpy changes.

Answer 7

• Fuel burned in oxygen in pressurised vessel
• Energy transferred to/from surrounding water; ΔT measured
• ΔH calculated using ΔE = mcΔT
• When calculating ΔH, pressure variations caused by constant volume of closed vessel are accounted for

Question 8

Give the definition and symbolic representation of the standard enthalpy change of formation.

Answer 8

Δ_fH^⦵₂₉₈

The enthalpy change when one mole of a compound is formed from its elements, under standard conditions and in standard states.

Question 9

Give the definition and symbolic representation of the standard enthalpy change of reaction.

Answer 9

Δ_rH^⦵₂₉₈

The enthalpy change when molar quantities of reactants, of proportions stated in the equation, react under standard conditions and in standard states.

Question 10

Give 2 formulae which can be used to calculate Δ_rH^⦵.

Answer 10

ΣΔ_fH^⦵_products - ΣΔ_fH^⦵_reactants

ΣH_{bonds formed} - ΣH_{bonds broken}

Question 11

What is the definition and symbolic representation of the standard enthalpy change of combustion?

Answer 11

Δ_cH^⦵₂₉₈

The enthalpy change when one mole of a substance is combusted completely in oxygen, in standard conditions and states.

Standard conditions + states are impossible in practice, so the substance is burnt in non-standard conditions, then adjustments are made to account for this

Question 12

What 3 things should you always do when writing the equation to represent the standard enthalpy change of combustion?

Answer 12

• Balance it
• Show one mole of substance combusting
• Include standard state symbols

Question 13

Give the definition and symbolic representation of the standard enthalpy change of neutralisation.

Answer 13

Δ_neutH^⦵₂₉₈

The enthalpy change when one mole of hydrogen ions react with one mole of hydroxide ions to form one mole of water, under standard conditions and in solutions of molar concentration.

Question 14

Answer 14

A

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol NaOH = mol H₂O = 2 x 10 x 10^-3 = 0.02 mol

Energy to form 0.02 mol H₂O = 20 x 4.18 x y

Energy to form 1 mol H₂O (as in definition) = 4180 y J = 4.18 y kJ

Question 15

Define specific heat capacity.

Answer 15

The amount of energy required to raise the temperature of 1 g of a substance by 1^oC.

Question 16

The following is Hess’ law. What assumption is made?

The total enthalpy change during the complete course of a chemical reaction is consistent, whether the reaction’s course is over one or several steps.

Answer 16

Standard initial conditions are used.

Question 17

Draw an enthalpy cycle for the formation of 1 mole of methane.

Answer 17

Question 18

Answer 18

Question 19

1. Write an equation for the formation of butane from its elements in their standard states.
2. Draw an enthalpy cycle showing the relationship between the formation of butane from its elements and their combustion.
3. Use this and the enthalpies below to calculate a value for the standard enthalpy change of formation of butane.

Answer 19

Question 20

A student measured the enthalpy change of the reaction:

CuSO₄(s) + 5H₂O(l) → CuSO₄ • 5H₂O(s)

Experiment 1: dissolve 3.99 g (0.0250 mol) anhydrous copper sulfate in 50 cm³ water. Temperature rose from 19.52^oC to 27.40^oC.

Experiment 2: add 6.24 g (0.0250 mol) hydrated copper sulfate to 48 cm³ water. Temperature fell from 19.56^oC to 18.28^oC.

Why can’t the enthalpy change of this reaction be measured directly?

Answer 20

Water begins dissolving products or reactants before reaction goes to completion.

Question 21

A student measured the enthalpy change of the reaction:

CuSO₄(s) + 5H₂O(l) → CuSO₄ • 5H₂O(s)

Experiment 1: dissolve 3.99 g (0.0250 mol) anhydrous copper sulfate in 50 cm³ water. Temperature rose from 19.52^oC to 27.40^oC.

Experiment 2: add 6.24 g (0.0250 mol) hydrated copper sulfate to 48 cm³ water. Temperature fell from 19.56^oC to 18.28^oC.

Show by calculation that 48 cm³ water was the correct volume to use in the 2nd experiment.

Answer 21

In both experiments, the student is forming CuSO₄(aq). In experiment 2, their aim is to add the correct amount of water to form a solution containing 50 cm³ of water, because this is the volume used in experiment 1.

Mol H₂O in CuSO₄ • 5H₂O = 5 x 0.025 = 0.125

Mass H₂O = 0.125 x 18 = 2.25 g so vol H₂O = 2.25 cm³

50 - 2.25 = 47.75 cm³, which is 48 cm³ to precision with which volume was measured.

Question 22

A student measured the enthalpy change of the reaction:

CuSO₄(s) + 5H₂O(l) → CuSO₄ • 5H₂O(s)

Experiment 1: dissolve 3.99 g (0.0250 mol) anhydrous copper sulfate in 50 cm³ water. Temperature rose from 19.52^oC to 27.40^oC.

Experiment 2: add 6.24 g (0.0250 mol) hydrated copper sulfate to 48 cm³ water. Temperature fell from 19.56^oC to 18.28^oC.

Calculate the enthalpy change in each experiment, stating any assumptions you make.

Answer 22

ΔH = mcΔT

E₁ = 50 x 4.18 x (19.52 - 27.4) x 10^-3 = -1.64 kJ

ΔH₁ = -1.64 / 0.025 = -65.9 kJ mol^-1

E₂ = 48 x 4.18 x (19.56 - 18.28) x 10^-3 = +0.257 kJ

ΔH₂ = +0.257 / 0.025 = +10.3 kJ mol^-1

Assumptions:

• No heat loss from apparatus
• Calorimeter absorbs negligable heat
• c_water x volume_water = c_solution x volume_solution

Question 23

A student measured the enthalpy change of the reaction:

CuSO₄(s) + 5H₂O(l) → CuSO₄ • 5H₂O(s)

Experiment 1: dissolve 3.99 g (0.0250 mol) anhydrous copper sulfate in 50 cm³ water. ΔH₁ = -65.9 kJ mol^-1

Experiment 2: add 6.24 g (0.0250 mol) hydrated copper sulfate to 48 cm³ water. ΔH₂ = +10.3 kJ mol^-1​

1. Draw a Hess’ cycle connecting the enthalpy changes in the two experiments with that of the equation.
2. Use this to calculate the enthalpy change for the reaction.

Answer 23

Δ_fH = ΔH₁ - ΔH₂ = -65.9 - 10.3 = -76.2 kJ mol^-1

Question 24

Answer 24

B

The amount doubles, but so does the volume of water to be heated

Question 25

Define average bond enthalpy.

Answer 25

The average energy change when one mole of a bond is broken, producing separate, gaseous atoms.

Question 26

Is bond-breaking an exothermic or endothermic process?

Answer 26

Endothermic

Question 27

Is bond-making an exothermic or endothermic process?

Answer 27

Exothermic

Question 28

Explain what is meant by “equilibrium bond length”.

Answer 28

Equilibrium between:

• nuclei-electron attraction
• nucleus-nucleus repulsion

Question 30

Why are bond enthalpies always positive values?

Answer 30

• Represent enthalpy change of breaking covalent bonds
• Bond breaking = endothermic

Reactants gain energy, so positive

Question 31

Why do bond enthalpy values for a given bond differ?

Answer 31

• Imprecise averages; specific value depends on compound in which bond is found
• Reaction may not be performed at standard conditions

Nothing to do with accuracy of procedure. Calorimeters + pure fuel sources used.

Question 32

Is a bond with a higher bond enthalpy longer or shorter, and why?

Answer 32

• Shorter
• More energy needed to break bond means greater electron-nucleus attraction

Question 33

Draw an enthalpy cycle and use the following bond enthalpies to work out a value for the enthalpy change of combustion of methane.

C-H = +413 kJ mol^-1

O=O = +498 kJ mol^-1

C=O = +805 kJ mol^-1

O-H = +464 kJ mol^-1

Answer 33

ΔH₂ = 4 x 413 + 2 x 498 = +2648 kJ mol^-1

ΔH₃ = -(2 x 805 + 4 x 464) = -3466 kJ mol^-1

ΔH_c or ΔH₁ = 2648 - 3466 = -818 kJ mol^-1

If you put 2O₂ in place of 4O, you need to subtract 2 x 498 when calculating ΔH₃

Question 34

Why may some reactions not need continuous heating to proliferate?

Answer 34

• Heating required initially to break bonds since this is endothermic
• New bonds form as old ones break; this is exothermic
• So bond-forming may provide enough energy to break further bonds; reaction proliferates