# DF defg: Energetics Flashcards

1
Q

What is thermochemistry?

A

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

2
Q

Draw an enthalpy level diagram for an exothermic reaction.

A

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.

3
Q

Enthalpy change is measured in what units?

A

kJ mol-1

4
Q

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

A

ΔrH298

5
Q

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

A
• 298 K (25oC)
• 100 kPa (1 atm)
• 1 mol dm-3 (molar)
6
Q

How are enthalpy changes often measured in the laboratory?

A
• Energy involved in reaction transferred to/from water surrounding reaction vessel
• ΔT of water measured
• ΔH calculated using ΔE = mcΔT
7
Q

Explain how a bomb calorimeter accurately measures enthalpy changes.

A
• 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
8
Q

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

A

ΔfH298

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

9
Q

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

A

ΔrH298

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

10
Q

Give 2 formulae which can be used to calculate ΔrH.

A

ΣΔfHproducts - ΣΔfHreactants

ΣHbonds formed - ΣHbonds broken

11
Q

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

A

ΔcH298

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

12
Q

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

A
• Balance it
• Show one mole of substance combusting
• Include standard state symbols
13
Q

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

A

ΔneutH298

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.

14
Q
A

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

15
Q

Define specific heat capacity.

A

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

16
Q

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.

A

Standard initial conditions are used.

17
Q

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

A
18
Q
A
19
Q
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.
A
20
Q

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 19.52oC to 27.40oC.

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

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

A

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

21
Q

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 19.52oC to 27.40oC.

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

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

A

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.

22
Q

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 19.52oC to 27.40oC.

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

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

A

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
• cwater x volumewater = csolution x volumesolution
23
Q

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.
A

ΔfH = ΔH1 - ΔH2 = -65.9 - 10.3 = -76.2 kJ mol-1

24
Q
A

B

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

25
Q

Define average bond enthalpy.

A

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

26
Q

Is bond-breaking an exothermic or endothermic process?

A

Endothermic

27
Q

Is bond-making an exothermic or endothermic process?

A

Exothermic

28
Q

Explain what is meant by “equilibrium bond length”.

A

Equilibrium between:

• nuclei-electron attraction
• nucleus-nucleus repulsion
30
Q

Why are bond enthalpies always positive values?

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

Reactants gain energy, so positive

31
Q

Why do bond enthalpy values for a given bond differ?

A
• 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.

32
Q

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

A
• Shorter
• More energy needed to break bond means greater electron-nucleus attraction
33
Q

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

A

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

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

ΔHc 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

34
Q

Why may some reactions not need continuous heating to proliferate?

A
• 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