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Section 3 Energetics
3.1 Energy changes in chemical reactions
Conservation of energy
This law states that energy must be conserved in all processes.
Endothermic and exothermic reactions
An exothermic reaction is one in which stored chemical energy is converted to heat energy.
We often say that ‘heat is given out’.
An endothermic reaction is one in which heat energy is converted to chemical energy. Heat
energy is absorbed and this can cause a fall in temperature. Dissolving ammonium chloride
in water is an example of an endothermic reaction.
3.2 Standard enthalpy changes
Standard states of a substance
When we measure the enthalpy of a substance, we must state the temperature, pressure, and
physical state of the substance. It is usual to compare the enthalpies of substances in their
standard states.
The standard state of a substance is the pure substance in a specified state (solid, liquid or gas)
at 1 atmosphere and usually at 25 °C.
Associated with any physical or chemical change is a standard enthalpy change, denoted by
∆H°.
There are many types of enthalpy changes
(1) Standard enthalpy of formation
It is the heat absorbed when 1 mole of a substance is formed from its elements in their
standard states.
e.g. the enthalpy of formation of NaCl : Na(s) + ½Cl2(g) → NaCl(s)
Write down the equation for the enthalpy of formation of calcium carbonate.
By definition, all elements in their standard states are assigned a value of zero for their
standard enthalpies of formation. The following diagram illustrates how the value of ∆H°F
may be negative (for sodium chloride) or positive (as for ethyne). If the enthalpy absorbed is
negative, enthalpy (heat) is released, and the reaction is exothermic.
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(2) Standard enthalpy of combustion, ∆H°C
It is the heat absorbed when 1 mole of a substance is completely burnt in oxygen at 1 atm.
Since heat is usually evolved in such a reaction, ∆H°C will be negative, e.g.
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)
(3) Standard enthalpy of neutralisation, ∆H°neut
It is the heat absorbed when an acid and base react to form 1 mole of water under
standard conditions, e.g.
HCl(aq) + NaOH(aq) → NaCl(s) + H2O(l)
(4) Standard enthalpy of reaction, ∆H°R
It is the heat absorbed in a reaction at 1 atm between the number of moles of reactants
shown in the equation for the reaction.
In the reaction 4H2O(g) + 3Fe(s) → Fe3O4(s) + 4H2(g)
The enthalpy of reaction is related to 4 moles of steam and 3 moles of iron.
(5) Standard enthalpy of dissolution is the heat absorbed when 1 mole of a substance is
dissolved at 1 atm in a stated amount of solvent, e.g.
NaCl(s) → NaCl(aq)
(6) Standard enthalpy of atomisation, ∆H°atom
It is the enthalpy absorbed when a substance decomposes to form 1 mole of gaseous
atoms, e.g. ½Cl2(g) → Cl(g)
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Experimental Methods to determine enthalpy changes
(1) Neutralisation reactions between acids and bases can be determined using a simple
calorimeter.
Typical Result
When 100 g of water at 94.0 °C were added to a
calorimeter at 17.5 °C, the temperature rose to
80.5 °C.
250 cm3 of sodium hydroxide (0.4000 mol
dm-3) were added to 250 cm3 of hydrochloric
acid (0.400 mol dm-3) in the calorimeter. The
temperature of the solutions was 17.5 °C
initially and rose to 20.1 °C.
Calculate
(a) the heat capacity of the calorimeter
(b) the standard enthalpy of neutralisation
(2) Method to determine the enthalpy of combustion
Typical Result
When ethanol (C2H5OH) was burnt in the
apparatus shown on the left, the results were
m1 = 1.50 g, m2 = 500 g, t = 19.5 °C.
Find the enthalpy of combustion of ethanol.
(C=12, H=1, O=16)
Compare this with the listed value of -1368 kJ
mol-1. Explain the difference
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3.3 Hess’s Law
Sometimes a reaction can take place via two or more different routes. A question arises as
whether the enthalpy changes for the reaction differs if a reaction follows different routes.
Hess’s Law states that if a reaction can take place by more than one route, the overall change
of enthalpy is the same, whichever route is followed.
Hess’s Law is a special case of the Law of Conservation of Energy.
Finding the standard of enthalpy of formation of a compound indirectly
Some reactions are difficult to study, and the standard enthalpy of reaction must be found
indirectly.
(a) Enthalpy of Formation of Ethyne
For example, 2C(s) + H2(g) → C2H2(g)
This reaction involving the formation of ethyne from its constituent elements cannot be
achieved under standard conditions. Nevertheless., by measuring the enthalpy of combustion
of the reactants and products separately, the enthalpy of formation of ethyne can be calculated.
(a) First, write down an equation for the enthalpy of formation of ethyne.
(b) When ethyne is burnt, carbon dioxide and water are produced. Write an equation to
represent the combustion.
(c) Similarly when carbon and hydrogen are burnt separately, carbon dioxide and water are
formed. Construct an enthalpy diagram linking up the enthalpy of formation of ethyne,
enthalpy of combustion of ethyne and enthalpy of formation of carbon dioxide and water.
(d) Derive a mathematical relationship to relate the above energy terms.
(f) Given that the enthalpy of formation of water and carbon dioxide are -286 and -393 kJ
mol-1 respectively, and that the enthalpy of combustion of ethyne is -1300 kJ mol-1,
determine the enthalpy of formation of ethyne.
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(b) Enthalpy of Formation of Calcium Carbonate
Ca(s) + C(s) + 3/2 O2(g) → CaCO3(s)
The elements Ca, C and O2, under standard conditions, do not easily react to form calcium
carbonate. Thus, the enthalpy of formation of CaCO3 could not be obtained directly.
(i)
Ca is allowed to react with HCl, write an equation for this.
(ii)
CaCO3 is allowed to react with HCl, write an equation for this.
(iii)
Draw an energy cycle relating the equations (i), (ii) and the heat of formation of
calcium carbonate.
(iv)
Derive a mathematical relationship for all the energy terms appearing in (iii).
show how the enthalpy of formation of calcium carbonate could be obtained.
Hence
Bond enthalpies
When covalent bonds are broken, energy has to be supplied.
For example, in a complete dissociation
CH4(g) → C(g) + 4H(g); ∆H = +1662 kJ mol-1
Dividing the ∆H value by 4 gives an average value for the C-H bond enthalpy to be 416 kJ
−
mol-1. This value is often called the bond energy and is abbreviated as E (C-H).
Using bond energy to find enthalpy of reaction
Calculate the standard enthalpy of ethane, using the following information.
−
E (C=C) = 348 kJ mol-1
∆Hatom (carbon) = 718 kJ mol-1
−
E (C-H) = 416 kJ mol-1
∆Hatom(hydrogen) = 218 kJ mol-1
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Bond enthalpies, bond length and bond order
The following table gives the values of some bond energy terms kJ mol-1
H-H
436
O-H
464
C-C
347
H-Cl
431
C=C
611
H-Br
364
C≡C
837
H-I
299
(a) Compare the values for H-Cl, H-Br and H-I, what explanation can you give to the trend
in bond energy ?
(b) Compare the value of C-C, C=C and C≡C, what explanation can you give for the trend ?
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Energetics of formation of ionic compounds
The Born-Haber cycle is a technique of applying Hess’s law to the standard enthalpy changes
which occur when an ionic compound is formed.
The formation of sodium chloride from its constituent elements can be considered to occur by
means of the following steps.
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Given the following standard enthalpies, construct a Born-Haber Cycle, and use it to find the
standard lattice enthalpy of cadmium(II) chloride.
Cd(s) → Cd(g)
Cd(g) → Cd2+(g) + 2e
I2(s) → I2(g)
I2(g) → 2I(g)
I(g) + e → I-(g)
Cd(s) + I2(s) → CdI2(s)
∆H = + 113 kJ mol-1
∆H = + 2490 kJ mol-1
∆H = + 19.4 kJ mol-1
∆H = + 151 kJ mol-1
∆H = - 314 kJ mol-1
∆H = - 201 kJ mol-1