enthalpy changes
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Enthalpy changes. The enthalpy change for a process is the heat energy exchanged with the surroundings at constant pressure. Enthalpy change is given the symbol ∆ H . The units are kJ mol –1. - PowerPoint PPT PresentationTRANSCRIPT
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Enthalpy changes
The enthalpy change for a process is the heat energy exchanged with
the surroundings at constant pressure.
Enthalpy changes are frequently measured under standard conditions, i.e. 298 K and 100 kPa. If an enthalpy change is measured under standard conditions, the symbol ө is used in superscript, ∆Hө.
Enthalpy change is given the symbol ∆H. The units are kJ mol–1.
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Enthalpy changes
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Examples of enthalpy changes
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Bond enthalpies
When a chemical reaction takes place, bonds are broken in the reactants and bonds are formed in the products. Breaking bonds is an endothermic process. Making bonds is an exothermic process.
The mean bond enthalpy is the average (mean) bond dissociation enthalpy for a particular bond in
a range of different compounds.
Precisely, it is the average enthalpy change for breaking 1 mole of a particular bond in a range of
different compounds in the gas phase.
The enthalpy change for a reaction can be calculated by working out the enthalpy changes for bonds made and bonds broken during the reaction using mean bond enthalpies.
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Calculations using bond enthalpies
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What is a Born–Haber cycle?
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Constructing a Born–Haber cycle
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Born–Haber cycle for MgCl2
Mg(s) + Cl2(g)
Mg(g) + Cl2(g)
Haө (Mg) = +150
Mg(g) + 2Cl(g)
2 ×Haө (Cl) =
2 × (+121)
Mg+(g) + 2Cl(g) + e–
Hi(1st)ө (Mg) = +736
Hi(2nd)ө (Mg) = +1450
Mg2+(g) + 2Cl(g) + 2e–
Mg2+(g) + 2Cl–
(g)
2 ×Heaө (Cl) =
2 × (–364)
MgCl2(s)
Hlө (MgCl2) = –2493
Hfө (MgCl2)
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Can you construct a Born–Haber cycle?
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Born–Haber cycle questions
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Lattice formation enthalpies
When an ionic lattice is formed, the oppositely charged ions are attracted to each other. The stronger the attraction, the higher the lattice formation enthalpy.
Two factors increase the attraction and therefore the lattice formation enthalpy:
high charge
small size.
N3–O2–F–
Li+Na+K+
increasing lattice formation enthalpy
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Polarization and lattice enthalpy
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Fill in the missing words: lattice enthalpy
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Enthalpies of solution and hydration
The standard enthalpy of solution (Hsol
ө) is the enthalpy change when one mole of an ionic
compound is dissolved in water to produce aqueous ions.
Na+(g) Na+
(aq)
The standard enthalpy of hydration (Hhydө) is the
enthalpy change when one mole of gaseous ions is converted to one mole of aqueous ions.
NaCl(s) Na+(aq) + Cl–
(aq)
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Calculating enthalpies of solution
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Enthalpy of solution calculations
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Factors affecting enthalpy of hydration
The size of the enthalpy of hydration depends on:
incr
easi
ng s
ize
The charge on the ion. The larger the charge on the ion, the larger the enthalpy of hydration.
The size of the ion. The smaller the ion, the larger the enthalpy of hydration.
Li+
Na+
K+
–519
–406
–322
F–
Cl–
Br –
–506
–364
–335
Ion Hhyd (kJ mol–1) Ion Hhyd (kJ mol–1)
Ion Hhyd (kJ mol–1) Ion Hhyd (kJ mol–1)
Fe2+ Fe3+–1950 –4430
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What is a spontaneous reaction?
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Entropy
Entropy is a measure of disorder, and is given the symbol S. The units of S are: J K–1
mol–1.
ordered disordered
low entropy high entropy
regular arrangement of particles
random arrangement of particles
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Entropy change for reactions
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Predicting entropy changes
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Calculating entropy changes
Standard entropy changes for any chemical reaction or physical change can be calculated using the following simple expression:
Remember the following points:
entropies of elements are not zero like Hf values, so they should be included in calculations.
the units of entropy, S, are J K–1 mol–1
S = Sөproducts – Sө
reactants
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Calculating entropy changes
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Entropy change calculations
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Entropy changes in the surroundings
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Entropy changes in the surroundings
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Gibbs free energy
Whether a reaction is spontaneous depends on:
The change in a quantity called the Gibbs free energy provides a measure of whether a reaction is spontaneous. The Gibbs free energy change is given the symbol G and can be calculated for a reaction using the expression:
A reaction will be spontaneous if G < 0.
the temperature.
the enthalpy change of the system
the entropy change of the system
G = H – TS
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How to calculate ∆G
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Calculating ∆G
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Feasibility of reactions
Even if G is positive at room temperature, there may be a higher temperature at which a reaction becomes feasible.
If S is positive, there may be a point at which TS is big enough to outweigh the enthalpy factor.
G = H – TS
positive
negative
positive
negative
positive
positive
negative
negative
makes TS > H
makes G more negative
no effect: G always positive
unlikely to make TS > H
yes, above a certain temp.
always
never
usually
H S As temp. increases… Feasible?
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Finding the temperature
Consider the reduction of aluminium oxide with carbon:
If G = 0, then T = H / S
The temperature at which this reaction becomes feasible can be calculated. This will be when G = 0.
As both H and S are positive, G will become negative if TS > H.
H = +1336 kJ mol–1
Al2O3(s) + 3C(s) 2Al(s) + 3CO(g)
S = +581 J K–1 mol–1
T = 2299 K
T = 1336 / (581/1000)
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When is a reaction feasible?
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Solubility
In the same way that reactions are only feasible if G < 0, a substance will be soluble in water at a specific temperature if Gsol < 0.
Gsol = Hsol – TSsol
positive
negative
positive
negative
positive
positive
negative
negative
Hsol Ssol G / feasibility
feasible if T is large enough to make G negative
always feasible
never feasible
usually feasible (T is unlikely to be large enough to make G positive)
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Thermodynamics vs. kinetics
Just because a reaction is spontaneous does not mean that it appears to happen. It may be that the reaction is so slow or has such a high activation energy that it is not generally observed.
A reaction that is thermodynamically feasible is not necessarily kinetically feasible.
This reaction has a negative value for G, so is feasible, but a piece of carbon does not spontaneously burn if left on the table. Energy would need to be put in for the reaction to begin.
S = +3 J K–1 mol–1H = –394 kJ mol–1
C(s) + O2(g) CO2(g)
Consider the reaction involving the combustion of carbon:
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Glossary
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What’s the keyword?
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Multiple-choice quiz