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1)A -> B H +B->A H + or -?

2) What happens to the equilibrium if temperature is increased?

3) What happens to Kc if the temperature is increased?

Free energy

L.O.:Predict which chemical reactions are spontaneous.Calculate the temperature at which some reactions become feasible.

The free energy change G is the balance between enthalpy, entropy and temperature for a process.

FREE ENERGY & ENTROPYFREE ENERGY & ENTROPY

SPONTANEOUS CHANGESSPONTANEOUS CHANGES

- occur in one particular direction and not the other

- take place without the need for work

Exothermic reactions are usually spontaneous

- this is because they go from higher to lower enthalpy

FREE ENERGY & ENTROPY FREE ENERGY & ENTROPY

FREE ENERGY (G)FREE ENERGY (G)

A reaction is only spontaneous if it can do work - it must generate free energy

A negative G indicates a reaction capable of proceeding of its own accord





G < 0 (- ive) Spontaneous reaction

G > 0 (+ ive) Non-spontaneous reaction(will be spontaneous in reverse direction)

G = 0 The system is in equilibrium





G < 0 (- ive) Spontaneous reaction

G > 0 (+ ive) Non-spontaneous reaction(will be spontaneous in reverse direction)

G = 0 The system is in equilibrium

ENTROPY (S)ENTROPY (S)

• Entropy (symbol S) is a measure of the disorder of a system• The more the disorder, the greater the entropy• If a system becomes more disordered, the value of S is positive• Values tend to be in JOULES - not kJ

S = Sfinal - Sinitial


Free energy, enthalpy and entropy are related ... G = H - TS

Specialcase For a reversible reaction at equilibrium the value of G is zero

If G = ZERO then H = T S

and S = HT

This occurs during changes of state (melting, boiling etc)


Free energy, enthalpy and entropy are related ... G = H - TS

Specialcase For a reversible reaction at equilibriumreversible reaction at equilibrium the value of G is zero

If G = ZERO then H = T S

and S = H T

Worked ExampleCalculate the entropy change when water turns to steam at 100°C (373K).The enthalpy of vaporisation of water is +44 kJ mol-1

S = H = + 44 kJ mol -1 = + 118 J K -1 mol -1

T 373 K (+ive as gases have more disorder)

Entropy change values are much smaller than enthalpychange values; they tend to be in Joules rather than kJ

IS A REACTION SPONTANEOUS?IS A REACTION SPONTANEOUS?

(1)(1) H2(g) + F2(g) —> 2HF(g)

H - ive highly exothermic processS 0 same number of gas moleculesG - ive must be negative

(taking 0 away from a negative number)

Specific examplesSpecific examples

G = H - TS


(3)(3) NH4NO3(s) —> NH4+(aq) + NO3¯(aq)

H +ive endothermic (the solution goes colder)S +ive more disorder as lattice breaks upG - ive if T is high or the value of S is big enough

Specific examplesSpecific examples

G = H - TS


What is the sign of the entropy change in the following reaction ? Give reasons for your decision. What is the sign of G?

a) CaCO3(s) —> CaO(s) + CO2(g)

G = H - T S




increase in the number of gas molecules

Endothermic

Because S is positive; TS must be positive

H is positive (very endothermic reaction)

ForG to be NEGATIVE, TS must be bigger than H

This is more likely with a higher temperature

The reaction is more likely to be spontaneous if heated

++

Temperature is ALWAYS positive

+G = H - T S



b) NH3(g) + HCl(g) —> NH4Cl(s)

G = H - T S




decrease in the number of gas molecules

Exothermic

Because S is negative; TS must be negative

H is negative

For G to be negative TS must be less negative than H

This is more likely if the value of T is lower

The higher the temperature the less likely that the reaction will proceed

+

Temperature is ALWAYS positiveG = H - T S


What is the sign of the entropy change in the following reactions ? Give reasons for your decision. What is the sign of G?

c) Na(s) —> Na(g) (Equivalent to Enthalpy of Atomisation)

G = H - T S



c) Na(s) —> Na(g) (Equivalent to Enthalpy of Atomisation)


Endothermic


H is positive

For G to be NEGATIVE, TS must be bigger than H

This is more likely with a higher temperature so the reaction

Solids are more likely to become gases if heated

++

Temperature is ALWAYS positive

+G = H - T S



d) C6H12(l) + 9O2(g) —> 6CO2(g) + 6H2O(g)

G = H - T S



d) C6H12(l) + 9O2(g) —> 6CO2(g) + 6H2O(g)


COMBUSTIONis exothermic


H is negative

Taking a +ive value away from a –ive value must give a –ive value

G must be NEGATIVE so the reaction is SPONTANEOUS

++




e) C(s) + O2(g) —> CO2(g)

G = H - T S



e) C(s) + O2(g) —> CO2(g)

no change in the number of gas molecules

COMBUSTIONis exothermic

S is very small (could be –ive or +ive)

TS will therefore not be a large number

H will be have a relatively large negative value

Taking a small +ive value away from a –ive value must give a –ive value

G must be NEGATIVE so the reaction is SPONTANEOUS

very small+


ANSWERSANSWERS

What is the sign of the entropy change (What is the sign of the entropy change (S) in the following reactions ?S) in the following reactions ?Give reasons for your decision.



c) Na(s) —> Na(g)

d) C6H12(l) + 9O2(g) —> 6CO2(g) + 6H2O(g)

e) C(s) + O2(g) —> CO2(g)

ANSWERSANSWERS

What is the sign of the entropy change (What is the sign of the entropy change (S) in the following reactions ?S) in the following reactions ?Give reasons for your decision.



c) Na(s) —> Na(g)

d) C6H12(l) + 9O2(g) —> 6CO2(g) + 6H2O(g)

e) C(s) + O2(g) —> CO2(g)

+

-

+

+

‘0’

more gas molecules

fewer gas molecules

goes from solid to gas

more gas molecules

‘similar’ moles of gas


A reaction should be spontaneous if A reaction should be spontaneous if G is negative, G is negative, so ...

• Work out if it is exothermic (H -ive) or endothermic (H +ive)

• Is there an increase in disorder ? If YES then S will be positive.

• Is the temperature high or low ? This can affect the value of TS°

General examples

• If H is –ive and S is +ive then G must be negative

• If H is +ive and S is -ive then G must be positive

Working out the temperature at which a reaction becomes feasible.

Because the free energy change ∆G depends on T∆S, the effect of entropy becomes more important at higher temperatures. If ∆H is –ve and ∆S is positive, the reaction will be spontaneous at all temperatures.If ∆H is +ve and ∆S is negative, the reaction will not be spontaneous at any temperature. However if ∆H and ∆S are both +ve, the reaction will be spontaneous only above a certain temperature. If ∆H and ∆S are both –ve, the reaction will be spontaneous only below a certain temperature. The critical temperature is the temperature at which ∆G = 0, so ∆H = T∆S. T = ∆H/∆S

L.O.:

•Explain that the tendency of a process to take place depends on temperature, T, the entropy change in the system, ∆S, and the enthalpy change, ∆H, with the surroundings.

•Explain that the balance between entropy and enthalpy change is the free energy change, ∆G.

•State and use the relationship ∆G = ∆H – T∆S.

•Explain how endothermic reactions are able to take place spontaneously.

Date post:	01-Jan-2016
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