# 5.1.8 Reaction Feasibility

### Reaction Feasibility

• The Gibbs equation can be used to calculate whether a reaction is feasible or not

ΔG = ΔHreaction – TΔSsystem

• When ΔG is negative, the reaction is feasible and likely to occur
• When ΔGis positive, the reaction is not feasible and unlikely to occur

#### Worked Example

Determining the feasibility of a reaction

Calculate the Gibbs free energy change for the following reaction at 298 K and determine whether the reaction is feasible.

2Ca (s) + O(g) → 2CaO (s)         ΔH = -635.5 kJ mol-1

S[Ca(s)] = 41.00 J K-1 mol-1

S[O2(g)] = 205.0 J K-1 mol-1

S[CaO(s)] = 40.00 J K-1 mol-1

Step 1: Calculate ΔSsystem

ΔSsystem= ΣΔSproducts – ΣΔSreactants

ΔSsystem= (2 x ΔS [CaO(s)]) –  (2 x ΔS [Ca(s)] + ΔS[O2(g)])

= (2 x 40.00) – (2 x 41.00 + 205.0)

= -207.0 J K-1 mol-1

Step 2: Convert ΔSto kJ K-1 mol-1

= -207.0 J K-1 mol-1÷ 1000 = -0.207 kJ mol-1

Step 3: Calculate ΔG

ΔG = ΔHreaction – TΔSsystem

= -635.5 – (298 x -0.207)

= –573.8 kJ mol-1

Step 4: Determine whether the reaction is feasible

Since the ΔG is negative the reaction is feasible

### Reaction Feasibility: Temperature Changes

• The feasibility of a reaction can be affected by the temperature
• The Gibbs equation will be used to explain what will affect the feasibility of a reaction for exothermic and endothermic reactions

#### Exothermic reactions

• In exothermic reactions, ΔHreaction is negative
• If the ΔSsystem is positive:
• Both the first and second term will be negative
• Resulting in a negative ΔG so the reaction is feasible
• Therefore, regardless of the temperature, an exothermic reaction with a positive ΔSsystem will always be feasible
• If the ΔSsystem is negative:
• The first term is negative and the second term is positive
• At very high temperatures, the –TΔSsystemwill be very large and positive and will overcome ΔHreaction
• Therefore, at high temperatures ΔGis positive and the reaction is not feasible
• Since the relative size of an entropy change is much smaller than an enthalpy change, it is unlikely that TΔS > ΔH as temperature increases
• These reactions are therefore usually spontaneous at normal conditions

#### Endothermic reactions

• In endothermic reactions, ΔHreaction is positive
• If the ΔSsystem is negative:
• Both the first and second term will be positive
• Resulting in a positive ΔG so the reaction is not feasible
• Therefore, regardless of the temperature, endothermic with a negative ΔSsystem will never be feasible
• If the ΔSsystem is positive:
• The first term is positive and the second term is negative
• At low temperatures, the –TΔSsystemwill be small and negative and will not overcome the larger ΔHreaction
• Therefore, at low temperatures ΔGis positive and the reaction is not feasible
• The reaction is more feasible at high temperatures as the second term will become negative enough to overcome the ΔHreaction resulting in a negative ΔG
• This tells us that for certain reactions which are not feasible at room temperature, they can become feasible at higher temperatures
• An example of this is found in metal extractions, such as the extraction if iron in the blast furnace, which will be unsuccessful at low temperatures but can occur at higher temperatures (~1500 oC in the case of iron)

A summary table of free energy conditions ### Free Energy Vs Temperature Graphs

• Rearranging the Gibbs equation allows you to determine the temperature at which a non-spontaneous reaction become feasible

ΔG = ΔHreaction – TΔSsystem

• For a reaction to be feasible ΔGmust be zero or negative

0 = ΔH – TΔS

ΔH = TΔS

T = ΔH / ΔS

#### Worked Example

At what temperature will the reduction of aluminium oxide with carbon become spontaneous?

Al2O3(s) + 3C(s)     2Al(s) + 3CO(g)             ΔH = +1336 kJ mol-1        ΔS = +581 J K-1mol-1

If ΔG= 0 then ,   T = ΔH / ΔS

T = 1336 ÷ (581/1000)

T = 2299 K

#### Graphing the Gibbs Equation

• The Gibbs equation can be expressed as the equation for a straight line

ΔG = ΔH – TΔSꝋ

ΔG– ΔST  + ΔH

y = mx + c

• A graph of free energy versus temperature ( in K) will give a straight line, with slope -ΔS and y-intercept, ΔH.
• The variation of ΔGagainst T for the synthesis of ammonia has been plotted below:

N2 (g) + 3H2 (g)⇌ 2NH3 (g) Graph of free energy versus temperature for the synthesis of ammonia

• From this graph you should be able to see some key features:
• The x-intercept shows you where the reaction ceases to be spontaneous, in this case at 460 K (187 oC)
• Above this temperature ΔGis positive so the reaction is not feasible
• However, you may recall that the operating conditions of the Haber process are higher than this temperature, but this graph takes no account of the use of a catalyst which affects the energetics of the system, nor does it take into account anything about the rate of reaction or the fact that it is an equilibrium and removal of the ammonia as soon as it is formed also tips the balance in favour of the product
• The y-intercept shows you reaction is exothermic which you can see from the enthalpy of formation; the value is approximately -46 kJ mol -1
• An exothermic equilibrium reaction would be favoured by lower temperatures- this is seen by the value of ΔGꝋ  becoming increasingly negative as the temperature falls

#### Exam Tip

You will notice that the line on the graph does not continue below 240 K. The simple reason for this is that at this point the ammonia will have reached it boiling point and so the gradient would change because it is now liquid ammonia.

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