Feasible Reactions (Edexcel A Level Chemistry): Revision Note
Feasible Reactions
Graphing the Gibbs Equation
The Gibbs equation can be expressed as the equation for a straight line
ΔGꝋ = ΔHꝋ - TΔSꝋ
ΔGꝋ = - ΔSꝋT + ΔHꝋ
y = mx + c
A graph of free energy versus temperature (in K) will give a straight line, with slope -ΔSꝋ and y-intercept, ΔHꝋ.
The variation of ΔGꝋagainst T for the synthesis of ammonia has been plotted below:
N2 (g) + 3H2 (g)⇌ 2NH3 (g)
![5-1-8-free-energy-versus-temperature](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2023/05/5-1-8-free-energy-versus-temperature.png)
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 ΔG is 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
Gibbs Free Energy & the Equilibrium Constant
The equilibrium constant, Kc, gives no information about the individual rates of reaction
It is independent of the kinetics of the reaction
The equilibrium constant, Kc, is directly related to the Gibbs free energy change, ΔGꝊ, according to the following (van't Hoff's) equation:
ΔGꝊ = -RT lnK
ΔGꝊ= Gibbs free energy change (kJ mol–1)
R = gas constant (8.31 J K-1 mol-1)
T = temperature (Kelvin, K)
K = equilibrium constant
This relationship between the equilibrium constant, Kc, and Gibbs free energy change, ΔGꝊ, can be used to determine whether the forward or backward reaction is favoured
The relationship between the equilibrium constant, Kc, and Gibbs free energy change, ΔGꝊ
![The relationship between the equilibrium constant, Kc, and Gibbs free energy change, ΔGꝋ, downloadable IB Chemistry revision notes](https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=3840/https://cdn.savemyexams.com/uploads/2022/01/17.1.2-The-relationship-between-the-equilibrium-constant-Kc-and-Gibbs-free-energy-change-ΔGꝋ.png)
At a given temperature, a negative ΔG value for a reaction indicates that:
The reaction is feasible / spontaneous
The equilibrium concentration of the products is greater than the equilibrium concentration of the reactants
The value of the equilibrium constant is greater than 1
As ΔG becomes more negative:
The forward reaction is favoured more
The value of the equilibrium constant increases
Limitations of using ΔG
∆Go can only be used to predict the feasibility of a reaction under standard conditions
Under non-standard conditions, ∆G must be calculated
It is important to note that just because a reaction is feasible does not mean that it will occur at an observable rate
While ∆G can be used to determine the feasibility of a reaction, it does not take into account the kinetics of the reaction i.e. rate of reaction
There might be a large energy barrier (Ea) which the reacting species have to overcome before a reaction can occur
Some reactions are feasible since ∆G is negative, but kinetically not feasible since it just occurs too slowly
Such reactions are feasible but very slow
An example is the decomposition of hydrogen peroxide at 298 K
H2O2 (l) → H2O (l) + ½O2 (g) ∆G = -117 kJ mol-1
This reaction has a very large Ea so must be catalysed using manganese dioxide, MnO2
If the reaction was left for long enough, the hydrogen peroxide would eventually decompose, however the addition of the MnO2 allows the reaction to take place via an alternative route with a lower Ea
Although the value for ∆G indicates the reaction is feasible, it does not take into account the kinetics of the reaction
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