Gibbs free energy is also known as Gibbs function, Gibbs energy, or free energy. Gibbs free energy is a thermodynamic quantity. Gibb's free energy was discovered by an American scientist Josiah Willard Gibbs in the 1870s. His formulation of free energy and his contribution to statistical mechanics laid the principles for thermodynamics and chemistry.
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Gibbs free energy change is a state function and an extensive quantity that depends on the number or quantity of the substance. Gibbs free energy is the amount of energy that is available to do work. This energy is associated with the chemical reaction that can be used to work at constant temperature and pressure.
It is a thermodynamic quantity that is used to predict the spontaneity of the reaction means whether the reaction can occur or not. It can only be concerned with changes in G values rather than its absolute value.
It was introduced to relate H, and S and to explain spontaneity. According to J. Willard Gibb's Free energy of a system is defined as the maximum amount of energy available to a system during a process that can be converted into useful work.
or It is the thermodynamic quantity, especially characterizing the system, the decrease in whose value during a process is equal to the useful work done by the system.
It is denoted by G and it is given mathematically as follows:
G=H−TS
Here,
H = Enthalpy
T = Absolute Temperature
S = Entropy
Also, we learned that
H=E+PVG=E+PV−TS
Therefore, Free energy change at constant temperature and pressure is given as:
ΔG=ΔE+PΔV−TΔS
As ΔH=ΔE+PΔV
So, ΔG=ΔH−TΔS
At standard conditions that is, 298 K and 1 atm pressure
ΔG∘=ΔH∘−TΔS∘
It is called the Gibbs equation and it is used to explain the criterion of spontaneity, driving force, etc.
It is a state function and an extensive property.
For a general reaction, it can be given as follows:
pA+qB→rC+sDΔG∘=∑ΔGP∘−∑ΔGR∘=[(r∑GC∘+s∑ΔGD∘)−(p⋅∑ΔGA∘+q∑ΔGB∘)]
This requires the exact same treatment as ΔH or ΔS
G=H−TSdG=dH−TdS−SdT→(1)
Now,
dH=dE+PdV+VdP→(2)
Using equations (1) and (2), we can write
dG=dE+PdV+VdP−TdS−SdT→(3)
Now,
dE=dq+dw;dq=TdS;dw=−PdV
Putting these values in the above expression (3), we have
dG=VdP−SdT
Note: Remember this important formula for small changes in dG values
Example.1 The incorrect expression among the following is :
1) ΔGsystem ΔStotal =−T
2)isothermal process Wreversible =−nRTlnVfVi
3) (correct) lnK=ΔH∘−TΔS∘RT
4) K=e−ΔG∘/RT
Solution
As we learned,
ΔG∘=−RTlnK
and ΔG∘=ΔH∘−TΔS∘
Thus, we can write
ΔH∘−TΔS∘=−RTlnKlnK=−(ΔH∘−TΔS∘RT)
Hence, the answer is the option (3).
Example.2 For a particular reversible reaction at temperature T,ΔH and ΔS were found to be both +Ve.
If Te is the temperature at equilibrium, the reaction would be spontaneous when-
1) T=Te
2) Te>T
3) ( correct) T>Te
4)Te is 5 times T
Solution
Gibb's free energy (Δ G)
ΔG=ΔH−TΔS
At equilibrium ΔG=0
Hence, ΔG=ΔH−TeΔS=0
∴ΔH=Te⋅ΔS or Te=ΔHΔS
A spontaneous reaction ΔG must be negative which is possible only if ΔH<TΔS
This will happen when T > Te.
Hence, the answer is the option (3).
Example.3 Standard entropy of X2,Y2 and XY3 are 60, 40, and 50 J K-1 mol-1, respectively. For the reaction,
12X2+32Y2→XY3,ΔH=−30 kJ to be at equilibrium, the temperature will be
1)1000 K
2)1250 K
3)500 K
4) (correct)750 K
Solution
For a reaction to be in equilibrium
ΔG=0ΔH−TΔS=0∴ΔH=TΔS
For reaction
12X2+32Y2→XY3,ΔH=−30 kJ
Calculating ΔS for the above equation, we get
ΔS=50−[12×60+32×40]=−40JK−1 mol−1
Thus at equilibrium,
T×(−40)=−30×1000 T=−30×1000−40=750 K
Hence, the answer is the option (4).
Example.4 ΔfG0at 500 K for substance ‘S’ in liquid state and gaseous state are +100.7 kcal mol −1 and +103 kcal mol−1 , respectively. Vapour pressure (in atm) of liquid ‘S’ at 500 K is approximately equal to : (R=2 cal K−1 mol−1 ):
1) (correct)0.1
2)1
3)10
4)100
Solution
ΔG of equilibrium ΔG0=−2.303RTlogKc At Equilibrium ΔG=0 and Q=KcΔGReaction o=103−100.7=2.3kcal=2.3×103calΔG∘=−2.303RTlogKp2.3×103=−2.303×2×500logKplogKp=−1 Kp=10−1=0.1 atm
Hence, the answer is an option (1).
Example. For the reaction
A(l)→2B( g)ΔU=2.1kcal,ΔS=20calK−1 at 300 K Hence ΔG in kcl is −−−−
1) (correct)-2.7 Kcal
2)3.3 Kcal
3)- 3.3 Kcal
4)2.7 Kcal
Solution
We know:
ΔG=ΔH−TΔSΔH=ΔU+2RT
Thus, we have:
ΔG=ΔU+2RT−TΔS
On putting the given values we get:
ΔG=2.1+2×2×300×10−3−300×20×10−3ΔG=2.1+4×300×10−3−300×20×10−3ΔG=1200×10−3−6000×10−3ΔG=2.1+1.2−6ΔG=3.3−6ΔG=−2.7Kcal
Hence, the answer is the option(1).
The spontaneity of the reaction is determined by the Gibbs free energy change. spontaneity means whether the reaction can occur or cannot occur. A reaction is said to be spontaneous when the value of free energy change is less than zero or negative. A reaction is said to be nonspontaneous when the value of free energy change is more than zero or positive.
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