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Question

For a reaction,
P+QR+S. The value of Ho is30 kJ mol1 and S is100 J K1mol1. At what temperature the reaction will be at equilibrium?

A
27 oC
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B
52 oC
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C
30 oC
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D
45 oC
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Solution

The correct option is A 27 oC
We know, G=HTS,
where ΔG = change in Gibb's free energy
ΔH = change in enthalpy = 30,000 J mol1
ΔS = change in entropy=100 JK1mol1
At equilibrium, G=0 for a reaction,
So, H=TS
30000=T×100
T=300 K or 27 oC
Hence, the reaction will be at equilibrium at 27 oC


Theory:
Gibbs Free Energy : A system parameter that predicts the spontaneity of a chemical reaction . It is the free energy at the constant pressure. Gibbs free energy is also a state function. It is an extensive property that depends on the amount of the system.
ΔG=GfinalGinitial
Absolute value of Gibbs free energy cannot be determined because Gibbs energy can be given as G=HTS. Gibbs free energy is the function of both enthalpy as well as entropy.
Most of the chemical processes occur in closed or open systems, which undergo both enthalpy & entropy changes.
G=HTS
Differentiating the above equation
dG=dHTdSSdT

At constant temperature dT=0 and for considerable change dG will be Δ G
Δ G=ΔHTΔ S

Physical Interpretation of ΔG : This is also called non PV work. Magnitude of negative value of ΔG (process is spontaneous) is the maximum amount of non PV (non expansion) work (useful work eg: electrical work) that can be obtained from the system.

Magnitude of positive value of ΔG (process is non-spontaneous) is the minimum amount of work (at least amount of work) to be done on the system to make the process occur .



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