Question

anyhelp out there please

how to solve this exercise

Exercise 4 Given ΔSuniverse-Δ System + ΔSsurroundings, derive an expression for the Gibbs Free Energy, AG. I5 marks]

Answer 1

The Gibbs [free] energy (also known as the Gibbs function) is defined as

G = H – T S(4-1)

in which S refers to the entropy of the system. Since H, T and S are all state functions, so is G. Thus for any change in state, we can write the extremely important relation

ΔG = ΔH – T ΔS(4-2)Must know this!

How does this simple equation encompass the entropy change of the world ΔS_total, which we already know is the sole criterion for spontaneous change? Starting with the definition

ΔS_total = ΔS_surr + ΔS_sys(4-3)

we would first like to get rid of ΔS_surr. How can a chemical reaction (a change in the system) affect the entropy of the surroundings? Because most reactions are either exothermic or endothermic, they are accompanied by a flow of heat q_p across the system boundary. The enthalpy change of the reaction ΔHis defined as the flow of heat into the system from the surroundings when the reaction is carried out at constant pressure, so the heat withdrawn from the surroundings will be –q_pwhich will cause the entropy of the surroundings to change by –q_p / T = –ΔH/T. We can therefore rewrite Eq 4-3 as

ΔS_total = (– ΔH/T) + ΔS_sys(4-4)

Multiplying through by –T , we obtain

–T ΔS_total = ΔH – T ΔS_sys(4-5)

which expresses the entropy change of the world in terms of thermodynamic properties of the system exclusively. If –TΔS_total is denoted by ΔG, then we have Eq. 4-2 which defines the Gibbs [free] energy change for the process.