What describes the change in Gibbs free energy (∆G) of a spontaneous reaction?
1 answer:
<u>Answer:</u> The correct answer is Option A.
<u>Explanation:</u>
Gibbs Free energy is defined as amount of useful work that can be done in the system. It is equal to the enthalpy of the reaction minus the product of entropy of the reaction and absolute temperature.
Mathematically,
Where,
= Gibbs free energy
= Enthalpy of the reaction
= Entropy of the reaction
T = Absolute temperature
Sign Convention for Gibbs Free energy:
If , it is considered as a Spontaneous reaction.
If , it is considered as a Non-Spontaneous reaction.
If , reaction is in equilibrium.
Hence, the correct answer is Option A.
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The question is incomplete, the table of the question is given below
Answer:
I) xA= 0.34, yA= 0.55
ii) 76.2 mole % vapor
iii) Percentage of vapor volume = 98%
Explanation:
i) xA= 0.34, yA= 0.55
xA= 0.34, yA= 0.55
ii) 0.50 = 0.55 nv + 0.34 nL
Therefore, nV = 0.762 mol vapor and nL = 0.238 mol liquid
This shows 76.2 mole % vapor
iii) ρA= 0.791 g/cm3 and, ρE = 0.789 g/cm3
Therefore, ρ = 0.790 g/cm3
Now, we have:
MA = 58.08 g/mol and ME= 46.07 g/mol
So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol
1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor
Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3
Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3
Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %
