Free Energy and Thermodynamics

1-Consider the reaction: I 2 (g)+Cl 2 (g)⇌2ICl(g). K p = 81.9 at 25 ∘ C.
Calculate ΔG rxn for the reaction at 25 ∘ C under each condition:

Part A: standard conditions.Express your answer using one decimal place.
Part B: at equilibrium
Part C: P ICl = 2.55atm ; P I 2 = 0.315atm ; P Cl 2 = 0.219atm .Express your answer using one decimal place.

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2-Consider the following isomerization reactions of some simple sugars and values for their standard Gibbs free energy ΔG ∘:
reaction A: glucose-1-phosphate⟶ glucose-6-phosphate, ΔG ∘ =−7.28 kJ/mol
reaction B: fructose-6-phosphate ⟶, glucose-6-phosphate, ΔG =−1.67 kJ/mol

Part A
Calculate ΔG ∘ for the isomerization of glucose-1-phosphate to fructose-6-phosphate.
Express your answer with the appropriate units.

3-In photosynthesis, plants form glucose (C 6 H 12 O 6 ) and oxygen from carbon dioxide and water.
Part A :Write a balanced equation for photosynthesis.
Part B:Calculate ΔH ∘ rxn at 15 ∘ C .
Part C Calculate ΔS ∘ rxn at 15 ∘ C .
Part D:Calculate ΔG ∘ rxn at 15 ∘ C .
Part E :Is photosynthesis spontaneous? Solution Preview
Please see the attachment.

Free Energy and Thermodynamics
1-Consider the reaction: I 2 (g)+Cl 2 (g)⇌2ICl(g). K p = 81.9 at 25 ∘ C.
Calculate ΔG rxn for the reaction at 25 ∘ C under each condition:

Part A: standard conditions. Express your answer using one decimal place.
Part B: at equilibrium
Part C: P ICl = 2.55atm ; P I 2 = 0.315atm ; P Cl 2 = 0.219atm .Express your answer using one decimal place.

I2 (g) + Cl2 (g) <=> 2ICl (g)
Kp = [P(ICl)]^2 /[P(I2)][P(Cl2)] = 81.9 (this is possible because P is directly proportional to n the number of moles)
Now the two key equations are:
ΔG = ΔH – TΔS —-(1), ΔG = 0 at equilibrium
and ΔG = -RTlnKp = -2.303RTlogKp —— (2)
A. At standard condition, T = (25 + 273)K = 298 K into eqn (1) and get ΔG.
So ΔG = -RTlnKp = – 8.314 J mol-1 K-1 x 298 K x ln (81.9)
= – 8.314 x 298 x 4.405 J/mol
= 10913.70 J/mol
= 10. 9 kJ/mol

B. ΔG = 0 at equilibrium

C. Kp = [P(ICl)]^2 /[P(I2)][P(Cl2)]
= (2.55)2 / (0.315 x 0.219) = 6.5025 / …

Solution Summary
This solution contains step-wise calculation to a set of thermodynamics problems. This solution is well-explained and will clarify the basic concepts of thermodynamics.

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