# 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.

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?

https://brainmass.com/chemistry/general-chemistry/free-energy-thermodynamics-553850

#### 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.

2nd Law of Thermodynamics and Gibbs Free Energy

Please show all work.

Using the 2nd law of Thermodynamics and the definition of the Gibbs free energy, derive a relationship between ∆Suniv and ∆G.

View Full Posting Details