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    Free Energy and Thermodynamics

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    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?

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    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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