CHEM 11200 Lecture Notes - Lecture 12: Gibbs Free Energy

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lemonhedgehog555

19 Jul 2020

School

University of Chicago

Department

Chemistry

Course

CHEM 11200

Professor

Kozmin

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Document Summary

H = h products - h reactants. @ 25 degrees celsius and 1 atm (pressure) The amount of energy required to make one mole of a compound from its constituent elements. Note: o 2 does not have a heat of formation but o does because. The sum of the energies absorbed or given off in all the steps will be the same for the overall reaction. Energy is needed to break a bond. Energy is released when a bond is created. S = s products - s reactants. S is positive when order decreases and s is negative when order increases. Gases have more entropy than solids or liquids. Particles in solution have more entropy than solids. Two moles have more entropy than one mole. S forward and backward have the same magnitude. Energy that"s available to do useful work from a chemical reaction.

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I need help on this chemistry work. Thanks.

DATA ANALYSIS

Using the example, calculate the Gibbs free energy for the process of dissolving ammonium nitrate and the process of dissolving sodium hydroxide (equation: ÎG = ÎH â TÎS).

Use the standard temperature 298 K.

Use the experimental data to calculate enthalpy of dissolving (ÎH).

Use the following data to calculate the entropy change (ÎS):

	S (J/K mol)
NH₄NO₃(s)	151
NH₄⁺(aq)	113.4
NO₃^-(aq)	146.4
NaOH(s)	64.46
Na⁺(aq)	59.0
OH^-(aq)	- 10.7

After completing the calculations answer the following questions separately for each experiment:

Is the dissolving exothermic or endothermic?

Does the entropy decreases or increases during dissolving?

Is the Gibbs free energy positive or negative? Is the process of dissolving spontaneous?

Using data in the appendix in your textbook, calculate Gibbs free energy for the reaction of dissolving

lithium phosphate, Li₃PO₄, in water at 25 Â°C. Do the calculations in two ways:

1) Use standard enthalpy of formation (ÎH) and entropy (S) from the appendix in your textbook, and use the same formula as you did in the lab experiments: ÎG = ÎH â TÎS. Remember that both ÎH and ÎS has to include three lithium ions.

2) Use standard free energies of formation from Appendix 6 (ÎG) and equation

ÎG = âÎG _products â âÎG _reactants

Are the results similar in both calculations?

CHEM 11200 Lecture Notes - Lecture 12: Gibbs Free Energy

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