Calculate ΔG^∘ and K at 25^∘ C for the reactions in Exercises 35 and 39 .

Name: Problem 71, Chapter 18: Chemistry
Uploaded: 2021-02-13T16:58:24-08:00
Duration: 6 min 13 s
Channel: David Collins
Description: Calculate ΔG^∘ and K at 25^∘ C for the reactions in Exercises 35 and 39 .

Calculate ΔG^∘ and K at 25^∘ C for the reactions in...

Key Concepts

Standard Gibbs Free Energy Change (?G°)

Standard Gibbs free energy change represents the change in free energy for a reaction occurring under standard conditions, typically 1 atm pressure and 1 M concentration for all reactants and products. It indicates the thermodynamic favorability of a reaction; a negative ?G° signifies a spontaneous process under these conditions.

Equilibrium Constant (K)

The equilibrium constant quantifies the ratio of the concentrations or partial pressures of products to reactants when a system is at equilibrium. It serves as a measure of the extent of a reaction and is dimensionless when defined using activities.

Relationship between ?G° and K

There is a fundamental thermodynamic relationship linking ?G° and the equilibrium constant, expressed as ?G° = -RT ln K, where R is the universal gas constant and T is the absolute temperature. This equation bridges the thermodynamic favorability of a reaction with its position at equilibrium, allowing for computation of one parameter if the other is known.

Transcript

00:01 To answer this question, we simply need to know what the chemical reactions are, the number of electrons that are transferred based upon the balanced chemical reaction, and the cell potential, the standard cell potential for the chemical reactions that are found in problems 37 and 41.

00:27 The equations that allow us to calculate delta g standard and k are delta g standard is equal to n, which is the number of moles of electrons that are transferred in the balanced chemical reaction, multiplied by f, which is veridae's constant, multiplied by the standard potential for the chemical reaction.

00:48 Once we have delta g standard, then k is going to be equal to e to the negative delta g standard over rt, r being the universal gas constant at 8 .314, and t the kelvin temperature.

01:05 So for the first chemical reaction, this is what you would have balanced.

01:10 In question 37, and then in question 39, you would have identified that the cell potential is 0 .03 volts.

01:26 So this would have come from 37, and this from question 39.

01:32 So delta g standard then is going to be equal to negative n.

01:36 How do we identify n? we simply need to identify something that is oxidized or reduced.

01:42 It doesn't matter and count the number of electrons.

01:45 That were required.

01:47 Chromium goes from 2 plus.

01:50 Well, it's probably easier to look at chlorine going from 0 to 1 minus, and 6 of them are involved.

01:57 So we have 6 moles of electrons being transferred.

02:00 Multiply that by faraday's constant.

02:03 Multiply that by the cell potential.

02:08 This then gives us a delta to g of negative 17 ,400.

02:13 Our k value, equilibrium constant, will simply be e to the negative negative 17 ,400 divided by r, 8 .314, and t, which is 25 degrees celsius plus 273 or 298 kelvin.

02:31 And we get 1 ,100.

02:36 Now because we only have one significant figure here, you'll probably find in your textbook this rounded to negative 20 ,000.

02:43 Joules or 20 kilojoules and this to just 1 ,000.

02:49 For b, this is the balanced chemical reaction that you would have determined in 37b.

02:55 This would be the voltage that you would have calculated in 39b.

03:02 Delta g then is going to be equal to negative n, n being the moles of electrons transferred.

03:08 We have magnesium going from an oxidation state of zero to 2 plus, so that's two electrons, and also two electrons for the change of copper 2 plus to copper.

03:18 So n is going to be equal to 2...

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