Gibbs Free Energy Calculations - Video Tutorials & Practice Problems
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1
concept
The Gibbs Free Energy Formula
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44s
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Here, we're going to say that the Gibbs free energy formula allows us to calculate the value of delta G typically in units of kilojoules by using the change in the standard entropy change in our standard entropy and temperature in Kelvin values. Now, here we need to note that the change in the standard gives free energy change is under standard conditions. While delta G without the little not sign is under non-standard conditions. Here, when we're talking about standard conditions, we're gonna say that delta G zero equals delta H zero minus T delta S zero. And here we're going to say that under standard conditions, we're talking about a pressure of one atmosphere and a temperature of 25 °C.
2
example
Gibbs Free Energy Calculations Example
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1m
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Here, it says for a particular reaction, the change in the standard entropy equals negative 1 11.4 kilojoules. And the standard change in entropy equals negative 25 joules over Kelvin here, we need to calculate the standard change in our Gibbs free energy for this reaction at 298 Kelvin. And it says what can be said about the spontaneity, the reaction at this temperature. So to calculate our change in the standard Gibbs free energy, we're gonna say delta G equals delta H minus T delta S. Here. We need to make sure that the units are the same ones in kilojoules, but one's in jewels. Typically we have delta G in kilojoules. So remember that one kilojoule is equal to 1000 Jes. So this is equal to negative 0.025 kilojoules per Kelvin. Here we plug these numbers in. So that's gonna be negative 1, 11.4 kilojoules minus 298. Kelvin times negative 0.025 kilojoules over Kelvin Alvin's here, cancel out. So our answer at the end will be in kilojoules. When we do that, we get negative 103.95 kilojoules. So the standard change in our gibbs free energy here it is a value less than zero. That means that it is spontaneous. As written, remember if your delta G is less than zero, it's going to be a spontaneous reaction.
3
concept
∆Gº and Temperature Conditions
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32s
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Now, when dealing with temperature conditions, we say here that the change in our standard gives free energy uh formula can be used to approximate temperatures at which reactions are spontaneous or non spontaneous. Here, we're gonna say when our change in the standard gets free energy values is unknown. We can utilize some basic techniques that help us to determine if a reaction will be always spontaneous, always non spontaneous or whether it can be spontaneous at high or low temperatures. Now click on the next video and let's take a look at a typical question.
4
example
Gibbs Free Energy Calculations Example
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4m
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Says for the reaction where one mole of nitrogen gas reacts with three moles of hydrogen gas to produce two moles of ammonia gas. Our delta value is negative 92.4 kilojoules. Delta s uh not equals negative 198 joules per kelly is the reaction spontaneous under standard conditions. If not at which temperature will it be spontaneous? All right. So here we can follow the following steps. In order to find your answer for step one, we say using the Gibbs free energy formula set delta G not equal to zero. So here we're gonna set it equal to zero plug in given values of delta H and delta S and solve for temperature. The found temperature corresponds to equilibrium. All right. So we're gonna do what it says here. We're gonna say here that we have delta G equals delta H minus T delta S. We said delta G equal to zero. Here, we're gonna plug in the values here. So we're gonna have negative 92.4 kilojoules for delta H. We're solving for temperatures or temperature is unknown. S entropy is in joules dividing that by 1000 will get us kilojoules. So now we have that filled in. We're going to add 92.4 kilojoules to both sides. So when I do that, I'm gonna get 92.4 kilojoules. It was now we have a negative times the negative. We're gonna have equal to t times 0.198 kilojoules per Kelvin divide both sides here by the delta S value and will isolate our temperature. So here kilojoules will cancel out and we'll be left with Kelvin. So here our temperature equals 466.67 Kelvin. Now this is just the temperature to make delta G equal to zero, which means this is the temperature force to be at equilibrium. So, found temperature corresponds equilibrium. So what do we do? Well, we go to step two, here we say predict spontaneity by using the signs of delta H and delta S. And that's where a pundit square comes into play. Here. On the left side, we have plus delta H minus delta H at the top, we have plus delta S minus delta ss. The way we fill this in corresponds to our original equation. And it's set in a way where we're trying to manipulate uh temperature to help give us a delta G that's less than zero to make it spontaneous. Here, if delta H and delta S are both positive, then we'd be spontaneous at high temperatures. And the higher the temperature goes, the more spontaneous we become So, if spontaneous at high temperatures reaction will become spontaneous above our calculated temperature, if delta H is positive and delta S is negative, then we'll always be non spontaneous. So it doesn't matter what temperature we have when delta S is negative and delta S is positive, then you're spontaneous. No matter the temperature, your gibbs free energy will be less than zero. And then here when they're both negative, we're gonna say we're only spontaneous at low temperatures is spontaneous at low temperatures, reaction will be spontaneous below our calculated temperature. So let's go back to the original question. Here. We have delta S as negative and we have delta essence negative, which means we fall here, we're spontaneous and low temperatures. So that means here's our calculated temperature. It would mean that it's spontaneous below this temperature. So any temperature below 466.67 Kelvin would give us a spontaneous reaction. OK. So that would be the answer to this question.
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Problem
Problem
Calculate ∆G° for the following reaction: P4 (s) + 5 O2 (g) → P4O10 (s), ∆H° = −2940 kJ/mol, 25 °C.
Does the reaction favor reactants or products?
A
1419.3 kJ
B
−2653 kJ
C
−140.7 kJ
D
598.5 kJ
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Problem
Problem
Determine if reaction is spontaneous under standard conditions, if not at what temperature will it be spontaneous?
3 A (g) + 5 B (s) → 3 AB (s) + B2 (g) ∆H° = 112.7 kJ, ∆S° = 78.3 J/K.
A
1439.3 K
B
274.6 K
C
1.439 K
D
719.65 K
7
Problem
Problem
Nickel has ∆Hvap = 370.4 kJ/mol and ∆Svap = 123.3 J/mol•K. Will nickel boil at 2700 °C and 1 atm?
A
Yes, nickel will boil at 2700 °C and 1 atm.
B
No, the temperature is not high enough for nickel to boil.
8
concept
∆Gº of Reactions
Video duration:
1m
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Now, when it comes to gives free energy of a reaction, we're going to say it's similar to the change in the standard entropy of a reaction formula. Here we're going to say the change in our standard gives free energy of reaction can be calculated with free energy of formations. So delta G zero F these values will always be provided. That's because so many different compounds and elements have their own unique value. It'd be impossible for you to memorize all of them. So it will be given to you in a chart within the question in some way. Now, elements in natural states have a gift free energy of formation equal to zero much like the change in the standard entropy of reactions when they all. So in their natural state, they're equal to zero. Now, here we're gonna say Gibbs free energy of reaction formula is the change in the standard Gibbs free energy of a reaction equals the summation. So here this is gifts for energy of our reaction in kilojoules summation here is just signal or the sum of all your products, all your reactants and is the moles of your substance. So it would be moles of the standard gifts for energy or formation of products minus the summation of the moles of Gibbs free energy of reactants, products minus reactants. Here. Again, uh We're gonna say that standard Gibbs free energy formation is in units of kilojoules per mole. So we're gonna be utilizing this formula to help us figure out the change in the standard standard Gibbs free energy of a reaction, right? So we've seen this before with other variables. Now we're basically applying it to Gibbs free energy.
9
example
Gibbs Free Energy Calculations Example
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1m
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Here it says, determine the change in the standard gifts for your reaction or when we have nitric acid reacting with ammonia to produce ammonium nitrate solid. All right. So here we're given the standard gifts for energy of formations for each of these compounds. So remember that's just delta G reaction equals products minus reactants equals. So everything is a once one relationship. So we're gonna say one mole of our product of ammonium nitrate has a delta geo formation of negative 1 83.8 kilojoules per mole minus our pro our reactants, which is one mole of nitric acid, which has a value of negative 73.5 kilojoules for more plus one mole of ammonia, which has a value of negative one of 16.4 kilojoules per mole. Here we see that moles cancel out. So our answer at the end will be in killer. So when we plug this in, we're gonna get a gibbs free energy of reaction. Standard gibbs free energy of reaction equal to negative 93.9 kilojoules. So this would be our final answer.
10
Problem
Problem
Fe2O3 (s) + 3 H2 (g) → 2 Fe (s) + 3 H2O (g) is a redox reaction. What would be its Gibbs Free energy change under standard conditions? Is the reaction spontaneous at 25 °C?
A
513.6 kJ; No, the reaction is not spontaneous at 25 °C.
B
56.4 kJ; No, the reaction is not spontaneous at 25 °C.
C
−513.6 kJ; Yes, the reaction is not spontaneous at 25 °C.
D
56.4 kJ; Yes, the reaction is spontaneous at 25 °C.
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