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 see 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 enthalpy, change in our standard entropy, and temperature in Kelvin values. Now here we need to note that the change in the standard Gibbs 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 going to say that delta g zero equals delta h zero minus t delta s zero. And here we're going to say that under standard condition we're talking about a pressure of 1 atmosphere and a temperature of 25 degrees Celsius.
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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 enthalpy equals negative 111.4 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 of the reaction at this temperature. So to calculate our change in the standard Gibbs free energy, we're going to say delta G equals delta H minus T delta S. Here we need to make sure the units are the same, one's in kilojoules but one's in joules. Typically we have delta g in kilojoules, so remember that 1 kilojoule is equal to a 1,000 joules. So this is equal to negative 0.025 kilojoules per Kelvin. Here we plug these numbers in, so that's going to be negative 111.4 kilojoules minus 298 Kelvin times negative 0.025 kilojoules over Kelvin. Kelvins 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 0, that means that it is spontaneous as written. Remember, if your delta g is less than 0, 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 Gibbs free energy, 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 Gibbs 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.
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example
Gibbs Free Energy Calculations Example
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4m
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Here it says, for the reaction where 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas, Our delta h value is negative 92.4 kilojoules. Delta s, not equals negative 198 joules per kelvin. Is the reaction spontaneous under standard conditions? If not, at which temperature will it be spontaneous? Alright. So here we can follow the following steps in order to find our answer. For step 1 we say, using the Gibbs free energy formula, set delta G naught equal to 0. So here we're going to set it equal to 0. Plug in given values of delta h and delta s and solve for temperature. The found temperature corresponds to equilibrium. Alright, so we're going to do what it says here. We're going to say here that we have delta g equals delta h minus T delta s. We set delta g equal to 0. Here we're going to plug in the values here, so we're going to have negative 92.4 kilojoules for delta h. We're solving for temperatures when temperature is unknown. S, entropy as in joules, dividing that by a 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 equals now we have a negative times a negative, we're gonna have equal to t times 0.198 kilojoules per Kelvin. Divide both sides here by the delta s value and we'll 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 0, which means this is the temperature for us to be at equilibrium. So found temperature corresponds to equilibrium. So what do we do? Well, we go to step 2. Here we say predict spontaneity by using the signs of delta h and delta s, and that's where our punnett 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 s. The way we fill this in corresponds to our original equation, and it's set in a way where we're trying to manipulate, temperature to help give us a delta g that's less than 0, 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 all be less than 0. And then here when they're both negative, we're gonna say we're only spontaneous at low temperatures. If 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 S as negative, which means we fall here. We're spontaneous at 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. And so that will 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 Gibbs 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 Gibbs free energy of reaction can be calculated with free energy of formations. So delta g 0 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 we give it to you in a chart within the question in some way. Now elements in natural states have a Gibbs free energy of formation equal to 0, much like the change in the standard enthalpy of reactions. When they also are in their natural states, they're equal to 0. Now here we're gonna say Gibbs free energy of a reaction formula is the change in the standard Gibbs free energy of a reaction equals the summation. So here this is Gibbs free energy of our reaction in kilojoules. Summation here is just sigma or the sum of all your products, all your reactants. N is the moles of your substance. So it would be moles of the standard Gibbs free energy of formation of products minus the summation of the moles of Gibbs free energy of reactants, products minus reactants. Here again, 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 nth standard Gibbs free energy of a reaction. Right? So we've seen this before within 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 Gibbs free energy reaction, or when we have nitric acid reacting with ammonia to produce ammonium nitrate solid. Alright. So here we're given the standard Gibbs free energy of formations for each of these compounds. So So remember that's just delta G reaction equals products minus reactants equals, so everything's a one to one relationship, so we're going to say 1 mole of our product of ammonium nitrate has a delta G formation of negative 183.8 kilojoules per mole, minus our reactants, which is 1 mole of nitric acid, which has a value of negative 73.5 kilojoules per mole, plus 1 mole of ammonia, which has a value of negative 16.4 kilojoules per mole. Here we see that moles cancel out, so our answer at the end will be in kilojoules. So when we plug this in, we're going to 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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