Entropy Calculations - Video Tutorials & Practice Problems
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concept
Total Entropy Formula
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Here when discussing entropy calculations, we need to take into account the entropy of the universe. Now, here we're gonna see the total entropy change in the universe is represented by the following equation or for formula below. Here, we're going to say that the total change in entropy, which is the same thing as the entropy of the universe equals the change in entropy of my system, which is the same thing as my reaction plus the change in entropy of my surroundings, which is everything else here. Units of this change in entropy are typically in units of jewels per Kelvin. Now here our change in entropy total can be either a positive number, a negative number or equal to zero. If it's positive, then it goes in line with the second law of thermodynamics which says that the entropy of the universe or total is ever increasing. The second law of thermodynamics is talking about the natural increase in entropy in the universe making it a natural process making it spontaneous. Now, if we're negative, we're the exact opposite of that. So it'll be non spontaneous. And then finally, for at zero, we're not spontaneous, we're not not spontaneous what we are, we're equilibrium. So these are the three types of conditions that exist depending on what we calculate for the change in entropy total or change in entropy of the universe.
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example
Entropy Calculations Example
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Here, it says calculate the total entropy change for reaction when the change in entropy of our surroundings equals 2.7 joules per Kelvin. And the standard entropy change of our reaction equals negative negative 450 kilojoules per Kelvin is this reaction spontaneous. All right. So we need the total entropy. So delta S total here will equal delta s of my surroundings plus delta S standard of our reaction. Now here typically we'll have the change in our entropy being joules per Kelvin. So I'll use this 2.7 jewels per Kelvin. Here I need to convert kilojoules to jewels. Remember 1 kg is 10 to the three. So that will come out to negative 450,000 jewels per Kelvin. When we add them together, we're going to get negative 449997 0.3 jewels per Kelvin because our change in total entropy is a negative value. That would mean that this is a non spontaneous reaction. So here we'd say it's non spontaneous for this particular example.
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concept
Entropy of Surroundings Formula
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When it comes to the entropy of our surroundings, we can say we can also calculate it if we know the temperature at which the reaction is taking place. And we're gonna say the conditions here will be under constant pressure and temperature. Now, if we take a look here, we have the entropy of surroundings formula, that is the change in the entropy of our surroundings is equal to negative change in the entropy of our reaction divided by temperature. Now, here we say that the entropy of our reaction is usually in units of kilojoules. And we say here that temperature as always will be in Kelvin. So again, we can calculate the real value of our change in troy of our surroundings through the utilization of the following formula.
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example
Entropy Calculations Example
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Here it says, determine the change in the entry of the universe for the fall reaction at 32 °C. So here they're giving us this chemical reaction. And we're gonna say that our entropy of the reaction change in entropy of our reaction is negative 140 kilojoules. And the entropy of our reaction is 3.6 joules per Kelvin. So remember here that our total uh change in entropy, which is the same thing as the entropy of our universe equals the change in the entropy of our surroundings plus the change in the entropy of our reaction. Here, we already know the change in entropy of our reaction. That's 3.6 joules per Kelvin. What we need to do is determine the change in the entropy of our surroundings, which remember is equal to negative delta H of our reaction divided by temperature. And Kelvin. So we're gonna take that number and plug it in. And we're gonna say here since entropy change of our reaction is in joules, I'm gonna convert these kilojoules into jewels by multiplying them by 1000 temperature. We add 2 73.15 to this 32 °C. And that gives me 305.15 Kelvin. So we have a negative of a negative. Remember the negative is already there and then we have an additional negative here. The changing the entropy of my surroundings will be a positive value. It comes out to be positive 458.79 jewels per Kelvin. We take that and we plug it here into our formula and now we're gonna have the change in the entropy of our universe or total that comes out to a positive 4 62.39 joules per Kelvin. So this would be our final answer.
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concept
Entropy of the System
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In this video, we'll take a look at how to calculate the entropy of our system, which represents our chemical reaction here. We're going to say that each substance has a standard molar entropy represented by it's not associated with it. Now, these values will be al will always be provided in some way. That's because there's so many compounds, so many substances each with their own unique molar entropy. There's no way you can memorize all of them. Now, this is important unlike standard molar entropies, remember R delta H not for substances in their natural state. We're gonna say that your entropy standard molar entropy does not equal zero. It's always gonna be a value greater than zero. Now, here we have the entropy of reaction formula here. It says that the change in the standard entropy of our reaction which is our system equals. So when we talk about standard entropy of our reaction, that's typically in units of joules per Kelvin. Here we have Sigma which is summation and this would be N times standard entropy of our products minus Sigma mole standard entropy of our reactants. Now remember Sigma just stands for some of so we're taking all the products lumping them together, taking all the reactants, lumping them together. And here represents the moles of substance. And then we're gonna say here, entropy with not equals your standard mole entropy of a substance. And this will be in joules over moles times K. When we talk about standard conditions, remember that we're talking about a temperature of 25 °C and a pressure of one atmosphere. So basically, we're saying here that the entropy standard entropy of our reaction is equal to products minus reactants. We're gonna utilize this formula to help us solve for the entropy of our system in the following questions. So let's pay attention and see how we utilize this formula.
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example
Entropy Calculations Example
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Here, it says to calculate the change in the standard entropy of our reaction for the following reaction at 25 °C. Here we have two moles of nitrogen monoxide reacting with one mole of oxygen gas to produce two moles of nitrogen dioxide gas. Here, we're told that the end of reaction is equal to negative 1, 14.14 kilojoules. All right. So here they want us to calculate this and remember change in the standard entropy of our reaction is just equal to products minus reactants. Here, we're given the standard molar entropies of each of the compounds found within this reaction. We look at our products. I'm gonna say we have two moles of nitrogen dioxide. And according to our chart, each one has a standard moet entropy of 240.1 joules over moles times K. I'm gonna plug that in moles cancel out minus my reactants. We have two moles of nitrogen monoxide. Each one is 2 10.8 jewels over most times K again, moles cancel out plus we have one mole of oxygen gas with a value of 205.2 joules over moles times K. So again, moles cancel out here, we can see that the moles can ups or units at the end will be in jewels per Kelvin. When we plug this in, we get 1 46.6 jewels over Calvin, this would be our final answer. Now, you might also notice that I gave us the entropy of the reaction here. We don't even need to use it because the formula to calculate the entropy of our system doesn't need this entropy value. So again, sometimes professors give you additional information and that doesn't mean you have to use it. In this case, I gave you entropy of the reaction. And it's just import is it just telling us at which the reaction is releasing energy? But it's not important to answer the question here.
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Problem
Problem
For the following reaction at 27 °C, calculate ∆S°rxn, ∆Ssurr, and ∆Stot. Determine if reaction is favorable.
Fe2O3 (s) + 3 H2 (g) → 2 Fe (s) + 3 H2O (g) ∆Hrxn = 98.8 kJ