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We are ready to begin our 7th learning objective in which way to learn to

Â calculate the Delta S

Â of the system. Se have previously learned to calculate the Delta S

Â of the the surroundings from Delta H. Here we are going to do it for the system.

Â We have led up to this by learning about the third law of thermodynamics.

Â We were, actually, in this section going to be talking about doing it in

Â standard state conditions.

Â Here's our equation. The Delta s for a reaction is going to be the sum of the

Â entropies of products minus the sum the entropies

Â of the reactants. This equation we very, very similar to one that you saw

Â when you dealt with thermochemistry and learn to calculate the Delta

Â H of a reaction using

Â the sum of the Delta H of formation

Â of products minus the sum of

Â the Delta H of formation of reactants.

Â So we went to tables we were able to obtain those values

Â and we always had to look at delta H but now that we know about the

Â third law of thermodynamics we know we can talk about

Â just entropy. We don't have to look at Delta S but just entropies of the

Â products and reactants.

Â But the process is so very very similar to what you did with Delta H of a reaction.

Â Because of that similarity I will let you just go ahead and work a problem.

Â In this problem you are going to determine the Delta

Â S of the reaction that you see here.

Â Go ahead and work it and then we will comment.

Â Did you get 4

Â as your answer? Number 4 is your answer which is, 20 joules per Kelvin.

Â If so then you work the question correctly. If not

Â I'll go through some common mistakes. One common mistake is that students do not

Â to incorporate that 2.

Â They forget about the coefficients. So to get the Delta S we need to go

Â 2 moles times the entropy of HCl. So that is one mistake that

Â students make. Another common mistake as we continue on to this equation,

Â is they forget to

Â subtract the sum of these two things. So it is the S

Â of the H_2 and you need to add that together with the

Â s of the Cl_2.

Â We have those that we would add together and then subtract.

Â A common mistake is student will subtract the first and then add the second but you cannot.

Â Subtract the first and then add the second.

Â You will not want to do that. You want to subtract the sum both the those

Â Now in that

Â previous problem. and we'll look back at it here,

Â we had 2 moles of gas on the left

Â and we have 2 moles and gas on the right.

Â What would happen if we had a reaction in which the number of moles and gas is increasing

Â from reactants to products? So maybe we start with 1 mole of gas in the left and

Â we have 3 moles of gas on the right.

Â That would lead to an increased number moles of gas. What would you expect

Â to be the sign of delta S in that case?

Â Well, if you said positive you'd be correct because we know entropy

Â of gases far higher than entropies

Â of solid and liquids. So if you're increasing how much gas you have in your reaction chamber

Â then you would have an increase in entropy. Now in their previous example

Â we have the same number of moles of gas but we are seeing an increase in entropy. It is

Â not a big increase, twenty joules per mole,

Â but it is an increase. What do you think would lead to that increase in entropy?

Â Well. its molecular complexity.

Â We have 2 moles of gas that are elements.

Â Then we have two moles of gas that are compounds. So we have increased the molecular

Â complexity so we should see an increase

Â in the entropy. Now let's go back to consider what would happen if we had a

Â decrease in the number of moles a gas.

Â Maybe your reaction had 4 moles of gas on the left hand side of the equation. The

Â reaction sign and you are producing 2 moles on the right.

Â What would you expect?

Â Well in this case it would be a negative change because the

Â moles of gas is decreasing.

Â One more problem for you to work and here we are going to calculate the Delta G of a reaction

Â by going back to thinking of a equation that you learned.

Â That the Delta G over a reaction would be the Delta

Â H at the reaction minus

Â T, the temperature, delta S of the

Â other reaction. Now that we have learned how to calculate the delta S of the reaction

Â and if we are given the Delta H of a reaction. We are forced to calculate it, we can get the

Â Delta G of a reaction

Â and then determine whether a reaction is spontaneous or not.

Â So calculate it.

Â Did you come up with a -190.6 kilojoules?

Â If so you did it correctly. If not here are some other common mistakes.

Â Did you remember to work in Kelvin? I gave it to you in Celsius

Â but you have to be working in Kelvin, in that equation.

Â If you didn't work in Kelvin then you probably answered B.

Â Did you watch your units? Another very common mistake of students'.

Â Here is kilojoules

Â our delta S is in joules, you have to make sure if we are going to have our answer in

Â kilojoules that you've converted

Â the entropy to kilojoules before you put it into the equation.

Â The last choice of you selected D maybe you didn't watch

Â sign here we have got a negative sign.

Â A minus there and if you add instead of subtract you may have come up with an

Â answer D.

Â So this is the end of the learning objective number 7 in which we

Â learned to calculate the delta S

Â of a reaction by knowing the entropies of the products and the reactants.

Â We also learned to predict the Delta s sign just by looking at the

Â moles of gas and whether it's increasing or decreasing.

Â Lastly we incorporated the Delta s for reaction

Â into the equation that we see on this slide and

Â obtain a value for the Delta G of a reaction.

Â We know that if the Delta G is negative as it is in this case the reaction would

Â be spontaneous in the forward direction.

Â