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Ch.5 - Thermochemistry

Chapter 5, Problem 55b

When a 6.50-g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee-cup calorimeter (Figure 5.18), the temperature rises from 21.6 to 37.8 °C (b) Using your result from part (a), calculate H (in kJ/mol KOH) for the solution process. Assume that the specific heat of the solution is the same as that of pure water.

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Hey folks, welcome back. So in this practice problem we're going to be looking at heat of the solution, which is the heat evolved or absorbed when a solid dissolves in water and can be determined using a coffee cup calorie meter. So now we're going to be using a coffee cup calorie meter here. So we find that when five g of copper sulfate is dissolved in 120 g of water, the temperature of the solution increases from 23 to 28.9 degrees Celsius. Then we have a separate experiment and the heat capacity of the calorie emitter is determined to be 1.65 joules per degree Celsius. What we need to calculate here is the heat of the solution of copper sulfate and killer joules per mole based on these findings, We're going to assume that the specific heat of the solution. So C is going to equal to the specific heat of water were not given that, but we can either memorize it or always just look that up. So specific heat of water is going to equal to 4.184 grams. I'm sorry jules, jules per gram times Celsius. Okay, so that is and that is the seed that we're going to be using. So what we need to find here is actually two cues to heat. So one here is going to be of the solution and then the other one is going to be of the calorie emitter. Okay, so the one of the solution is going to equal to Q equals M cat, right? Because it has a mass. The calorie meter does not have a mass. So it's just going to be C times the change in the temperature. Alright, so now let's go ahead and plug in those numbers for the cues and then just realize that for the heat of the solution key one plus Q two is going to give us the heat of the solution. Of course it's not going to be in the correct units, but we'll figure figure that out once we get there. So, let's just go ahead and find Q one first. Alright, so, the mass of the solution obviously is going to be uh the mass of the solid plus the mass of the water rates. So we have five g Plus g. It was going to give us 125. So, that's the total mass that we put in here. So 125, wow. Okay. Times the specific heat of water, we have 4.1 jules over grams times Celsius. Sorry, my plan is just not working right now. I'm trying I'm trying. Right And then times the change in temperature We're going to get through this. The change in temperature here is obviously 28.9 -23. Right? That gives us a change in temperature of 5. degree Celsius. Alright, so let's go ahead and multiply this degree Celsius here will cancel out. And so will the grams. Right? So, we're gonna have jewels as the Unit here for Q one. All right. So, the answer for this first one should be 3000 um 85 0. Jules. Okay, so that's Q one. Let's go ahead and find Q two. So, here, in the separate experiment we had heat capacity of the calorie meter as 1.65 and it is in no 65 jules over degree sources. Okay. And then again, times the same change in temperature which is 5. degrees Celsius. So degree Celsius cancel out. And we're going to have jewels again as the answer for the heat. So, the answer for Q two here should be 9. three five jules 35. All right. So now we have Q one and Q two. Let's go ahead and add them together. And that will give us the heat of the solution. Okay, So Q total is going to equal to 85.7 jules plus the other Q 9. three. Bye. All right. And that will give us a an answer of 95 47 44. Alright, so now we take this number here and we actually want to go ahead and convert that into killer jewels. Right, Because we do want the heat of dissolution to being killed, joules per mole. Now, head of the solution is going to be basically a delta H. So heat of the solution, We're looking for delta H of this whole reaction. And we know that delta H s are always in kilo joules per mole. Usually in killing jewels, but always per one mole. Now we do need to talk about whether this DELTA H will be positive or negative, right? Because here we've gotten a positive number but it's not necessarily going to be positive for the heat of the dissolution. So if we think about this, we look at the change in the temperature. So we went from 23 to 28.9 and that is the solution. So the solution is actually um increasing in temperature. So it's actually gaining heat, gaining energy the solution but the reaction is not right. So where is the solution getting uh this this heat from from the reaction. Which means that the actual reaction, which is the reaction of the solution has to be negative. It has to be an endo thermic because otherwise, where would that heat and energy be coming from to heat up the solution? Okay, so Delta H here will be actually negative. So even though again, we got a positive 30 95 the delta each year that we're trying to find of the solution. Sorry, my pen is just working really horrible and I have to write really slow. So we're going to take 95. jules. And we're going to go ahead and convert that. So again, this will be negative. So I'm going to go ahead and put a negative sign here, we're going to convert jewels and to cure jewels. Okay, so one killer jewel, It's going to be 1000 or 10 to the third jewels. Alright, so jules cancels out. Now we're going to have kilo jewels. Great. So this will be negative three point 095. Hello jules, But we want again per one mole. Now we actually have five g of copper sulfate. So we need to figure out how many moles is five g. And then we're going to divide this killer jewel by the amount of moles that we have. So copper sulfate, we have five g of it. So let's go ahead and figure this out. So five g of copper sulfate. We're gonna go ahead and Convert those two moles. So in one mole Of copper sulfate, how many g are there? If you find the molar mass, it's going to be .61 g. Right? So copper here weighs 63.55. This is all from the periodic table. And there's only one of them uh sulfur. There's only one sulfur and it weighs 32 point oh six. We have four oxygen. So four times its smaller mass, which is 16. That will give us 64 then you just add it all together and it gives you 100 and 61. Okay, so grams cancel out and we're going to be left over with mold. So let's go ahead and find this number. And the molds that we have is 0.3 or point oh molds of copper sulfate. So this is the number that we're gonna go ahead and divide the delta H by so point oh moles of copper sulfate. And Folks, our final final answer is going to be again negative rates. So it's going to be negative once you divide that, it's going to be 98, 98.80 eight. And the units are here in Kilo Jewels, Killer Joules Per one Mole. Alright finally we got to the end here is our final answer. If you we are done awesome. Alright, so let's go ahead and think about this. What did we do? So we had to basically two different processes going on. We had the heat of that solution that we needed to find in the heat of the caliber emitter. We need to find separate and then once we've added those heats together, that gives us the heat of the the solution. Okay, but heat of the solutions and kill jews from all because it is delta H of this reaction, that's basically what we're trying to find. So we also needed to figure out that the delta H here is actually negative even though the solution temperature is increasing. Where is that heat coming from from the actual reaction? This delta H. Here is the entropy of that reaction. So that's why it's negative. So once we had the delta H here and we figured out that it was negative sine, all we had to do is basically just converting jewels and two kg jewels, realizing that the delta H. Here's and killed joules per mole. So we had to find how many walls we had of copper sulfate by again, just converting grams into moles and then divide those two numbers together. And bam. That's our answer. Alright folks, thank you so much for watching and we'll see you in the next video.