Zinc metal reacts with hydrochloric acid according to the balanced equation: Zn(s) + 2 HCl(aq)¡ZnCl2(aq) + H2( g) When 0.103 g of Zn(s) is combined with enough HCl to make 50.0 mL of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.5 °C to 23.7 °C. Find ΔHrxn for this reaction as written. (Use 1.0 g>mL for the density of the solution and 4.18 J>g # °C as the specific heat capacity.)

Video transcript

Hi, everyone. I'm welcome. The next problem says zinc metal reacts with hydrochloric acid according to the balanced equation, zinc in the solid form plus two HCL aqueous becomes ZNCL two aqueous plus H two in the gas form. When 0.103 g of solid zinc is combined with enough hydrochloric acid HCL to make 50.0 mL of solution in a coffee cup calorimeter. All of the zinc reacts raising the temperature of the solution from 22.5 °C to 23.7 °C. Find delta H of reaction for this reaction as written. And then it says use 1.0 g per milliliters for the density of the solution and 4.18 joules per gram degrees Celsius as the specific heat capacity. And our answer choices are a positive 1.6 times 10 to the second kilojoules per mole B, positive 2.4 kilojoules per mole C negative 1.6 times 10 to the second KJ joules per mole or D negative 2.4 kilojoules per mole. Note that we have a little bit of a cheat here in that all of our answer choices have just two significant figures. So an important part of cal long calculations like this with lots of parts is keeping track of how many significant figures you have. So we'll point out where this comes from, but you don't have to actually think about it or figure it out because we have two significant figures in every single answer choice here. So let's think this through, we're given a reaction happening in a coffee up calorimeter in which the solution has a temperature raised. So the key we want to focus on here is that the heat released by the reaction which we know it's released because our temperature went up. So our Q solution or I should say RQ reaction. So the heat released by the reaction is equal to the negative of the heat absorbed by our solution because the heat lost by the reaction or pro excuse heat produced by the reaction is absorbed by the solution causing that rise in temperature. So how do we calculate what the heat absorbed by the solution is? Well, we would, we use the formula that Q of solution. So the heat absorbed by the solution is equal to the mass of the solution multiplied by the specific heat capacity times delta T or the change in temperature. So the mass of our solution, we have 50.0 mL of solution and we're given a value of density. So we'll just calculate that up here. We have 50.0 mL multiplied by 1.0 g per milliliter of solution. So very simple, we will have 50.0 g of our solution. So as we put in our numbers here, for a mass, we have 50.0 g. The specific key capacity of the solution was given to us as 4.18. That's joules over grams, degrees Celsius, put that in there. And then finally delta T. So what is delta T here? Well, we have that would be T final minus T initial, so that's going equal. Our final temperature was 23.7 °C while our initial temperature was 22.5 °C. So that change in temperature will be 1.2 degrees Celsius. So let's plug that into our equation. Our delta T is going to be 1.2 °C. So we see that our degrees Celsius cancel out. I'm going to do that with a red line, cancel out degrees Celsius, our grams cancel out and we're left with Jews as our units for our answer here. So we'll do this modification and that will give us the answer of 250 0.8 Jules as the heat absorbed by our solution. So therefore, we know that the heat of our reaction must equal negative 250.8 joules. So we have the heat, we're looking for the delta H of the reaction, the change in entropy, but we're in a coffee cup calorimeter. So we have a constant pressure. So we'd say at constant pressure. So again, we're at that constant pressure in the cal perimeter. So to get to delta H of the reaction, we can just use our Q value here. However, think about what's happening here. This heat is the result of the reaction of 0.103 g of zinc produced this much, it was completely consumed. But what is our question asking us? It says specifically find delta H reaction for this reaction as written, meaning the balanced equation we are provided by. And when we look at that equation, we have one mole of zinc in the equation as written. So we have the value of how much delta H we have or what the delta H is. When we've reacted 0.01 0.103 g of zinc, we need to convert that to say what is the delta H of reaction for one mole of zinc? So we need to use our molar mass to convert that. So we have 0.103 g of zinc. So let's look at the molar mass of zinc and that is 65 point 3 8 5 g of zinc. I'm writing that on the bottom, I'm writing out a conversion factor because I want my grams to cancel out. So I put that on the bottom with that fraction line above it. And then on top goes one mole of zinc. So I'm converting grams to mo my grams of zinc. Cancel out. I do my math and I'm left with one point 575 times 10 to the negative third moles of zinc. Let's tell you moles of zinc got reacted and I need to find out what is a delta H per mole. So that's a simple division. So the delta H of my reaction as written will equal. Well, we have the delta H four, our given amount. So I would say negative 250.8 jules divided by 1.575 times 10 to the minus third moles of zinc. But there and when I do that math, I'm now I'm going to finish up with my significant figures. Note that again, I didn't need to calculate this. But when I look at my entire calculation here, my smallest number of significant figures is in my delta T 1.2 °C. So there is my two but two sig fix. So don't forget to take that into account when you do these problems. So my answer here will be 1.6 times 10 to the fifth jules per mole of zinc. However, when I look at my answer choice here, I have it in kilojoules per M oh And note one very important thing, I have lost a negative sign. Always be careful. Sometimes it's useful when I have a negative number like I do in this problem to put parentheses around that negative number. That kind of helps set it aside and ensure you don't lose track of that little negative sign. So I need to change my value here two negative 1.6 times 10 to the fifth joules per mole zinc. So now I need to change that to kilojoules. Well, I've got 1000 joules per kilojoule. I want my jewels to cancel out. So on the bottom of my conversion factor 1000 jewels and on the top one kilojoule. So I divide by 1000. So that means subtracting three from my exponent and I get negative 1.6 times 10 squared 10 to the second power kilojoules per mole as my answer. Now, when I look at my answer, choice here, I have choice C negative 1.6 times 10 to the second kilojoules per mole note that I could have gotten the wrong answer just by losing that negative sign as choice A is positive 1.6. So be careful. Don't lose your negative sign. So choice A is incorrect just because it's the wrong sign. Note that a good way to check yourself on your sign when you're talking about an exothermic reaction, which we know we are because our solution warmed up, we should end up with a negative delta H for the reaction the reaction has produced heat. So you can always double check your sign in that way, when you know, for instance, in this case that your solution is warmed up, see you in the next video.