Thermochemical Equations

now recall that start geometry deals with the numerical relationship between compounds and a balanced chemical equation. Now, with a thermal chemical equation, we're going to deal with chemical reactions that include an entropy of reaction, which is Delta h r x n so delta h of reaction here with thermal chemical equations, we're gonna be introduced to our thermal chemical start geometric chart. Now here the chart uses the given quantity of a compound to determine the unknown quantity of another compound. So here, with the thermal chemical equation, we have our balanced chemical equation, and to the side of it, you'll see your Delta h of reaction. It's our job to make a connection between your entropy of reaction and either moles, grams molecules. What have you in terms of the chemical reaction? So in a thermal chemical equation, what we're trying to do is not a multiple comparison. We're trying to do a Delta H two mole comparison, and that's the key difference with the thermal chemical equation. Now that we've seen this, let's move on words and talk about mawr with thermal chemical equations

Now, if you've seen my stock geometric chart understood geometry or solution, chemistry or gas Toki aama tree. This should be pretty familiar to you, but if this is the first time you're seeing this documentary chart, let's go through it. All right. So what we need to realize here is that with thermo chemical equations, sometimes it's pretty common to be given the Delta h off our chemical reaction. So we'll start out with Delta H of Given. And like I said before in Sochi, a mature. We're used to doing a multiple comparison, which can still happen here. But the more important thing is that we establish a connection between the Delta h of reaction and one of the moles for one of the compounds within our chemical reaction. So here we go from Delta H of Given, two moles of Given and what we can say here is that besides going from Delta H given, two moles have given we can go from grams of given two moles of given or we can go from ions. Adams formula units or molecules, have given two moles of given. Once we get there, we have to go to moles of are unknown to do. This requires a leap of faith, in a sense, because you're going from an area where you know information to an area where you know nothing at all. So we call this to jump when you make this jump. In order to do it correctly, you have to do a multiple comparison and use the coefficients in the balanced equation. From this point, you are at most of unknown, and from here you can go in any way you want. You can go from most of unknown toe ions, Adams formula units or molecules. You can go to grams, or you can go to a new Delta h of unknown. So just realized with the thermo chemical equation will have a balanced chemical equation, which introduces the idea of Stoke eom. A tree before to be a thermo chemical equation will also have the Delta H of reaction present

here, it says. Consider the following thermal chemical reaction. We have two moles of magnesium solid reacting with one mole of oxygen gas to produce two moles of magnesium oxide solid. It gives us an entropy of reaction equal to negative. 12 04 killing jewels here were asked how many grams of magnesium oxide are produced doing entropy? Change of negative 3. 75. Killer Joel's So we have to do here is we have to convert the given quantity into moles of Given. All right, so they're giving us negative 3 killer jewels of energy And what we need to do is establish a relationship between magnesium oxide and this value of negative 3 75. Well, according to my balanced equation for Step two, it says, we need to do a multiple comparison to convert. Moles have given it to moles of unknown here because it's a thermal chemical equation. It's gonna be moles of given relating to Delta h of reaction. So we're gonna say for every two moles of magnesium oxide, the energy involved or entropy involved is negative. 12 04 killed jewels per mole. So we just found our moles in terms of magnesium oxide steps, three says, If necessary, convert the moles into desired units. Here they want grams, not moles. So we're gonna do one more step and say, for every one mole of magnesium oxide, the mass of magnesium oxide. One magnesium is 24.31 g, according to periodic table one. Oxygen is 16 g. Multiplying and add those numbers together gives us 31 Gramps. But moles cancel out. And now I'm gonna have 25.11 g of magnesium oxide step forwards and needed here because in Step four, if we have to calculate more than one final amount, then we must compare them to determine the theoretical yield. Here. We're only given one given amount of negative 3 75 killer Joel's and using. That helps us to determine the final answer of 25.11 magnesium oxide grants. Magnesium oxide