Acid spills are often neutralized with sodium carbonate or sodium hydrogen carbonate. For neutralization of acetic acid, the unbalanced equations are 112 CH3CO2H1l2 + Na2CO31s2 S CH3CO2Na1aq2 + CO21g2 + H2O1l2 122 CH3CO2H1l2 + NaHCO31s2 CH3CO2Na1aq2 + CO21g2 + H2O1l2 (c) How much heat in kilojoules is absorbed or liberated in each reaction? See Appendix B for standard heats of for- mation; ΔH°f = - 726.1 kJ>mol for CH3CO2 Na(aq).

Video transcript

Hi everyone for this problem, it reads acid spills are frequently neutralized using sodium carbonate or sodium hydrogen carbonate. The unbalanced equations that show the neutralization of Ben's OIC acid are the following determine the amount of heat absorbed or released in kila jewels by the neutralization of 0.500 gallons of formic acid were given, the density is equal to 1.22 g per mil liter. And below that were given the standard heats of four Formation. So the question that we want to answer for this problem is the amount of heat absorbed or released in Kila Jules. Okay. And we're told that we have .500 gallons of formic acid. So the first thing that we're going to want to do is we're going to there were given gallons of formic acid and we're gonna first want to go from gallons of formic acid to moles of formic acid. So let's go ahead and start there. We'll start with what's given. And what's given is 0. gallons a formic acid. Okay, so we're going to want to go from gallons to Leaders and the conversion that we'll use is one gallon is equal to 3. 854 Leaders. Okay, starting there, we'll see that our gallons of formic acid canceled and now we're in Leaders now we want to go from Leaders, two mL. Okay. And the reason that we want to do that is because the density is given in grams per milliliters. So we need to go from Leaders two mL and the conversion for that is and one leader There is 1000 MIla leaders. Okay, so now our leaders cancel and we're left in units of male leaders. Now we can use the density that was given to go from milliliters two g. Okay, so let's go ahead and write the density that was given for formic acid and that is in one millimeter There is 1.22 g of formic acid. So our milliliters cancel and we're left with grams. And remember we want to go to moles. And so the last thing that we're going to need to do is go from grams to moles. And we're gonna do this using the molar mass of formic acid. Okay, so in one mole Of formic acid there is 46. g. Okay, so now we see that our grams cancel and we're left with moles. So now we can go ahead and do this calculation. And when we do this calculation, what we're going to get is 50. moles of formic acid. So now we know how much molds of formic acid we have from the starting point of gallons. Okay, so now what we're going to want to do next is we're going to want we're going to calculate the entropy for the reaction. Okay, so our entropy for the reaction is going to equal the standard heats of formation of the products. Let's move this down. It's going to equal the standard heats of formation of the products minus the standard heats of formation of the reactant. And before we can do this first, we're going to need to balance the two equations that were given. Okay, because we're told that the equations are unbalanced. So we need to balance the two equations. Okay, so let's just go ahead and write out our first equation. And this should say equations. Okay, So when we write out our first equation and our first equation we have. Okay, so this is our unbalanced equation. Let's go ahead and write out everything that we have. We'll start with our reactant since and our products. Okay, so looking at our equation, we have carbon, We have hydrogen, sodium and oxygen on the reactant side. For carbon, there are two for hydrogen, there are two for sodium, there are two and for oxygen there are five for our products. For carbon, there are two for hydrogen, there are three for sodium, there is one and for oxygen there is five. So we can see here that these numbers do not match. And so we're going to need to find a way to balance this out. Okay, so we're going to balance this out by adding a two here and to hear. Okay, so once we do that, what we now get is we're going to have three carbons for the react mints, four hydrogen, we'll have to sodium and now we'll have seven oxygen's and for the products, what we're going to have is three carbons for hydrogen, two sodium and seven oxygen's. So this is our balanced equation. So now we can go ahead and calculate the entropy change for this reaction. Remember we said it's going to be the standard heats of formation of the products minus the standard heats of formation of the reactant. And were given all of those values here. So we're going to reference these values here to do this calculation. So let's go ahead and write this out. So our standard heat, our standard entropy change for the reaction is going to equal. So let's start off with our products. It's the sum of our products. Looking at our balanced equation, we have two more. We have two moles of this product and its entropy. It's standard heat of formation based off of the values given is negative 2 93. kg jewels per mole Plus we have one mole of carbon dioxide gas and its standard heat of formation is negative 393.5 kg joules per mole. And the last product is water as a liquid. We only have one mole of it. Okay, and it's standard heat of formation is negative two 85. killer joules per mole. Okay, so that's the sum of our products and we're going to minus the sum of our reactant. So now let's take a look at our reactant, we have two moles of formic acid and its standard heat of formation is negative 425.0 kg jewels per mole. Plus we have one mole of this reactant and its standard heat of formation is negative 1130.7 Killer joules per mole. Okay, so now we're going to do this calculation and when we do this calculation, we're going to get the entropy change for this reaction is equal to 714.8 kg jewels. For we need to look at how many moles of formic acid we have based off the balance equation. This is 714.8 kg jewels for two moles of formic acid. Okay, so that is how much heat we have for two moles formic acid. So now we need to determine whether this heat is going to be released or absorbed. So let's go ahead and we know how many moles of formic acid we have because we calculated that up above we have 50.1650 moles of formic acid. So, given this amount of moles of formic acid, we're going to calculate the heat. Okay, so he can be represented by Q. And so our heat is We have our moles. We're going to write 50.1650 moles of formic acid. That's what we calculated. Now, we're going to use this standard heat of formation here. So that and we're going to set it up as a dimensional analysis. Okay, so we're going to do For every two moles of formic acid. There is 714.8 kg joules of heat. So now we'll do this calculation. And what we're going to get is 17, .971 Killer Jewels. Okay, and this is going to be absorbed. So let's go ahead and write this out in the scientific notation. So, we're going to say sodium carbonate absorbs 1.79 times 10 to the 4th kila jewels. And this is going to be our answer. All right, So this is our answer for sodium carbonate. Now we need to do the same thing for sodium, hydrogen carbonate. Alright, so let's go ahead and do that. Now, what we're going to need to do is we're going to need to balance our second equation. So let's go ahead and rewrite out our second equation, just like we did the first one. Alright, so let's do that here. Our second equation. Okay, so this is our second equation. Alright, so we're going to count our reactant since and our products. Alright, so we have carbon, hydrogen, sodium and oxygen on the reactant side. For carbon. There are two for hydrogen. There are three for sodium, there is one and for oxygen there are five on the product side for carbon. There are two for hydrogen, there are three for sodium, there is one and for oxygen there is five. So we see here that our reactant and our products match. And so this equation is balanced. Now we can go ahead and solve for the entropy change for our reaction. Okay, so the entropy change for our reaction is going to equal the standard heats of formation of the products minus the standard heats of formation of the reactant. So let's go ahead and start with our products. Okay, so our products we have this is our first one. We have one mole of it. And it's standard heat of formation is negative 293.3 kg jewels Permal. And remember we're referring to the values that we're given up above. Okay, for our second reactant, I mean for our second product, excuse me, we have one mole. And the standard heat of formation is negative 0.5 kg joules per mole. And for our last product we have one mole of it. And it's standard heat of formation is negative 85.8 kg jewels Per mole. Okay, so that's the sum of our products and we're going to minus the sum of our reactant. Okay, so for our reactant, we have one mole of formic acid. And its standard heat of formation is negative 425.0 killer jewels per mole. And our second reactant we have one mole of it. And it's standard heat of formation is negative 950.8 kg joules per mole. Alright, so we wrote this out correctly with all the standard pizza formation and the moles that match. So let's go ahead and do the calculation. So the standard entropy change for this reaction is going to equal 534.9 kg jewels Permal. And remember we need to know how many moles of formic acid we have. So based off our balanced equation, this is gonna be for one mole of formic acid. Okay, so let's go ahead and calculate the heat. So our heat Q. We're going to start off with the moles of formic acid that we calculated above which is 50. moles Formic acid. And now we're going to go ahead and plug in our standard entropy change for the reaction. So for one mole of formic acid There is 534.9 kg jewels. Okay, So our moles of formic acid cancel. And now we're left with killer jewels. And when we do the calculation, we get 26,833. kila jules. And this is absorbed. Alright, so let's go ahead and write this out in scientific notation. And so we can write as our final answer for this sodium hydrogen carbonate absorbs 2.68 times 10 to the four kg jewels. So this is the final answer for our sodium hydrogen carbonate. Okay, so our sodium carbonate absorbs 1.79 times 10 to the four kg jewels, and our sodium hydrogen carbonate absorbs 2.68 times 10 to the four kg jewels. That is it for this problem, I hope this was helpful.