ShowMe: Example Calorimetry Problem
- OK, so here's an example problem for calorimetry. This is a constant pressure calorimeter, we are adding 75 milliliters of 0.750 molar sulfuric acid to 75 milliliters of 0.370 molar sodium hydroxide. This reaction caused the temperature of the whole solution to rise from 24 degrees to 26.52 degrees celsius. So we're supposed to calculate delta H per mole of water produced.
OK. So what we need to think about here, one, we do need to figure out how much heat was produced in this reaction, and what was absorbed by the calorimeter. Of course, the same quantity just in opposite signs. So, it's easiest to start by calculating the heat that the water absorbed in the calorimeter.
OK, so we're adding 75 milliliters of sulfuric acid to 75 milliliters of sodium hydroxide solution so we're gonna have 150 grams of water to start off with. And that just came from 1 gram per milliliter density of water. And we have a total of 150 milliliters that we're adding together. OK, so that's 150 grams of water in the calorimeter.
OK. So let's go ahead and calculate the heat that was absorbed by this water in the calorimeter. So we're gonna use q equals MCS delta T. We have our 150 grams of water, we know the heat capacity for water, we just look it up, 4.184 joules per gram degrees C. And our delta T, which is 26.52, that's the final temperature, minus 24.0 degrees C.
So when we multiply that out, we get 1581.6 joules. So let's go ahead and convert that to kilojoules right now. So what we need to do for that, we take 1581.6 joules, we have 1000 joules, 1 kilojoule, so we're gonna end up with 1.582 kilojoules. So let's hold onto that number for a little while. Let's go to the next part of the problem.
So the next part of the problem requires us to figure out one, how much of each reactant we're starting with, and two, which one is the limiting reactant. So we have sulfuric acid, and we have 75 milliliters of it, not gonna show the conversion but that's 0.075 liters. And the molarity is 0.750 for the sulfuric acid. So let's figure out how many moles of sulfuric acid we have. We have 0.750 moles per liter, multiplied by 0.075 liters, and we get 0.05625 moles of sulfuric acid.
Alright, so now let's go ahead and do the same thing for sodium hydroxide. And it's 0.370 moles per liter to begin with. We also have 75 milliliters. Let's go ahead and multiply that one by 0.075 liters. And for sodium hydroxide, we end up with 0.02775 moles of sodium hydroxide. Ok, now, how many moles of water are we gonna be able to make? So we need to check for the limiting reactant on this. Alright, so let's go to the next one.
OK, so checking for limiting reactant. So let's start off with our, actually the first thing we need to do-we haven't done this yet-let's go ahead and write an equation for mixing sodium hydroxide and sulfuric acid together and let's see what we get. So let's go ahead and do that. So sulfuric acid, plus sodium hydroxide, gives what? So we're mixing a strong acid and a strong base and we're always gonna get water and salt. And the salt is just gonna be the anion from whatever is left. So in this case it's gonna be Na2SO4.
And, let's see, so we need to make sure that we are balanced, so I think we need a 2 in front of sodium hydroxide, and so we have 4 hydrogens total on the reactant side so we need to put a 2 here by this water. Alright, so I believe we are balanced. Take a look at that and make sure.
OK, so let's go ahead and figure out how much water, cause that's our product, we can make from each of these moles of sulfuric acid and sodium hydroxide. So let's start with the sulfuric acid. So we have 0.05625 moles of sulfuric acid. Excuse my terrible handwriting on this thing. Alright, so it requires 1 mole of sulfuric acid to 2 moles of water--it produces 2 moles of water I should say. OK, and we end up with 0.1125 moles of water.
Alright so let's do the same thing for sodium hydroxide, we started off with 0.02775 moles of sodium hydroxide, and we require 2 moles sodium hydroxide to make 2 moles of water. So we end up with 0.02775 moles of water. OK, so sodium hydroxide is the limiting reactant. So this is all the water that we can make.
Alright so let's go ahead and do the last step now so now we're gonna calculate delta H, ok. And remember, delta H is just the heat at constant pressure, and we're in a constant pressure calorimeter. And so we made 0.02775 moles of water. We had 1.58 kilojoules of heat produced. OK, now, 'produced,' that means we need a negative sign right? Because the calorimeter temperature went up, and the heat came from this acid base reaction. So let's add a negative sign, and divide by the number of moles, and then we end up with 57.0 kilojoules. Negative. Ok, kilojoules per mole. And we have 3 sig figs from our original problem, so this is the final answer.