Worked example: Measuring enthalpy of reaction using coffee-cup calorimetry

0:00- [Instructor] A constant pressure calorimeter can be used to find the change in entropy for a chemical reaction. Let's look at the chemical reaction between an aqueous solution of silver nitrate and aqueous solution of sodium chloride to form a precipitate of silver chloride and an aqueous solution of sodium nitrate.

0:18Let's say we have 25.0 milliliters of a 0.100 molar solution of silver nitrate and 25.0 milliliters of a 0.100 molar solution of sodium chloride. Both solutions are initially at a temperature of 25.000 degrees Celsius. Next, we add our two solutions to our calorimeter which has made of two coffee cups.

0:44And since the top coffee cup is loose fitting, this reaction is under the constant pressure of the atmosphere. So this is constant pressure calorimetry. After the two aqueous solutions mix, the reaction occurs and we watch the thermometer in the calorimeter. In this case, the temperature of the solution increases and the final temperature, the highest one reached in our experiment is 25.781 degrees Celsius.

1:12So the change in the temperature of the solution would be the final temperature minus the initial temperature, which is 25.781 minus 25.0 which is equal to positive 0.781 degrees Celsius. The total volume of solution would be 25 plus 25, which is equal to 50.0 milliliters of solution.

1:41If we assume the density of the solution is one gram per milliliter, 50.0 milliliters is equal to 50.0 grams. Next we need to solve for the heat gained by the water and we can use the Q is equal to MC delta T equation. So we're solving for the heat, which is symbolized by Q, M is the mass of our solution which we saw was 50.0 grams, So we can plug that in.

2:09We can assume that the specific heat of the solution is the same as the specific heat of water, which is 4.18 joules per gram degrees Celsius and the change in the temperature of the solution was 0.781 degrees Celsius. So we can plug that in as well. Grams cancels out, degrees Celsius cancels out and we find that Q is equal to positive 1.63 times 10 to the second joules.

2:47The positive sign means that energy was gained by the water. But let's think about the distinction between system and surroundings. The system consists of the reactants and products for our particular reaction and the surroundings are everything else which includes the water.

3:05So in this case, since the temperature of the surroundings increased, right? We saw an increase in the temperature, that means that heat flowed from the system to the surroundings and so the surroundings increased in energy and that's what we see with this positive sign here.

3:21If we assume a perfect transfer of heat from the system to the surroundings, if the surroundings gained energy, that means the system lost energy. So if we're thinking about the heat transferred for the reaction, it's the same in magnitude, 1.63 times 10 to the second joules.

3:40However, we need to put a negative sign in here which indicates that energy was given off by the reaction. The heat that's transferred under constant pressure is equal to the change in the entropy of the reaction, delta H. However, let's find the change in entropy, delta H in terms of kilojoules per mole of silver chloride for our units.

4:05Since silver chloride is one of our products, we first need to find moles of our reactants and we're gonna do that using the molarity equation which says that molarity is equal to moles divided by liters. For our silver nitrate solution, the concentration was 0.100 molar and trying to solve for moles so that's X.

4:26The volume of our silver nitrate solution was 25.0 milliliters which is 0.0250 liters. So we solve for X and we get 0.00250. So that's how many moles of silver nitrate that we started with and it's the exact same calculation for sodium chloride as well. So this is also how many moles of sodium chloride we have.

4:57Next, we go back to our balanced chemical equation and we can see we have coefficients of one in front of silver nitrate, in front of sodium chloride and in front of silver chloride. Therefore, we're also going to produce 0.00250 moles of silver chloride. Next we're gonna calculate the change in the entropy, delta H, for our reaction.

5:21The heat that was transferred was negative 1.63 times 10 to the second joules and we're gonna divide that by the moles of silver chloride which was 0.00250 moles of silver chlorides. This is equal to negative 65,200 joules per mole of silver chloride and we could convert that into kilojoules.

5:53And so this is equal to negative 65.2 kilojoules per mole of silver chloride. We could stop right here and give this as our final answer, but let's keep going and convert two kilojoules per mole of reaction. First, let's rewrite this. We have negative 65.2 kilojoules per mole of silver chlorides and what it means by kilojoules per mole of reaction is how the reaction is written in the balanced equation down here.

6:31So if we look at the balanced equation, there's one mole of silver chloride for how the reaction is written. So we can write a conversion factor of one mole of silver chloride per one mole of reaction and writing it this way for the conversion factor, the moles of silver chloride would cancel out and give us negative 65.2 kilojoules per mole of reaction.

7:06So this could also be our final answer. Finally, the negative sign means that we have an exothermic reaction. The reaction gave off energy and this value when you do a constant pressure calorimetry experiment is often a little bit lower than the actual value because in reality there's not always a perfect transfer of heat from the reaction to the water.

7:31Often some of the energy is lost to the environment.