# Video: Calculating the Heat of Reaction for a Balanced Chemical Reaction Equation Given the ΔHs for a Set of Related Reactions

Consider the heats of reaction for the following individual reactions. B + D → 2C, ΔH = −600 kJ/mol. A + 2D → E, ΔH = −500 kJ/mol. 2B + 3A → F, ΔH = +70 kJ/mol. Find the heat of reaction for the following reaction. F + 8D → 4C + 3E [A] −2770 kJ [B] −2650 kJ [C] −1050 kJ [D] −350 kJ [E] +2650 kJ

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### Video Transcript

Consider the heats of reaction for the following individual reactions. Find the heat of reaction for the following reaction. F plus 8D reacting to form 4C plus 3E. A) Negative 2770 kilojoules, B) Negative 2650 kilojoules, C) Negative 1050 kilojoules, D) Negative 350 kilojoules, or E) Positive 2650 kilojoules.

H is called the enthalpy, which is often related to the heat that’s absorbed or released during a reaction. If the change in enthalpy for a reaction is positive, that means that the reaction absorbed heat or thermal energy during the course of the reaction. We call a reaction that has a positive change in enthalpy, and therefore absorbs heat, endothermic. If the change in enthalpy for a reaction is negative, that means that heat or thermal energy was released during the reaction. If a reaction has a negative change in enthalpy, meaning that it released heat, we call that reaction exothermic.

This question is asking us to consider the heats of reaction, which, as we just discussed, is going to be related to the change in enthalpy for a reaction. In particular, we’re tasked with calculating the heat of reaction for this reaction at the bottom, but we’re not given the change in enthalpy for this reaction in particular. So, how are we going to find it? To solve this problem, we’re gonna use what’s called Hess’s law, which says that the change in enthalpy for an overall reaction is equal to the sum of the individual enthalpy changes for each step.

So, we’re gonna think of the above reactions as individual steps in a bigger reaction that when we combine them, we’ll create the reaction that we’re interested in. We’re wanting to create the reaction F plus 8D reacts to form 4C plus 3E. So, let’s start by looking at the reaction that I’ve labelled one. This reaction is the only one that has C as a product, and the reaction that we’re wanting to construct also has C as a product. So, let’s use this reaction the way it’s written. We’ll ultimately want 4C, though, so let’s multiply it by two. So, if we multiply that, we get 2B plus 2D reacting to form 4C. I’ve just made a note, this is two times reaction one so we can remember it for later.

Reaction two is A plus 2D reacting to form E. The reaction that we’re looking to build has 3E as a product. Reaction two is the only reaction that forms E. So, let’s use this one the way it’s written again. Since we want to form 3E in the reaction we’re looking to build, let’s multiply this by three. That will give us 3A plus 6D reacting to form 3E, which, again, is three times reaction two. Reaction three is 2B plus 3A reacting to form F. And the reaction that we want to build, we want F to be a reactant, not a product. So, let’s use this reaction the reverse of the way it was written, which I’ll make a note of by saying that’s the negative of reaction three.

Now, we’ve used all three of the reactions that we were given, so let’s see if we constructed the reaction that we’re looking for. We can do this by adding together things that are on the same side of the reaction arrow. So, reactants get added to reactants and products get added to products, and we can cancel things that are on opposite sides. So, for example, we had 2B as a reactant, but in a different reaction, we had 2B as a product. So, those two can cancel. If we add everything together, we get F plus 8D reacting to form 4C plus 3E. Which is the reaction that we were looking to construct.

Now, we need to find the change in enthalpy for this reaction, so we’ll know the heat of the reaction. The first thing that we did was use two of the first reaction. The first reaction has an enthalpy change of negative 600 kilojoules per mole. So, each mole of reaction or, in other words, each time we do the reaction the way it’s written, the change in enthalpy is negative 600 kilojoules. Since we used two of these, we need to multiply this negative 600 by two. The next thing that we did was use three of the second reaction.

The second reaction has a change in enthalpy of negative 500 kilojoules per mole. So, we need to multiply this by three since we used three of them. The third thing that we did was use one of the third reaction, but we flipped the way that it was written. So, we’ll need to flip the sign of the change in enthalpy for the third reaction. And since we used one, we’ll just multiply it by one. Now, what we need to do is add everything together. This gives us negative 2770 kilojoules, which matches answer choice A. So, the heat of reaction for the reaction F plus 8D reacting to form 4C plus 3E is negative 2770 kilojoules.