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### Course: AP®︎/College Chemistry > Unit 6

Lesson 5: Introduction to enthalpy of reaction# Enthalpy of reaction

The enthalpy change that accompanies a chemical reaction is referred to as the enthalpy of reaction and is abbreviated Δ

*H*_*rxn*. The value of Δ*H*_*rxn*depends on how the balanced equation for the reaction is written and is typically given in units of kJ/mol-rxn. Created by Jay.## Want to join the conversation?

- At2:45why is 1/2 the conversion factor? The reactant has 2 moles and the product has 3 moles.(2 votes)
- When Jay mentions one mole of the reaction, he means the balanced chemical equation. And in the balanced chemical equation there are two moles of hydrogen peroxide. That's why the conversion factor is (1 mol of rxn/2 mol of H2O2).

Hope that helps.(4 votes)

- I always understood that to calculate the change in ▲H° for a rxn or if you wanted to calculate any change such as ▲S or ▲G or anything, you did products minus reactants. But I came across a formula for ▲H of reaction(not the standard one with the ° symbol) and it said that it was equal to Σ bond energy of bonds broken + Σbond energy of bonds formed. When do I know when to use the ▲H formula and when the ▲H° formula? Thanks!(1 vote)
- 1. standard enthalpy (with the little circle) is the enthalpy, but always under one atmosphere of pressure and 25 degrees C.

2. I believe both equations can work for both types of enthalpy. To figure out which formula to use, you just look at the data you're given. Technically I think you can do either formula if you're given all the information. They should both give you the same delta h value.

3. For the bonds broken and formed formula, it tends to give slightly less accurate results because the bond energy is affected by surrounding atoms.

I'm not perfect with chemistry, so apologies if I got something wrong!(3 votes)

- i dont understand how the conversion factor was derived, please help!(1 vote)
- Since we are observing the reaction for h2o2, and we are given the enthalpy change for the reaction given (which includes 2 mols of h2o2), we know we must convert it to reflect the 1 mol of h2o2 desired. There are 2 moles h2o2, we want the enthalpy change to be for 1 mol h2o2, so we divide by 2. Dr. Skipper demonstrated it in a way that is analogous to dimensional analysis since that is already familiar from stoichiometry and is more difficult to mess up.(1 vote)

## Video transcript

- [Instructor] The change in enthalpy for a chemical reaction delta H, we could even write delta
H of reaction in here is equal to the heat transferred during a chemical reaction
at constant pressure. So delta H is equal to qp. Let's say we are performing
a chemical reaction, an aqueous solution under
constant atmospheric pressure. The reactants and products
of that chemical reaction make up the system and
everything else makes up the surroundings. When heat flows from the
surroundings to the system, the system or the reaction absorbs heat and therefore the change in enthalpy is positive for the reaction. This is called an endothermic reaction. If heat flows from the
system to the surroundings, the reaction gave off energy. Therefore the change in enthalpy for the reaction is negative and this is called an exothermic reaction. As an example of a reaction,
let's look at the decomposition of hydrogen peroxide to form
liquid water and oxygen gas. The change in the
enthalpy for this reaction is equal to negative 196 kilojoules. The negative sign means
the reaction is exothermic. And for the units, sometimes
you might see kilojoules. Sometimes you might see
kilojoules per mole, and sometimes you might see
kilojoules per mole of reaction. What kilojoules per mole of reaction is referring to is how
the equation is written. So if we look at this balanced equation, there's a two as a coefficient
in front of hydrogen peroxide and therefore two moles
of hydrogen peroxide are decomposing to form two moles of water and one mole of oxygen gas. So when two moles of
hydrogen peroxide decompose, 196 kilojoules of energy are given off. Next, let's calculate
how much heat is released when 5.00 grams of hydrogen
peroxide decomposes at a constant pressure. The first step is to
find out how many moles of hydrogen peroxide that we have. So we take the mass of hydrogen peroxide which is five grams and we divide that by the
molar mass of hydrogen peroxide which is 34.0 grams per mole. Grams cancels out and this gives us 0.147 moles of hydrogen peroxide. Next, we take our negative 196 kilojoules per mole of reaction and we're gonna multiply
this by a conversion factor. When we look at the balanced
equation for how it's written, there are two moles of hydrogen peroxide. So for our conversion factor for every one mole of
reaction as it is written, there are two moles of hydrogen peroxide. So two moles of H2O2. Now the of reaction will cancel out and this gives us negative 98.0 kilojoules per one mole of H2O2. So two moles of hydrogen peroxide would give off 196 kilojoules of energy. And one mole of hydrogen
peroxide would give off half that amount or
98.0 kilojoules of energy. Next, we take our 0.147
moles of hydrogen peroxide. So let me just go ahead and write this down here really quickly. So we have 0.147 moles of H202. And remember, we're trying to calculate, we're trying to calculate
the amount of heat that was released. So next we multiply that
by negative 98.0 kilojoules per mole of H202, and moles
of H2O2 will cancel out and this gives us our final answer. So the heat that was
released when 5.00 grams of hydrogen peroxide decompose
at constant pressure, this turns out to be equal
to negative 14.4 kilojoules.