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In this article, we will be discussing the** enthalpy of reaction, **which quantifies this exchange of heat.

- This article covers the
**enthalpy reaction** - First, we will define the enthalpy of a reaction
- Next, we will describe the rules of enthalpy
- Then, we will explain
**standard enthalpy** - We will then cover the enthalpy of reaction formula and
**Hess's Law** - Lastly, we will walk through examples of how to calculate the enthalpy of a reaction.

## Enthalpy of Reaction Definition

Before we define the enthalpy of reaction, we should first look at the definition of **enthalpy**.

**Enthalpy **is the total heat content of a system. The formula definition is the internal energy plus the product of pressure and volume: $$H=U+PV\,\,\text{where: H is enthalpy, U is internal energy, P is pressure, and V is volume}$$

**enthalpy of reaction.**

The **enthalpy of reaction (ΔH _{rxn}) **is the change in enthalpy due to a chemical reaction. The general formula is: $$\Delta H_{rxn}=H_{final}-H_{inital}=q\,\,\text{; where q is heat}$$

If ΔH_{rxn} **> **0, the reaction is **endothermic **(the system pulls in heat from its surroundings)

If ΔH_{rxn} **< **0, the reaction is **exothermic **(the system releases heat into its surroundings)

Here is a diagram showing the difference between these two reaction types:

- In an
**endothermic reaction**, heat is absorbed from the surroundings for the reaction to proceed. Since the system ends with more heat than it started with, the change in enthalpy is positive.

- For
**exothermic reactions**, the products have a lower enthalpy than the reactants. This means that the "extra" heat is released into the surroundings, and the change in enthalpy is negative.

We also will sometimes label the change in enthalpy as something other than ΔH_{rxn} based on the type of reaction that is happening. These are:

- E
**nthalpy of combustion (ΔH**change in enthalpy due to a combustion reaction (substance burns in oxygen)._{comb}): **Enthalpy of fusion (ΔH**change in enthalpy due to fusion (melting) of 1 mol of a substance._{fus}):**Enthalpy of vaporization (ΔH**change in enthalpy due to vaporizing (liquid to gas) of 1 mol of a substance._{vap}):**Enthalpy of solution (ΔH**change in enthalpy due to a set amount of a solute dissolving in a given amount of solvent._{sol}_{n}):

Enthalpy of reaction can be related to the** activation energy**, which is the energy required for a reaction to proceed. Reactions with a negative enthalpy change will have a smaller activation energy than those with positive enthalpy changes. The activation energy is equal to the enthalpy change between the reactants and the **activated complex**, which is the "intermediate" between reactants and products.

## Enthalpy Change of Reaction

There are a few "rules" to follow when looking at enthalpy change.

First: when a reaction/process is reversed, the sign of the reaction enthalpy changes. For example, when ice melts, the reaction has a positive enthalpy change (it is absorbing heat to melt). However, when water freezes, the enthalpy change is negative since it is releasing heat (i.e. it is cooling).

Second: enthalpy is proportional to the amount of reactants. When you look up enthalpy values for reactions, it is assumed that it is for 1 mol of reactants. So if you wanted to know the ΔH_{rxn} for 3 mols of reactant, then you would multiply the given value by three.

## Standard enthalpy of reaction

When we calculate a reaction's enthalpy, we utilize the **standard** **enthalpy of formation.**

**The ****standard ****enthalpy of formation (ΔH _{f}°) **is the enthalpy change for the formation of 1 mol of a compound from its elements. These elements are in their standard state, which is the most stable form of the element at 1 atm and 298 K. Here is an example: $$C_{(s)}+O_{2\,(g)}\rightarrow CO_{2\,(g)}\,\,\,\Delta H_f^\circ =-393.5 \frac{kJ}{mol}$$

Here is a table of some common standard enthalpies of formation:

Compound | ΔH°_{f} (kJ/mol) |

CO_{(g)} | -393.5 |

H_{2}O_{(g)} | -241.8 |

H_{2}O_{(l)} | -285.8 |

NaCl_{(s)} | -411.0 |

NH_{3 (g)} | -46.2 |

NO_{2 (g)} | 33.9 |

The standard enthalpy is dependent on the state (ex: solid/liquid/gas) of the molecule. As you'll see above, the standard enthalpy for water as a liquid and as a gas is different. In addition, the ΔH_{f}° for any element is 0. This is because it doesn't take any energy to form a naturally occurring compound (i.e there is no reaction).

When we calculate the reaction enthalpy using standard enthalpies, then it is the **standard reaction enthalpy**.

Now that we have a basic understanding of reaction enthalpy, let's learn how to calculate it.

## Enthalpy of reaction formula

When we calculate ΔH_{rxn}, we use this formula: $$\Delta H_{rxn}=\sum m\Delta H_{f\,(products)}^\circ- \sum n\Delta H_{f\,(reactants)}^\circ$$

where m, n are the molar amounts of the products and reactants, respectively.

However, if we are calculating the net enthalpy change for a series of reactions, we would use **Hess's law. **

**Hess's law **states that the net enthalpy of an overall reaction is equal to the sum of the enthalpies of its steps. The formula is: $$\Delta H_{net}=\sum \Delta H_{rxn}$$

## Calculating a molar heat of reaction from formation enthalpies

Now that we know our formulas, let's do some calculations!

What is ΔH_{rxn} for the combustion of methane (C_{3}H_{8}) given the following thermodynamic data:

$$C_3H_{8\,(g)}+5O_{2\,(g)}\rightarrow 3CO_{2\,(g)}+4H_2O_{(g)}$$

Compound | ΔH°_{f} (kJ/mol) |

C | -103.8 |

O_{2} | 0 |

CO_{2} | -393.5 |

H_{2}O | -241.8 |

Our first step is to adjust the enthalpy values to match the amounts of reactants/products specified in the reaction:

\(-393.5\frac{kJ}{mol}*3=-1,180.5\frac{kJ}{mol}\,CO_2\)

\(-241.8\frac{kJ}{mol}*4=-967.2\frac{kJ}{mol}\,H_2O\)

Now we can solve for ΔH_{rxn}:

\(\Delta H_{rxn}=\sum \Delta H_{f\,products}^\circ-\sum \Delta H_{f\,reactants}^\circ\)

\(\Delta H_{rxn}=(\Delta H_{f\,(CO_2)}^\circ+\Delta H_{f\,(H_2O)}^\circ)-(\Delta H_{f\,(C_3H_8)}^\circ+\Delta H_{f\,(O_2)}^\circ)\)

\(\Delta H_{rxn}=(-1,180.5\frac{kJ}{mol}-967.2\frac{kJ}{mol})-(-103.8\frac{kJ}{mol}+0)\)

\(\Delta H_{rxn}=-2,147.7\frac{kJ}{mol}+103.8\frac{kJ}{mol}\)

\(\Delta H_{rxn}=-2,043.9\frac{kJ}{mol}\)

Let's do one more example:

$$3Ca(OH)_{2\,(aq)}+2H_3PO_{4\,(aq)}\rightarrow Ca_3(PO_4)_{2\,(s)}+6H_2O_{(l)}$$

What is the ΔH_{rxn} for the above reaction given the following thermodynamic data:

Compound | ΔH°_{f} (kJ/mol) |

Ca(OH)_{2}_{(s)} | -986.1 |

Ca(OH)_{2}_{(aq)} | -1002.8 |

H_{3}PO_{4}_{ (aq)} | -113.7 |

Ca_{3}PO_{4 (s)} | -4120.8 |

H_{2}O_{ (l)} | -285.8 |

H_{2}O_{ (g)} | -241.8 |

Like before, we need to adjust the enthalpy values to match the molar amounts given in the chemical equation. You'll note that there are two values of ΔH°_{f} for both Ca(OH)_{2} and H_{2}O. You always need to pay attention to the states of your molecules when looking up thermodynamic data, since ΔH°_{f }depends on the state.

\(-1002.8\frac{kJ}{mol}*3=-3008.4\frac{kJ}{mol}\,Ca(OH)_2\)

\(-113.7\frac{kJ}{mol}*2=-227.4\frac{kJ}{mol}\,H_3PO_4\)

\(-285.8\frac{kJ}{mol}*6=-1,714.8\frac{kJ}{mol}H_2O\)

Now we can solve for ΔH_{rxn}:

\(\Delta H_{rxn}=\sum \Delta H_{f\,products}^\circ-\sum \Delta H_{f\,reactants}^\circ \)

\(\Delta H_{rxn}=(\Delta H_{f\,(H_2O)}^\circ+\Delta H_{f\,(Ca_3PO_4)}^\circ)-(\Delta H_{f\,(Ca(OH)_2)}^\circ+\Delta H_{f\,(H_3PO_4)}^\circ)\)

\(\Delta H_{rxn}=(-1,714.8\frac{kJ}{mol}-4120.8\frac{kJ}{mol})-(-3008.4\frac{kJ}{mol}-227.4\frac{kJ}{mol})\)

\(\Delta H_{rxn}=-5,835.6\frac{kJ}{mol}+3235.8\frac{kJ}{mol}\)

\(\Delta H_{rxn}=-2,599.8\frac{kJ}{mol}\)

### Hess's Law examples

Lastly, we are going to walk through an example of how to use Hess's law to calculate ΔH_{rxn}.

Calculate the ΔH_{rxn} of the following reaction using the list of reactions below:

$$2N_{2\,(g)}+5O_{2\,(g)}\rightarrow 2N_2O_{5\,(g)}$$ $$H_{2\,(g)}+1/2O_{2\,(g)}\rightarrow H_2O_{(l)}\,\,\Delta H_f^\circ=-285.8\frac{kJ}{mol}$$ $$N_2O_{5\,(g)}+H_2O_{(l)}\rightarrow 2HNO_{3\,(l)}\,\,\Delta H_{rxn}=-76.6\frac{kJ}{mol}$$ $$N_{2\,(g)}+3O_{2\,(g)}+H_{2\,(g)}\rightarrow 2HNO_{3\,(l)}\,\,\Delta H_{rxn}=-348.2\frac{kJ}{mol}$$

Our goal here is to arrange these three equations so that they add up to the top equation. Every action we do to an equation, we must also do to its enthalpy value.

For our first step, we will flip the second equation and multiply it by 2. This is so we have 2N_{2}O_{5(g)} on the right side. This means we have to multiply the enthalpy value by -2 (the negative is because of the flip). So our new array looks like this: $$H_{2\,(g)}+1/2O_{2\,(g)}\rightarrow H_2O_{(l)}\,\,\Delta H_f^\circ=-285.8\frac{kJ}{mol}$$ $$4HNO_{3\,(l)}\rightarrow 2N_2O_{5\,(g)}+2H_2O_{(l)}\,\,\Delta H_{rxn}=(-76.6\frac{kJ}{mol}*-2)=153.2\frac{kJ}{mol}$$ $$N_{2\,(g)}+3O_{2\,(g)}+H_{2\,(g)}\rightarrow 2HNO_{3\,(l)}\,\,\Delta H_{rxn}=-348.2\frac{kJ}{mol}$$

For our next step, we will flip the top equation and multiply it by 2. This is so the H_{2}O molecules will cancel, since there are none present in the main equation. (If a species is on both the reactant and product side, it will cancel). $$2H_2O_{(l)}\rightarrow 2H_{2\,(g)}+O_{2\,(g)}\,\,\Delta H_f^\circ=(-285.8\frac{kJ}{mol}*-2)=571.6\frac{kJ}{mol}$$ $$4HNO_{3\,(l)}\rightarrow 2N_2O_{5\,(g)}+2H_2O_{(l)}\,\,\Delta H_{rxn}=153.2\frac{kJ}{mol}$$ $$N_{2\,(g)}+3O_{2\,(g)}+H_{2\,(g)}\rightarrow 2HNO_{3\,(l)}\,\,\Delta H_{rxn}=-348.2\frac{kJ}{mol}$$

Next, we will multiply the bottom equation by 2. This is for a few reasons: 1) to have 2N_{2} molecules on the left like the main reaction 2) to cancel out the H_{2} and HNO_{3} molecules and 3) to cancel out 1 mol of O_{2} so we have 5O_{2} on the left like the main equation. $$2H_2O_{(l)}\rightarrow 2H_{2\,(g)}+O_{2\,(g)}\,\,\Delta H_f^\circ=571.6\frac{kJ}{mol}$$ $$4HNO_{3\,(l)}\rightarrow 2N_2O_{5\,(g)}+2H_2O_{(l)}\,\,\Delta H_{rxn}=153.2\frac{kJ}{mol}$$ $$2N_{2\,(g)}+6O_{2\,(g)}+2H_{2\,(g)}\rightarrow 4HNO_{3\,(l)}\,\,\Delta H_{rxn}=(-348.2\frac{kJ}{mol}*2)=-696.4\frac{kJ}{mol}$$

Lastly, we add up the three reaction enthalpies to get the net reaction enthalpy: $$2N_{2\,(g)}+5O_{2\,(g)}\rightarrow 2N_2O_{5\,(g)}\,\,\Delta H_{rxn}=(571.6\frac{kJ}{mol}+153.2\frac{kJ}{mol}-696.4\frac{kJ}{mol})=28.4\frac{kJ}{mol}$$

The main thing to remember when using Hess's law is to change a reaction's enthalpy value to match whatever changes are done to the chemical equation.

## Enthalpy Reaction - Key takeaways

**Enthalpy**is the total heat content of a system.- The
**enthalpy of reaction (ΔH**is the change in enthalpy due to a chemical reaction. The general formula is: $$\Delta H_{rxn}=H_{final}-H_{inital}=q\,\,\text{where q is heat}$$_{rxn}) - If ΔH
_{rxn}**>**0, the reaction is**endothermic**(the system pulls in heat from its surroundings)If ΔH

_{rxn}**<**0, the reaction is**exothermic**(the system releases heat into its surroundings) **The****standard****enthalpy of formation (ΔH**is the enthalpy change for the formation of 1 mol of a compound from its elements. These elements are in their standard state, which is the most stable form of the element at 1 atm and 298 K._{f}°)When we calculate If ΔH

_{rxn}, we use this formula: $$\Delta H_{rxn}=\sum mH_{f\,(products)}^\circ- \sum nH_{f\,(reactants)}^\circ$$**Hess's law**states that the net enthalpy of an overall reaction is equal to the sum of the enthalpies of its steps. The formula is: $$\Delta H_{net}=\sum \Delta H_{rxn}$$

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##### Frequently Asked Questions about Enthalpy of Reaction

how to calculate enthalpy of reaction?

To calculate the enthalpy of a reaction, you subtract the sum of the enthalpies of your reactants from the sum of the enthalpies of your products.

What is enthalpy of reaction?

The **enthalpy of reaction (ΔH _{rxn}) **is the change in enthalpy due to a chemical reaction.

How to calculate change in enthalpy of a reaction?

If a reaction reverse, then the sign of the enthalpy value is flipped. If more moles of reactants are added, then the enthalpy value should be multiplied by the new molar amount.

What is the standard enthalpy of reaction?

The standard enthalpy of a reaction is calculated using the standard enthalpies of formation for the reactants and products. The standard is measured for 1 mol of a compound at 1 atm and 298 K.

How is activation energy related to enthalpy of reaction?

Activation energy is the energy required for a reaction to proceed, while the enthalpy of a reaction is how much heat energy is released/absorbed during the reaction. Reactions with a negative enthalpy change will have a smaller activation energy than those with positive enthalpy changes. The activation energy is equal to the enthalpy change between the reactants and the activated complex.

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