## Core Concepts

So what is Hess’s Law? In this tutorial, you will be introduced to **Hess’s Law**, as well as the equation that goes along with this concept. In addition, you will further master this concept by going through some example problems.

## Topics Covered in Other Articles

- What is a Chemical Reaction
- How to Balance Equations
- Gibbs Free Energy
- Specific Heat
- Tyndall Effect
- Net Ionic Equation
- Thermochemistry Introduction

## What is Hess’s Law?

Russian Chemist and Physicist Germain Hess developed the concepts of thermochemistry and physical chemistry. He introduced the concept known as Hess’s Law of Constant Heat of Summation or Hess’s Law for short.

This law has to do with net enthalpy in a reaction. Overall, it states that the total enthalpy change of a reaction is the sum of all the changes, no matter the number of steps or stages in the reaction (i.e. net enthalpy and the number of steps in a reaction are independent of each other). The ideas of this law are seen throughout science, such as in the principle of conservation of energy, or the first law of thermodynamics, and the statement that enthalpy is a state function.

There are some requirements that the reaction has to follow in order to use Hess’s Law. For example, if there are multiple steps to the reactions, each equation must be correctly balanced. Also, all the steps of the reaction must start and end at constant temperatures and pressures in order to keep reaction conditions constant.

## Hess’s Law Equation

To put this definition into mathematical terms, here is the Hess’s Law equation:

**∆H _{net}=∑∆H_{r}**

net enthalpy change = ∆H_{net}

the sum of all enthalpy change steps = ∆H_{r}

### Enthalpy Change

Enthalpy change, ∆H, can be defined as the amount of heat absorbed or released during a reaction. In each individual step of a multistep reaction, there is a beginning and end enthalpy value- the difference between them being the enthalpy change. This value can be either negative if the heat was absorbed, or positive if the heat was released. If you add up all the enthalpy changes of each reaction step(∆H_{r}), you have net enthalpy change, which is found by finding the difference between the final product enthalpy and the beginning reactant enthalpy (∆H_{net}). That is Hess’s Law!

## Hess’s Law Example Problems

Now that we understand the concept and equation of Hess’s Law, let’s expand on our knowledge with practice problems. These word problems may ask for some manipulation of reactions (i.e. changing the direction of equation, multiplication, division), but the general idea is the same for all Hess’s Law problems. Let’s go through some examples below!

### Example Problem 1

Find the net enthalpy change (∆H_{net}) of the reaction below, given the reaction steps and their ∆H values.

Overall reaction: N_{2}H_{4(l)} +H_{2(g)} → 2NH3 (g)

(i) N_{2}H_{4(l)} + CH_{4}O_{(l)} → CH_{2}O_{(g)} + N_{2(g)} + 3H_{2(g)} ∆H= – 37kJ/mol

(ii) N_{2(g)} + 3H_{2(g)} → 2NH_{3(g)} ∆H= -46kJ/mol

(iii) CH_{4}O_{(l)} → CH_{2}O_{(g)} + H_{2(g) } ∆H= -65kJ/mol

#### 1. Make sure the steps are (a) balanced (b) in the right direction and (c) result in the overall reaction.

To make sure all the steps given are necessary for the overall reaction, add the equations and cross off repeated compounds to make a overall equation.

However, if we do this step with the reactions as they are, we do not end up with the correct reaction because we have compounds on the wrong side as well as extra compounds. Because of this, we can analyze if one, or more than one, of the steps, go in the opposite direction.

Since reaction (i) is the only one with N2H4(l), which is a reactant in the overall equation, it is assumed that it is going in the correct direction. Next, reaction (ii) has the product 2NH_{3(g)} on the right side, so that equation remains the same as well. In the above attempt to find the overall equation, the hydrogen gas from equations (i) and (ii) cancel each other out, meaning the hydrogen gas from reaction (iii) is the only one left to make it to the overall equation, which belongs on the left. Because of this, we can flip the equation reactants and products to go the backward direction; however, because the reaction is going in the opposite way, the enthalpy also becomes the “opposite”. **If you change the direction of a reaction, the reciprocal of the enthalpy becomes the new enthalpy.**

The “new” equation steps look like this:

(i) N_{2}H_{4(l)} + CH_{4}O_{(l)} → CH_{2}O_{(g)} + N_{2(g)} + 3H_{2(g)} ∆H= – 37kJ/mol

(ii) N_{2(g)} + 3H_{2(g)} → 2NH_{3(g)} ∆H= -46kJ/mol**(iii) CH _{2}O_{(g)} + H_{2(g)} → CH_{4}O_{(l)} ∆H= +65kJ/mol**

With reaction (iii) switched the method of adding all the equations results in the correct overall reaction:

N_{2}H_{4(l)} +H_{2(g)} → 2NH3 (g)

#### 2. Find Net Enthalpy Change

Now that we have the official enthalpy values, we can use Hess’s Law equation to solve.

∆H_{net}=∑∆H_{r} = (-37 kJ/mol) + (-46 kJ/mol) + 65 kJ/mol = **-18kJ/mol**

### Example Problem 2

Find the net enthalpy change (∆H_{net}) of the reaction below, given the reaction steps and their ∆H values.

Overall Reaction: CS_{2(l)} + 3O_{2(g)} → CO_{2(g)} + 2SO_{2(g)}

(i) C_{(s)} + O_{2(g)} → CO_{2(g)} ∆H= -395 kJ/mol

(ii) S_{(s)} + O_{2(g)} → SO_{2(g)} ∆H= -295 kJ/mol

(iii) C_{(s)} + 2S_{(s)} → CS_{2(l)} ∆H= +90 kJ/mol

#### 1. Make sure the steps are (a) balanced (b) in the right direction and (c) result in the overall reaction.

First, using the same methods as above, we check if all the step reactions are going in the correct direction to make the correct reaction. Reaction (i) has the desired CO_{2(g)} product, which means it can remain unchanged. Reaction (iii) has CS_{2(l)} as a product, but is a desired reactant in the overall reaction; therefore, we flip this reaction and use the reciprocal ∆H value.

As for reaction (ii), the direction is correct because O_{2(g)} as a reactant and SO_{2(g)} as a product are both seen in the desired reaction; however, when adding the equations together, one O_{2(g)} and one SO_{2(g)} are missing (there is also an extra S_{(s)} that needs to be canceled out). This can be fixed by multiplying reaction (ii) by a factor of 2. **If you multiply(or divide) this, you also have to multiply (or divide) the ∆H value by the same coefficient.**

(i) C_{(s)} + O_{2(g)} → CO_{2(g)} ∆H= -395 kJ/mol**(ii) 2S _{(s)} + 2O_{2(g)} → 2SO_{2(g)} ∆H= -590 kJ/mol**

(iii) CS_{2(l)}→ C_{(s)} + 2S_{(s)} ∆H= -90 kJ/mol

With reactions (ii) and (iii) manipulated, the method of adding all the equations results in the correct overall reaction:

CS_{2(l)} + 3O_{2(g)} → CO_{2(g)} + 2SO_{2(g)}

#### 2. Find Net Enthalpy Change

Now that we have the official enthalpy values, we can use Hess’s Law equation to solve.

∆H_{net}=∑∆H_{r} = (-395 kJ/mol) + (-590 kJ/mol) + (-90 kJ/mol) = **-1075 kJ/mol**