Chapter 7. Energy and Chemistry

# 43 Hess’s Law

Learning Objectives

- Learn how to combine chemical equations and their enthalpy changes.

Now that we understand that chemical reactions occur with a simultaneous change in energy, we can apply the concept more broadly. To start, remember that some chemical reactions are rather difficult to perform. For example, consider the combustion of carbon to make carbon monoxide:

In reality, this is extremely difficult to do; given the opportunity, carbon will react to make another compound, carbon dioxide:

Is there a way around this? Yes. It comes from the understanding that chemical equations can be treated like algebraic equations, with the arrow acting like the equals sign. Like algebraic equations, chemical equations can be combined, and if the same substance appears on both sides of the arrow, it can be cancelled out (much like a spectator ion in ionic equations). For example, consider these two reactions:

If we added these two equations by combining all the reactants together and all the products together, we would get:

We note that 2CO_{2}(g) appears on both sides of the arrow, so they cancel:

We also note that there are 2 mol of O_{2} on the reactant side, and 1 mol of O_{2} on the product side. We can cancel 1 mol of O_{2} from both sides:

What do we have left?

This is the reaction we are looking for! So by algebraically combining chemical equations, we can generate new chemical equations that may not be feasible to perform.

What about the enthalpy changes? states that when chemical equations are combined algebraically, their enthalpies can be combined in exactly the same way. Two corollaries immediately present themselves:

- If a chemical reaction is reversed, the sign on Δ
*H*is changed. - If a multiple of a chemical reaction is taken, the same multiple of the Δ
*H*is taken as well.

What are the equations being combined? The first chemical equation is the combustion of C, which produces CO_{2}:

This reaction is two times the reaction to make CO_{2} from C(s) and O_{2}(g), whose enthalpy change is known:

According to the first corollary, the first reaction has an energy change of two times −393.5 kJ, or −787.0 kJ:

The second reaction in the combination is related to the combustion of CO(g):

The second reaction in our combination is the *reverse* of the combustion of CO. When we reverse the reaction, we change the sign on the Δ*H*:

Now that we have identified the enthalpy changes of the two component chemical equations, we can combine the Δ*H* values and add them:

Hess’s law is very powerful. It allows us to combine equations to generate new chemical reactions whose enthalpy changes can be calculated, rather than directly measured.

Example 7.14

# Problem

Determine the enthalpy change of:

from these reactions:

## Solution

We will start by writing chemical reactions that put the correct number of moles of the correct substance on the proper side. For example, our desired reaction has C_{2}H_{4} as a reactant, and only one reaction from our data has C_{2}H_{4}. However, it has C_{2}H_{4} as a product. To make it a reactant, we need to reverse the reaction, changing the sign on the Δ*H*:

We need CO_{2} and H_{2}O as products. The second reaction has them on the proper side, so let us include one of these reactions (with the hope that the coefficients will work out when all our reactions are added):

We note that we now have 4 mol of CO_{2} as products; we need to get rid of 2 mol of CO_{2}. The last reaction has 2CO_{2} as a reactant. Let us use it as written:

We combine these three reactions, modified as stated:

What cancels? 2C_{2}H_{2}, H_{2}, 2O_{2}, and 2CO_{2}. What is left is:

Which is the reaction we are looking for. The Δ*H* of this reaction is the sum of the three Δ*H* values:

# Test Yourself

Given the following thermochemical equations:

Determine Δ*H* for:

## Answer

+136 kJ

Key Takeaways

- Hess’s law allows us to combine reactions algebraically and then combine their enthalpy changes the same way.

Exercises

# Questions

- Define Hess’s law.
- What does Hess’s law require us to do to the Δ
*H*of a thermochemical equation if we reverse the equation? - If the Δ
*H*for C_{2}H_{4}+ H_{2}→ C_{2}H_{6}is −65.6 kJ, what is the Δ*H*for the reaction C_{2}H_{6}→ C_{2}H_{4}+ H_{2}? - If the Δ
*H*for 2Na + Cl_{2}→ 2NaCl is −772 kJ, what is the Δ*H*for the reaction 2NaCl → 2Na + Cl_{2}? - If the Δ
*H*for C_{2}H_{4}+ H_{2}→ C_{2}H_{6}is −65.6 kJ, what is the Δ*H*for the reaction 2C_{2}H_{4}+ 2H_{2}→ 2C_{2}H_{6}? - If the Δ
*H*for 2C_{2}H_{6}+ 7O_{2}→ 4CO_{2}+ 6H_{2}O is −2,650 kJ, what is the Δ*H*for the reaction 6C_{2}H_{6}+ 21O_{2}→ 12CO_{2}+ 18H_{2}O? - The Δ
*H*for C_{2}H_{4}+ H_{2}O → C_{2}H_{5}OH is −44 kJ. What is the Δ*H*for the reaction 2C_{2}H_{5}OH → 2C_{2}H_{4}+ 2H_{2}O? - The Δ
*H*for N_{2}+ O_{2}→ 2NO is 181 kJ. What is the Δ*H*for the reaction ? - Determine the Δ
*H*for the reaction Cu + Cl_{2}→ CuCl_{2}given these data:

- Determine Δ
*H*for the reaction 2CH_{4}→ 2H_{2}+ C_{2}H_{4}given these data:

- Determine Δ
*H*for the reaction Fe_{2}(SO_{4})_{3}→ Fe_{2}O_{3}+ 3SO_{3}given these data:

- Determine Δ
*H*for the reaction CaCO_{3}→ CaO + CO_{2}given these data:

# Answers

- If chemical equations are combined, their energy changes are also combined.

- Δ
*H*= 65.6 kJ

- Δ
*H*= −131.2 kJ

- Δ
*H*= 88 kJ

- Δ
*H*= −220 kJ

- Δ
*H*= 570 kJ

When chemical equations are combined algebraically, their enthalpies can be combined in exactly the same way.