Hess Law Practice Problems With Answers

Introduction

If you are looking for Hess Law practice problems with answers, you have come to the right place. This article will guide you through the fundamental concepts, step‑by‑step methods, and a series of realistic practice questions that you can solve on your own. By the end, you will feel confident applying Hess’s Law to any thermochemical equation, and you will have a ready‑made set of solutions to check your work.

Understanding Hess’s Law

Hess’s Law states that the total enthalpy change for a chemical reaction is the same regardless of the pathway taken, provided the initial and final states are identical. Put another way, enthalpy (H) is a state function—its change depends only on the start and end points, not on the steps in between. This principle is essential for calculating heat changes in reactions that are difficult to measure directly.

Some disagree here. Fair enough.

Key points:

• State function – italic term indicating that enthalpy is path‑independent.
• Additivity – you can combine two or more reactions to obtain the target reaction, and the enthalpy changes simply add up.

Steps to Solve Hess Law Practice Problems

1. Write the target equation – Identify the overall reaction you need to find the enthalpy for.
2. List the given thermochemical equations – Note each reaction’s ΔH value.
3. Manipulate the equations – Reverse a reaction (change sign of ΔH) or multiply it by a coefficient (multiply ΔH accordingly).
4. Add the equations – Combine them algebraically so that all intermediate species cancel, leaving only the target reaction.
5. Sum the ΔH values – The algebraic sum gives the enthalpy change for the target reaction.

Tip: Use a table to keep track of each manipulated equation and its corresponding ΔH; this prevents sign errors.

Sample Practice Problems with Answers

Below are five practice problems that illustrate different manipulations of Hess’s Law. Each problem is followed by a detailed solution.

Problem 1

Given:

1. C(s) + O₂(g) → CO₂(g)  ΔH₁ = –393.5 kJ
2. H₂(g) + ½ O₂(g) → H₂O(l)  ΔH₂ = –285.8 kJ

Find: ΔH for the reaction:

C(s) + O₂(g) → CO₂(g)  (ΔH = ?)

Solution:

The target reaction is identical to equation 1, so no manipulation is needed.

ΔH = ΔH₁ = –393.5 kJ

Problem 2

Given:

1. N₂(g) + O₂(g) → 2 NO(g)  ΔH₁ = +180.5 kJ
2. 2 NO(g) + O₂(g) → 2 NO₂(g)  ΔH₂ = –114.0 kJ

Find: ΔH for:

N₂(g) + 3/2 O₂(g) → 2 NO₂(g)

Solution:

Add equation 1 and equation 2:

(N₂ + O₂ → 2 NO)  ΔH₁ = +180.5 kJ
(2 NO + O₂ → 2 NO₂)  ΔH₂ = –114.0 kJ

Sum: N₂ + 3/2 O₂ → 2 NO₂  ΔH = +180.5 kJ + (–114.0 kJ) = **+66.

Problem 3

Given:

1. C(s) + O₂(g) → CO₂(g)  ΔH₁ = –393.5 kJ
2. C(s) + ½ O₂(g) → CO(g)  ΔH₂ = –110.5 kJ

Find: ΔH for:

CO(g) + ½ O₂(g) → CO₂(g)

Solution:

Reverse equation 2 (sign changes):

CO(g) + ½ O₂(g) → C(s)  ΔH = +110.5 kJ

Now add equation 1:

C(s) + O₂(g) → CO₂(g)  ΔH₁ = –393.5 kJ

Sum: CO(g) + ½ O₂(g) → CO₂(g)  ΔH = +110.5 kJ + (–393.5 kJ) = **–283 Still holds up..

Problem 4

Given:

1. 2 H₂(g) + O₂(g) → 2 H₂O(l)  ΔH₁ = –571.6 kJ
2. H₂(g) + ½ O₂(g) → H₂O(g)  ΔH₂ = –241.8 kJ

Find: ΔH for:

2 H₂(g) + O₂(g) → 2 H₂O(g)

Solution:

Multiply equation 2 by 2 to match the stoichiometry of equation 1:

2 H₂(g) + O₂(g) → 2 H₂O(g)  ΔH = 2 × (–241.8 kJ) = –483.6 kJ

Since this is exactly the target reaction, ΔH = –483.6 kJ.

Problem 5

Given:

1. Mg(s) + ½ O₂(g) → MgO(s)  ΔH₁ = –601.6 kJ
2. Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)  ΔH₂ = –878.