ASChemistryOCRRevision Notes3. Periodic Table & EnergyEnthalpy ChangesHess' Law

Hess' Law (OCR AS Chemistry A): Revision Note

Exam code: H032

Philippa Platt

Written by: Philippa Platt

Reviewed by: Caroline Carroll

Updated on

Hess’ Law & Enthalpy Cycles

  • Hess’s Law states that:

"The total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place as long as the initial and final conditions are the same."

  • This means that whether the reaction takes place in one or two steps, the total enthalpy change of the reaction will still be the same

Enthalpy cycle: reactants A + B can reach product C directly or by indirect route 1 via F + G or indirect route 2 via X then Y + Z, and the enthalpy change is the same by every route.

According to Hess’ Law, the enthalpy change of the direct route, going from reactants (A+B) to product (C), is equal to the enthalpy change of the indirect routes

Examiner Tips and Tricks

You do not need to learn Hess's Law word for word as it is not a syllabus requirement, but you do need to understand the principle as it provides the foundation for all the problem solving in Chemical Energetics

  • Hess’ Law is used to calculate enthalpy changes which can’t be found experimentally using calorimetry, e.g.:

3C (s) + 4H2 (g) → C3H8(g)

  • ΔfH (propane) can’t be found experimentally as hydrogen and carbon don’t react under standard conditions

Calculating ΔHr from ΔHf using Hess’s Law energy cycles

  • You can see the relationships on the following diagram:

    Enthalpy cycle for finding the enthalpy change of reaction from enthalpies of formation: reactants go to products with delta Hr; the elements form the reactants with delta H1 (indirect route) and the products with delta H2 (direct route).

The enthalpy change from elements to products (direct route) is equal to the enthalpy change of elements forming reactants and then products (indirect route)

  • The products can be formed:

    • Directly from the elements = ΔH2
      OR
      Indirectly from the elements = ΔH1 + ΔHr

  • Equation

ΔH2 = ΔH1 + ΔHr

  • Therefore for energy to be conserved,

ΔHr = ΔH2 – ΔH1

Examiner Tips and Tricks

There are two rules that you can use to help with your calculations using enthalpy changes of formation and combustion

  • Using enthalpy changes of formation, ΔHf

    • ΔrH = ∑ΔHf(products) - ∑ΔHf(reactants)

  • Using enthalpy changes of combustion, ΔHc

    • ΔrH = ∑ΔHc(reactants) - ∑ΔHc(products)

Calculating ΔHr  from ΔHf using Hess’s Law energy cycles

  • Knowing the enthalpy change of formation, ΔHf, allows us to determine the overall enthalpy change of a reaction, ΔHr

  • To do this, we follow these steps:

    1. Write the equation for the reaction

    2. Write the elements with the correct number of moles and state symbols underneath

    3. Draw upwards pointing arrows to each compound

    4. Write the appropriate values on the arrows and multiply by the number of moles

    5. In a cycle, go from the reactants to the products, changing the sign of the value if the arrow points in the opposite direction

Worked Example

Calculating the enthalpy change of reaction

Use the information in the table to calculate the enthalpy change for this reaction:

NH4NO3 (s)  +  ½C (s)  →  N2 (g)  +  2H2O (g)  +  ½CO2 (g)

Substance

C (s)

N2 (g)

H2O(g)

CO2 (g)

NH4NO3 (s)

∆Hfө / kJ mol–1

0

0

–242

–394

–365

Answer:

 Step 1: Write the balanced equation

Worked example step 1: the balanced equation NH4NO3(s) + 1/2 C(s) -> N2(g) + 2H2O(g) + 1/2 CO2(g).

Step 2: Write the elements with the correct number of moles and state symbols underneath

Step 2: the same equation with the elements in their standard states written underneath: N2(g) + 1.5O2(g) + C(s) + 2H2(g).

Step 3: Draw upwards pointing arrows to each compound

Step 3: upward arrows from the elements N2(g) + 1.5O2(g) + C(s) + 2H2(g) to NH4NO3(s), to 2H2O(g) and to 1/2 CO2(g).

Step 4: Write the appropriate values on the arrows and multiply by the number of moles

Step 4: values on the arrows: formation of NH4NO3 is -365 kJ mol-1, formation of 2H2O is 2 x -242 = -484, formation of 1/2 CO2 is 0.5 x -394 = -197.

Step 5: In a cycle, go from the reactants to the products, changing the sign of the value if the arrow points in the opposite direction

Step 5: the cycle goes from the reactants down to the elements (+365, arrow reversed) and up to the products (-484 - 197), so delta Hr = +365 - 484 - 197 = -316 kJ mol-1.

The elements needed are N2 (g) + 1.5O2 (g) + C (s) + 2H2 (g). The arrows up from these elements are: to NH4NO3 (s), ΔfH = –365; to 2H2O (g), 2 x (–242) = –484; to ½CO2 (g), ½ x (–394) = –197. Going from the reactants to the products reverses the –365 arrow:

ΔHrө = +365 - 484 - 197 = -316 kJ mol-1

  • The sign on -365 needs reversing as the cycle goes in the opposite direction to the arrow pointing upwards

  • There is no need to draw arrows from the elements to carbon and nitrogen as ΔHfө is 0 for elements

Examiner Tips and Tricks

Keep your enthalpy values inside their own brackets so that you don't accidentally lose a minus sign

Calculating ΔHf from ΔHc using Hess’s Law energy cycles

  • It can be difficult to find the enthalpy change of formation of compounds experimentally

  • However, many enthalpy changes of combustion can be measured experimentally so these can be used to find the enthalpy of formation

  • To do this, we follow these steps:

    1. Write the equation for the formation of the compound

    2. Write the combustion products below the equation

    3. Draw downward pointing arrows from each substance to its combustion products

    4. Write values on the arrows and multiply by the number of moles

    5. In a cycle, go from the reactants to the products, changing the sign of the value if the arrow points in the opposite direction

Worked Example

Calculating the enthalpy change of formation

Using the data provided, calculate the standard enthalpy change of formation, ΔHf, of propanone.

3C (s)  +     3H2 (g)   +      ½ O2 (g)       →        CH3COCH3 (l)

Substance

C (s)

H2 (g)

CH3COCH3 (l)

∆HCө / kJ mol–1

-394

-286

-1821

Answer:

  • Step 1: Write the balanced equation

Step 1: the balanced equation for the formation of propanone, 3C(s) + 3H2(g) + 1/2 O2(g) -> CH3COCH3(l).

  • Step 2:Write the combustion products below the equation

Step 2: the combustion products 3CO2 + 3H2O written below the formation equation.

  • Step 3: Draw downward pointing arrows from each substance to its combustion product

Step 3: downward arrows from 3C(s), 3H2(g) and CH3COCH3(l) to the combustion products 3CO2 + 3H2O.

  • Step 4: Write the appropriate values on the arrows and multiply by the number of moles

Step 4: values on the arrows: combustion of 3C is 3 x -394 = -1182, combustion of 3H2 is 3 x -286 = -858, combustion of propanone is -1821 kJ mol-1.

  • Step 5: In a cycle, go from the reactants to the products, changing the sign of the value if the arrow points in the opposite direction

Step 5: the cycle goes from the elements to the combustion products (-1182 - 858) and back up to propanone (+1821), so delta Hf = -1182 - 858 + 1821 = -219 kJ mol-1.

The combustion products are 3CO2 + 3H2O. The arrows down from the formation equation are: 3C (s), 3 x (–394) = –1182; 3H2 (g), 3 x (–286) = –858; CH3COCH3 (l), –1821. Going from the elements to propanone reverses the –1821 arrow:

ΔHfө = -1182 - 858 + 1821 = -219 kJ mol-1

  • The sign on -1821 needs reversing as the cycle goes in the opposite direction to the arrow pointing to the combustion products

Examiner Tips and Tricks

Don't forget to make sure the number of atoms for each element is balanced when drawing your cycle.

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Philippa Platt

Author: Philippa Platt

Expertise: Chemistry Content Creator

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.