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Hess's Law of Constant Heat Summation Chemistry Tutorial

Key Concepts

  1. Rearrange each equation so that it is balanced, with the reactants and products on the correct side.
    Check that you have the correct states, enthalpy changes will be different for species in the solid, liquid and gas states.
  2. Assign each rearranged equation the correct ΔH value.
    Remember, if you need to multiply each species in the chemical equation by 2, then the enthalpy change must also be multiplied by 2.
    If you reverse a chemical equation, then the sign of the enthalpy term also changes.
  3. Add the rearrranged equations together to give the overall equation for the reaction.
  4. Add the ΔH values for these equations to calculate ΔH for the overall reaction.

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Understanding Hess's Law of Constant Heat Summation

If I want to visit my neighbour in the house next door, I have a couple of choices for the route I could take. I could walk out my front door and then either:
(i) walk along the path around to my neighbour's front door, if I'm not in a hurry.
(ii) jump over the fence and run to my neighbour's front door, if I am in a hurry.

Either route will provide the desired result of arriving at my neighbour's front door.
The first route uses less energy per footstep, but I have to take more footsteps.
The second route seems to use a lot more energy, but it requires far fewer footsteps, so I could really be using exactly the same amount of energy overall but in a shorter period of time.

Chemical reactions can be considered in just the same way.

Consider the combustion, or burning in oxygen gas, of coal (solid carbon, C(s)).
There are a couple of different routes, or pathways, that could be used to produce carbon dioxide gas as the product of the reaction:

(i) The direct route: combust the carbon using excess oxygen gas

C(s) + O2(g) → CO2(g)

This reaction releases 393.5 kJ of energy per mole of CO2(g) produced.

C(s) + O2(g) → CO2(g)     ΔH = -393.5 kJ mol-1

(ii) The less direct route: first convert the carbon to carbon monoxide, then react this carbon monoxide with oxygen to produce carbon dioxide:

  1. Combust carbon in a limited supply of oxygen gas so that carbon monoxide is produced:

    C(s) + ½O2(g) → CO(g)

    This reaction releases 110.5 kJ of energy per mole of CO(g) produced.

    C(s) + ½O2(g) → CO(g)     ΔH = -110.5 kJ mol-1

  2. Combust all of this carbon monoxide in additonal oxygen to produce carbon dioxide gas:

    CO(g) + ½O2(g) → CO2(g)

    This reaction releases 283 kJ of energy per mole of CO(g) consumed, or, per mole of CO2(g) produced.

    CO(g) + ½O2(g) → CO2(g)     ΔH = -283 kJ mol-1

The big question is whether these two different reaction pathways, direct and indirect, are the same or different in terms of the overall chemical equation and enthalpy change.

First we can check to see if the overall chemical equations for the two reaction pathways is the same by adding together the chemical reactions for step 1 and 2 in the indirect pathway.
I can represent this two-step process as in the diagram below:

C(s) + ½O2(g) CO(g)
    +
    ½O2(g)
   
    CO2(g)

The combustion of 1 mole of C(s) used ½ mole of O2(g) to produce 1 mole of CO(g), BUT, all of this CO(g) is then combusted using another ½ mole O2(g) to produce 1 mole of CO2(g).
Overall the reaction combusted 1 mole of C(s) using 2 × ½ = 1 mole of O2(g), to produce 1 - 1 = 0 moles of CO(g), and, 1 mole of CO2(g).
In other words:

C(s) + O2(g) → CO2(g)


In other words, this two-step indirect method results in the same overall chemical equation as the direct one-step approach.

But, which reaction pathway, direct or indirect, produces the most energy per mole of CO2(g) produced?

  1. Direct route produces 393.5 kJ of energy per mole of CO2(g) produced:

    C(s) + O2(g) → CO2(g)     ΔH = -393.5 kJ mol-1

  2. Indirect route:

    First reaction produces 110.5 kJ of energy per mole of C(s) combusted.

    C(s) + ½O2(g) → CO(g)     ΔH = -110.5 kJ mol-1

    Second reaction produces 283 kJ of energy per mole of CO(g) combusted.

    CO(g) + ½O2(g) → CO2(g)     ΔH = -283 kJ mol-1

    Total amount of energy released by indirect pathway = 110.5 + 283 = 393.5 kJ mol-1

    Overall reaction: C(s) + O2(g) → CO2(g)     ΔH = -393.5 kJ mol-1

We can now see that the combustion of carbon via the direct 1-step pathway, or, via the indirect 2-step pathway, results in the same CO2(g) product, AND, releases exactly the same amount of energy per mole of CO2(g) produced.

This is what Hess's Law of Constant Heat Summation means for this reaction, the enthalpy change for the reaction in which carbon and oxygen gas react to produce carbon dioxide gas is the same, or constant, for the two different pathways we could use.

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Representing and Using Hess's Law of Constant Heat Summation

Chemists often use a diagram to show the different pathways for the same overall chemical reaction, and, the enthalpy changes associated with each pathway.

The diagram below represents the combustion of carbon in oxygen to produce carbon dioxide by the two pathways described in the previous section, the direct 1-step pathway and the indirect 2-step pathway:

Direct path:

C(s) + O2(g) → CO2(g)

393 kJ mol-1 of energy released, energy is a product of the reaction:

C(s) + O2(g) → 393 kJ mol-1 + CO2(g)

The reactants, C(s) and O2(g), are of higher enthalpy than the product, CO2(g)

C(s) and O2(g) are placed at the top of the diagram, CO2(g) is placed at the bottom of the diagram.

The arrow shows which direction the reaction is going (from C + O2 to CO2(g))

schematic diagram for the combustion of carbon to form carbon dioxide Indirect path:

(i) C(s) + ½O2(g) → CO(g)
110.5 kJ mol-1 of energy is released, energy is a product.
C(s) + ½O2(g) → 110.5 kJ mol-1 + CO(g)
C(s) and O2(g) have higher enthalpy than CO(g) and are placed at the top of the diagram.

(ii) CO(g) + ½O2(g) → CO2(g)
283 kJ mol-1 of energy is released, energy is a product.
CO(g) + ½O2(g) → 283 kJ mol-1 + CO2(g)
CO(g) and O2(g) have higher enthalpy than CO2(g) so are placed above CO2(g) in the diagram.

Usually, we treat Hess's Law of Constant Heat Summation like a set of simultaneous equations in mathematics when we use it to calculate the enthalpy change for an "overall chemical reaction" resulting from multiple reaction pathways.

For the indirect 2-step pathway for the production of CO2(g) as shown in the diagram above:

First we write each equation, along with its enthalpy change, on a separate line:

Reaction 1: C(s) + ½O2(g) CO(g)     ΔH = -110.5 kJ mol-1
Reaction 2: CO(g) + ½O2(g) CO2(g)     ΔH = -283 kJ mol-1

Next we draw a line under this set of equations:

Reaction 1: C(s) + ½O2(g) CO(g)     ΔH = -110.5 kJ mol-1
Reaction 2: CO(g) + ½O2(g) CO2(g)     ΔH = -283 kJ mol-1
 

Now we look for substances that occur as a reactant in one equation but as a product in the other equation, for example, CO(g) is a product in reaction 1 but is a reactant in reaction 2, so these "cancel out" and we draw a line through them:

Reaction 1: C(s) + ½O2(g) CO(g)     ΔH = -110.5 kJ mol-1
Reaction 2: CO(g) + ½O2(g) CO2(g)     ΔH = -283 kJ mol-1
 

Now we can add together all the substances on the "reactant" side of the equation and all the substances on the "product" side of the equation.
Note that ½O2(g) + ½O2(g) = 1O2(g) on the reactant side of the equation!

Reaction 1: C(s) + ½O2(g) CO(g)     ΔH = -110.5 kJ mol-1
Reaction 2: CO(g) + ½O2(g) CO2(g)     ΔH = -283 kJ mol-1
 
Overall reaction: C(s) + O2(g) CO2(g)    

And, we can add together the enthalpy change terms for each chemical reaction as well...

Reaction 1: C(s) + ½O2(g) CO(g)     ΔH = -110.5 kJ mol-1
Reaction 2: CO(g) + ½O2(g) CO2(g)     ΔH = -283 kJ mol-1
 
Overall reaction: C(s) + O2(g) CO2(g)     ΔH = -110.5 + -283
        = -393.5 kJ mol-1

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Worked Example Using Hess's Law to Calculate the Enthalpy Change of a Reaction

Question: Calculate ΔH for the reaction:
    NH3(g) + HCl(g)NH4Cl(s)
Given that:
    ½N2(g) + 1½H2(g)NH3(g)     ΔH = -46.1 kJ mol-1
    ½H2(g) + ½Cl2(g)HCl(g)     ΔH = -92.3 kJ mol-1
    ½N2(g) + 2H2(g) + ½Cl2(g)NH4Cl(s)     ΔH = -314.4 kJ mol-1

Solution:

(Based on the StoPGoPS approach to problem solving.)

  1. What is the question asking you to do?

    Calculate the enthalpy change for the reaction
    ΔH(overall reaction) = ? kJ mol-1

  2. What data (information) have you been given in the question?

    Extract the data from the question:

    (i) Final overall equation for reaction:
        NH3(g) + HCl(g)NH4Cl(s)

    (ii) Reactions to be used to construct the pathway from reactants to products:
        ½N2(g) + 1½H2(g)NH3(g)     ΔH = -46.1 kJ mol-1
        ½H2(g) + ½Cl2(g)HCl(g)     ΔH = -92.3 kJ mol-1
        ½N2(g) + 2H2(g) + ½Cl2(g)NH4Cl(s)     ΔH = -314.4 kJ mol-1

  3. What is the relationship between what you know and what you need to find out?
    (i) Rearrange each balanced equation with reactants and products on the correct side, that is, NH3(g) and HCl(g) on the left, NH4Cl(s) on the right hand side.

    (ii) Assign correct ΔH values to each equation.

    (iii) Add the rearranged equations together.

    (iv) Add the ΔH values together.

  4. Perform the calculations
    (i) Rearrange each balanced equation with reactants and products on the correct side, that is, NH3(g) and HCl(g) on the left, NH4Cl(s) on the right hand side.

    NH3(g) ½N2(g) + 1½H2(g)  
    HCl(g) ½H2(g) + ½Cl2(g)  
    ½N2(g) + 2H2(g) + ½Cl2(g) NH4Cl(s)  

    (ii) Assign correct ΔH values to each equation.

    NH3(g) ½N2(g) + 1½H2(g)     ΔH = +46.1 kJ mol-1
    HCl(g) ½H2(g) + ½Cl2(g)     ΔH = +92.3 kJ mol-1
    ½N2(g) + 2H2(g) + ½Cl2(g) NH4Cl(s)     ΔH = -314.4 kJ mol-1

    (iii) Add the rearranged equations together.

    NH3(g) ½N2(g) + 1½H2(g)     ΔH = +46.1 kJ mol-1
    HCl(g) ½H2(g) + ½Cl2(g)     ΔH = +92.3 kJ mol-1
    ½N2(g) + 2H2(g) + ½Cl2(g) NH4Cl(s)     ΔH = -314.4 kJ mol-1

    NH3(g) + HCl(g) NH4Cl(s)  

    (iv) Add the ΔH values together.

    NH3(g) ½N2(g) + 1½H2(g)     ΔH = +46.1 kJ mol-1
    HCl(g) ½H2(g) + ½Cl2(g)     ΔH = +92.3 kJ mol-1
    ½N2(g) + 2H2(g) + ½Cl2(g) NH4Cl(s)     ΔH = -314.4 kJ mol-1

    NH3(g) + HCl(g) NH4Cl(s) ΔH = +46.1 + +92.3 + -314.4
        = -176 kJ mol-1

  5. Is your answer plausible?
    First, NH3 produces N2(g) and H2(g), which is the reverse of the first reaction given so:
    NH3(g) → ½N2(g) + 1½H2(g) ΔH = -(-46.1) = +46.1 kJ/mol

    Then HCl(g) must produce H2(g) and Cl2(g), the reverse of the reaction given, so:
    HCl(g) → ½H2(g) + ½Cl2(g) ΔH = -(-92.3) = +92.3 kJ/mol

    Adding these two equations together we get:
    NH3(g) + HCl(g) → ½N2(g) + [ 1½H2(g) + ½H2(g) ] + ½Cl2(g)
    NH3(g) + HCl(g) → ½N2(g) + 2H2(g) + ½Cl2(g) ΔH = +46.1 + 92.3 = +138.4 kJ/mol

    We can add this equation above directly to the reaction given for the formation of NH4Cl:
    NH3(g) + HCl(g) ½N2(g) + 2H2(g) + ½Cl2(g)   ΔH = +138.4 kJ/mol
    ½N2(g) + 2H2(g) + ½Cl2(g) → NH4Cl(s)   ΔH = -314.4 kJ/mol
    NH3(g) + HCl(g) → NH4Cl(s)   ΔH = +138.4 + -314.4 = -176 kJ/mol

    Since this value for ΔH is the same as the one calculated above, we are reasonably confident that our answer is correct.

  6. State your solution to the problem "enthalpy change for overall reaction":

    ΔH = -176 kJ mol-1

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