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Standard Heat of Formation (Standard Enthalpy of Formation) Chemistry Tutorial

Key Concepts

  1. Write the balanced chemical equation for the chemical reaction

    Remember to include the state (solid, liquid, gas, or aqueous) for each reactant and product.

  2. Write the general equation for calculating the standard enthalpy (heat) of reaction:

    ΔHo(reaction) = ΣHof(products) - ΣHof(reactants)

  3. Substitute the values for the standard enthalpy (heat) of formation of each product and reactant into the equation.

    Remember, if there are 2 moles of a reactant or product, you will need to multiply the enthalpy term by 2, if molar enthalpies (heats) of formation are used.

  4. Solve the equation to find the standard enthalpy of reaction (standard heat of reaction).

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Standard Enthalpy of Formation or Standard Heat of Formation

The standard heat of formation (standard enthalpy of formation) of a compound is defined as the enthalpy change for the reaction in which elements in their standard states produce products.

At 25°C and 1 atm (101.3 kPa), the standard state of any element is solid with the following exceptions:

liquids gases
monatomic gases
(Group 18 or Noble Gases)
diatomic gases
bromine, Br2(l)

mercury, Hg(l)

helium, Heg

neon, Ne(g)

argon, Ar(g)

krypton, Kr(g)

xenon, Xe(g)

radon, Rn(g)

hydrogen, H2(g)

nitrogen, N2(g)

oxygen, O2(g)

fluorine, F2(g)

chlorine, Cl2(g)

Consider a water molecule, H2O.
A molecule of water contains the elements hydrogen (H) and oxygen (O).

The standard heat of formation of liquid water would be defined as the enthalpy change when the elements hydrogen and oxygen in their standard states to produce liquid water.
If the conditions set are 25°C and 101.3 kPa (1 atm), then the standard states for reactants and products are:

The chemical reaction for the standard heat of formation per mole of liquid water (standard enthalpy of formation of liquid water) is:

H2(g) + ½O2(g) → H2O(l)

If you looked up the standard enthalpy of formation of liquid water in tables (at 25°C and 1 atm), the value would be given as:

ΔHfo = -285.8 kJ mol-1

This means that when molecular hydrogen gas reacts with molecular oxygen gas, 285.8 kJ of energy will be released for every mole of liquid water that is produced.

If 10 moles of liquid water was produced from molecular hydrogen gas and molecular oxygen gas, then 10 × 285.8 = 2858 kJ of energy would be released.

Similarly, if 0.1 moles of liquid water was produced from molecular hydrogen gas and molecular oxygen gas, then 0.1 × 285.8 = 28.58 kJ of energy would be released.

The values for the standard enthalpy of formation for a number of different compounds at 25°C is given in the table below:

Compound ΔHfo
kJ mol-1
Compound ΔHfo
kJ mol-1
Compound ΔHfo
kJ mol-1
HCl(g) -92.3 NH3(g) -46.1 CH4(g) -74.8
H2O(g) -241.8 NH4Cl(s) -314.4 CH3OH(l) -239.0
H2O(l) -285.8 NaCl(s) -412.1 C2H2(g) +226.8
H2O2(l) -187.6 Na2O(s) -415.9 C2H4(g) +52.3
H2S(g) -20.6 O3(g) +143 C2H6(g) -84.6
H2SO4(l) -814 SO2(g) -296.8 CO(g) -110.5
    SO3(g) -395.7 CO2(g) -393.5

Note that the standard heat of formation (enthalpy of formation) of some compounds is positive and for others it is negative:

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Calculating Heat of Reaction Using Standard Heat of Formation Data

Consider the reaction in which gaseous hydrogen chloride (HCl(g)) reacts with gaseous ammonia (NH3(g)) to produce solid ammonium chloride (NH4Cl(s)) at 25°C.

NH3(g) + HCl(g) → NH4Cl(s)

From the table of values for Standard Enthalpy of Formation at 25°C given in the previous section, we find that the standard enthalpy of formation of ammonium chloride (NH4Cl(s)) is -314.4 kJ mol-1.
That is, the formation of 1 mole of solid ammonium chloride (NH4Cl(s)) from the elements nitrogen, hydrogen and chlorine in their standard states releases 314.4 kJ of energy.
Therefore we can write a chemical equation to represent this reaction as shown below:

½N2(g) + 2H2(g) + ½Cl2(g) → NH4Cl(s)     ΔHfo = -314.4 kJ mol-1

But where will these elements come from in order to react?
They can come from the reactant molecules breaking apart.

Hydrogen chloride molecules could break apart to provide molecules of hydrogen and chlorine gas according to the following chemical equation:

HCl(g) → ½H2(g) + ½Cl2(g)

But how much energy will be absorbed or released to do this?

From the table of values for Standard Enthalpy of Formation at 25°C we find that the enthalpy of formation of HCl(g) is -92.3 kJ mol-1.
We can represent this formation reaction as:

½H2(g) + ½Cl2(g) → HCl(g)     ΔHfo = -92.3 kJ mol-1

This reaction is just the reverse of the one we want!
So we can reverse the reaction, AND, reverse the sign of the enthalpy change as well!

HCl(g) → ½H2(g) + ½Cl2(g)     ΔHo = +92.3 kJ mol-1

Similarly, ammonia gas (NH3(g)) could break apart to provide the nitrogen and hydrogen we need for the overall reaction to occur:

NH3(g) → ½N2(g) + 3/2H2(g)

This reaction is the reverse of the heat of formation (enthalpy of formation reaction) shown below:

½N2(g) + 3/2H2(g) → NH3(g)

From the table of values for Standard Enthalpy of Formation at 25°C we find that the enthalpy of formation of NH3(g) is -46.1 kJ mol-1.
That is:

½N2(g) + 3/2H2(g) → NH3(g)     ΔHfo = -46.1 kJ mol-1

So, the reaction to break apart ammonia molecules into hydrogen gas and ammonia gas is the reverse of this equation, AND, we must remember to reverse the sign of the enthalpy change as well!

NH3(g) → ½N2(g) + 3/2H2(g)     ΔHo = +46.1 kJ mol-1

The enthalpy change required to produce the elements hydrogen, nitrogen and chlorine in their standard states is the sum of the enthalpy change for breaking apart hydrogen chloride molecules and for breaking apart ammonia molecules:

HCl(g) ½H2(g) + ½Cl2(g)   ΔHo = +92.3 kJ mol-1
NH3(g) ½N2(g) + 3/2H2(g)   ΔHo = +46.1 kJ mol-1

HCl(g) + NH3(g) 2H2(g) + ½Cl2(g) + ½N2(g)   ΔHo = 92.3 + 46.1
  = +138.4 kJ mol-1

Now we can use this H2(g), Cl2(g) and N2(g) to produce NH4Cl(s).
Recall from the beginning of this section that this reaction, the formation of NH4Cl(s) from its elements in their standard states, releases 314.4 kJ mol-1 of energy.
So now we can add together two chemical equations and their associated enthalpy terms; 1 equation for the breaking apart of reactant molecules into elements, and, 1 equation for the elements coming together to form product molecules.
This is shown below:

HCl(g) + NH3(g) 2H2(g) + ½Cl2(g) + ½N2(g)   ΔHo = +138.4 kJ mol-1
2H2(g) + ½Cl2(g) + ½N2(g) NH4Cl(s)   ΔHo = -314.4 kJ mol-1

HCl(g) + NH3(g) NH4Cl(s)   ΔHo = +138.4 + -314.4
    = -176.0 kJ mol-1

Look at what we have done here:
(a) We found the value of the standard heat of formation of the product
      ΔHfo(product)
(b) We have added together the standard heat of formation of each reactant molecule, AND, reversed the sign:
      ΔHo(reactants) = -ΣΔHfo(reactants)
(c) We added the two values together to find the enthalpy change for the reaction, ΔHo(reaction):
      ΔHo(reaction) = ΔHfo(product) + -ΣΔHfo(reactants)

This is a very quick way to use standard heat of formation data to calculate the enthalpy change of a chemical reaction:

ΔHo(reaction) = ΣΔHfo(products) -ΣΔHfo(reactants)

Using the formation of solid ammonium chloride (NH4Cl(s)) from the reactants hydrogen chloride gas (HCl(g)) and ammonia gas (NH3(g)) as an example of the application of this equation:

  reactants products
Chemical
Equation:
HCl(g) + NH3(g) NH4Cl(s)
ΔHfo
kJ mol-1:
-92.3   -46.1   -314.4
ΔHo(reaction)
kJ mol-1
= -ΣΔHfo(reactants) + ΣΔHfo(products)
= -(-92.3 + -46.1) + -314.4
  = -(-138.4) + -314.4
  = +138.4 + -314.4
  = -176.0

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Worked Example of Calculating Heat of Reaction from Standard Heat of Formation Values

Question: Calculate the standard enthalpy (heat) of reaction for the oxidation of ammonia gas to produce nitrogen dioxide gas and water vapour given the following standard enthalpies (heats) of formation:
    NH3(g)     ΔHfo = -46 kJ mol-1
    NO2(g)     ΔHfo = +34 kJ mol-1
    H2O(g)     ΔHfo = -242 kJ mol-1
    O2(g)       ΔHfo = 0 kJ mol-1 (by definition)

Solution:

(Based on the StoPGoPS approach to problem solving.)

  1. What is the question asking you to do?

    Calculate the enthalpy change for the oxidation of ammonia
    ΔHo(reaction) = ? kJ mol-1
    (in kJ per mole of ammonia consumed in the reaction)

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

    Extract the data from the question:

    ammonia gas + oxygen gas → nitrogen dioxide gas + water vapor
    ΔHfo(NH3(g)) = -46 kJ mol-1
    ΔHfo(NO2(g)) = +34 kJ mol-1
    ΔHfo(H2O(g)) = -242 kJ mol-1
    ΔHfo(O2(g)) = 0 kJ mol-1 (by definition)
  3. What is the relationship between what you know and what you need to find out?
    General equation: ΔHo(reaction) = ΣΔHfo(products) - ΣΔHfo(reactants)

    (i) Write a balanced chemical equation for the oxidation of 1 mole of ammonia gas:

    ammonia gas + oxygen gas → nitrogen dioxide gas + water vapor

    (ii) Re-write the general form of the equation for ΔHo(reaction) to apply to the specific reaction, that is, to the oxidation of 1 mole of ammonia gas.

    (iii) Substitute the standard enthalpy of formation values into the equation

    (iv) Solve this equation to find the enthalpy change for the oxidation of ammonia gas in kJ per mole of ammonia gas consumed.

  4. Perform the calculations

    (i) Write a balanced chemical equation for the oxidation of 1 mole of ammonia gas:

    word equation: ammonia gas + oxygen gas nitrogen dioxide gas + water vapor
    per 2 mol NH3(g): 2NH3(g) + 7/2O2(g) 2NO2(g) + 3H2O(g)
    per 1 mol NH3(g): NH3(g) + 7/4O2(g) NO2(g) + 3/2H2O(g)

    (ii) Re-write the general form of the equation for ΔHo(reaction) to apply to the specific reaction, that is, to the oxidation of 1 mole of ammonia gas.

    ΔHo(reaction) = ΣΔHfo(products) - ΣΔHfo(reactants)
    ΔHo(reaction) = [ ΔHfo(NO2(g)) + 3/2 × ΔHfo(H2O(g)) ] - [ ΔHfo(NH3(g)) + 7/4 × ΔHfo(O2(g)) ]

    (iii) Substitute the standard enthalpy of formation values into the equation

    ΔHo(reaction) = [ ΔHfo(NO2(g)) + 3/2 × ΔHfo(H2O(g)) ] - [ ΔHfo(NH3(g)) + 7/4 × ΔHfo(O2(g)) ]
    ΔHo(reaction) = [ +34 + 3/2 × -242 ] - [ -46 + 7/4 × 0 ]

    (iv) Solve this equation to find the enthalpy change for the oxidation of ammonia gas.

    ΔHo(reaction) = [ +34 + 3/2 × -242 ] - [ -46 + 7/4 × 0 ]
      = [ +34 + -363 ] - [ -46 + 0 ]
      = [ -329 ] - [ -46 ]
      = -329 + 46
      = -283 kJ mol-1    

  5. Is your answer plausible?
    Use the direct application of Hess's Law to calculate the enthalpy change; that is, reactants break apart to produce elements in their standard states, products are formed from elements in their standard states:

    Overall reaction: NH3(g) + 7/4O2(g) → NO2(g) + 3/2H2O(g)

    reactants products ΔHo
    kJ mol-1
        NH3(g) ½N2(g) + 3/2 H2(g) -(1 × -46) = +46
        7/4O2(g) 7/4O2(g)     -(7/4 × 0) = 0
    ½N2(g) + O2(g) NO2(g)     1 × +34 = +34
    3/2H2(g) + 3/4O2(g) 3/2H2O(g)     3/2 × -242 = -363

    NH3(g) + 7/4O2(g) NO2(g) + 3/2H2O(g) 46 + 34 - 363 = -283

    Since this value for ΔHo(reaction) is the same as the value we calculated previously, we are reasonably confident that our answer is correct.

  6. State your solution to the problem "enthalpy change for oxidation of ammonia":

    ΔHo(reaction) = -283 kJ mol-1

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