Standard level
Balanced chemical equations can be treated in very similar ways to mathematical equations, as they represent ratios between the various components in the reaction.
Syllabus ref: R1.2.2Reactivity 1.2.2 - Hess’s law states that the enthalpy change for a reaction is independent of the pathway between the initial and final states.
- Apply Hess’s law to calculate enthalpy changes in multistep reactions.
Guidance
Tools and links
The law of conservation of energy
"Energy can be neither created nor destroyed; it can be transformed from one form of energy to another".
The significance of this law is that any energy change resulting from a physical, or chemical process, must be the same regardless of how many steps are taken from the initial to the final situation.
This was recognised by Germain Henri Hess, a Swiss-born Russian chemist,
who formulated the eponymous law.
Hess' law
In any process the energy change between two situations is independent of the route taken between them.
In some ways it is similar to a journey taken from one town to another. The route between the two towns can never affect the real distance between them.
The beauty of this concept is that if we cannot measure the energy change using one route, we can always select other routes and add up all of the energy changes to obtain the results for the original route.
Energy calculations
In the previous section we saw how energy cycles and diagrams can be used to determine energy changes for processes. Energy changes can also be calculated mathematically by manipulation of equations.
We know that energy is an extensive characteristic of matter, i.e. the quantity of energy depends on the amount of matter involved. The equation for the formation of ammonia as written below releases -92 kJ of energy.
N2 + 3H2 → 2NH3
ΔH
= -92 kJ
However, if the equation is written to form 1 mole of ammonia, exactly half the quantity of energy is released.
½N2 + 1½H2 →
NH3ΔH
= -46 kJ
It should be apparent that the first equation is simply two times the second equation. Manipulation of equations in this way is perfectly valid and very useful in the application of Hess' law.
Equation manipulation
Chemical equations represent a physical reality and consequently can have any of the four rules of number operations applied. The main thing to bear in mind is that whatever is done to an equation must also be done to the energy change value.
Simply rearranging an equation does not affect the energy change, but performing any of the four rules of number, mutiplying, dividing, adding or subtracting does affect the energy.
The operations that can be carried out are summarised below.
| operation | energy change |
|---|---|
| rearrangement | has no effect on the energy change |
| reversing the equation | changes the sign of the energy change |
| mutiplication | the energy change is mutiplied |
| division | the energy change is also divided |
| adding two equations | the energy changes are added together |
| subtracting two equations | the energy changes are subtracted |
Operations not affecting the energy change
Rearrangement of an equation by moving a component from one side to the other, while changing its sign, does not affect the energy of the reaction.
|
Example: The first ionization energy of sodium can be represented by the following equation: Na(g) → Na+(g) + 1e ΔH = +496 kJ This equation can be rearranged to: Na(g) - 1e → Na+(g) ΔH = +496 kJ Notice that this rearrangement does not affect the energy change. |
Operations affecting the energy change
1 Reversing the equation
Reversing an equation also reverses the sign of the energy change. For example, the enthalpy of vaporisation of water, ΔH(vap), is represented by the following equation:
| H2O(l) →
H2O(g)ΔH
= +40.65 kJ |
The reverse process is the enthalpy change of condensation:
| H2O(g) →
H2O(l)ΔH
= -40.65 kJ |
Reversing the equation has changed the sign of the enthalpy change from positive to negative.
2 Adding together two equations
When two (or more) equations are added together the energy is also added. The equation produced may not have any physical reality in that it may be chemically impossible, but this does not alter the energetic validity of the procedure.
|
2C(s) + 2O2(g) →
2CO2(g) H2(g) + 2C(s) + 2½O2(g) → 2CO2(g) + H2O(l) ΔH = -1072.8 kJ |
3 Subtracting two equations
The energy changes are subtracted, taking care to note that subtraction of a negative value is the same as addition, i.e. (-)- = +. In the example below the second equation 2 is subtracted from the first equation 1.
The energy change calculation is -1072.8 - -1410.0 = -1072.8 + 1410.0 = +337.2.
| 1 H2(g) + 2C(s) + 2½O2(g)
→ 2CO2(g) +
H2O(l) ΔH
= -1071.8 kJ 2 C2H2(g) + 2½O2(g) → 2CO2(g) + H2O(l) ΔH = -1410.0 kJ H2(g) + C(s) - C2H2(g) → zero ΔH = +337.2 kJ |
The 'zero' indicates that there is nothing on the right hand side of the equation. However, notice that the term - C2H2(g) is negative and so rearrangement, by changing this component to the right hand side, gives:
H2(g) + C(s) → C2H2(g)ΔH
= +337.2 kJ
This is the equation representing the enthalpy of formation of ethyne, ΔHf.
Worked examples
Q431-01 The enthalpy of formation of water is -286 kJ. What is the enthalpy change for the combustion of hydrogen under standard conditions?Answer
|
The equation for the enthalpy of formation of water is: H2(g) + ½O2(g) → H2O(l) This is exactly the same equation as for the standard enthalpy of combustion of hydrogen. Therefore the enthalpy change for the combustion of hydrogen is also -286 kJ |
Q431-02 The enthalpy of formation of carbon dioxide is -394 kJ. What is the enthalpy change of the following reaction?
2CO2(g) → 2C(s) + 2O2(g)
Answer|
The equation for the enthalpy of formation of carbon dioxide is: C(s) + O2(g) → CO2(g) The equation we require is double this in the reverse direction. Hence we must multiply the energy of formation by two and reverse the sign. Enthalpy change = -(-394 x 2) = +788 kJ |
Q431-03 Using the following information:
| C(s) + O2(g) → CO2(g) ΔH = -394 kJ. |
Find the energy change when 1.8 g of carbon dioxide are formed
Answer|
1.8 g of carbon dioxide is equal to 1.8/44 moles = 0.0409 moles From the equation the enthalpy change when 1 mole of carbon dioxide is formed = -394 kJ Therefore the enthalpy change when 0.0409 moles are formed = -394 x 0.0409 = -16.1 kJ |
Q431-04 Consider the reaction:
4Al(s) + 3O2(g) →
Al2O3(s)
ΔH = -3339 kJ. |
Calculate the amount of heat evolved when 10.0 g of aluminum burns in oxygen.
Answer|
In the above equation 4 moles of aluminium releases -3339 kJ, thus 1 mole releases -834.8 kJ 10.0g of aluminium = 10.0/27 moles = 0.370 mol. Energy released by 0.370 mol = -834.8 x 0.370 = -309 kJ |
Q431-05 Consider the reaction:
| K(s) + ½Cl2(g) →
KCl(s); ΔH = -439
kJ. |
How much energy is released when 10 g of KCl(s) is formed.
Answer|
10 g of KCl(s) represents 10/74.5 mol = 0.134 mol 1 mol of KCl produced releases -439 kJ, therefore formation of 0.134 mol releases -439 x 0.134 = -58.9 kJ |
Q431-06 When concentrated sulfuric acid is added to water, heat is liberated. The heat of dilution is about -75.3 kJ/mole. How much heat would be released if 3.4 g of H2SO4 were added to excess water?
Answer
|
3.4 g of H2SO4 = 0.0347 mol Thus heat released = 0.0347 mol x -73.5 = -2.62 kJ |
Q431-07 Given the reaction:
| C8H18(l) + 12½O2(g) →
8CO2(g) + 9H2O(l)
ΔH = -5473 kJ. |
Determine the number of grams of octane required to produce 100 kJ of energy.
Answer|
From the equation, 1 mole of octane produces -5473 kJ. To produce 100 kJ you need 100/5473 mol = 0.0183 mol 1 mol of octane has a mass of 114 g, hence 0.0183 x 114 = 2.09 g are needed. |
Q431-08 When concentrated sulfuric acid is added to water, heat is liberated. The enthalpy of dilution is -75.3 kJ/mol. Predict the mass of sulfuric acid needed to release 100 kJ of energy when added to excess water.
Answer
|
1 mole of sulfuric acid releases -75.3 kJ. Moles needed to release 100 kJ = 100/75.3 mol = 1.33 mol Mass = Mr x mol = 98 x 1.33 = 130 g |
Q431-09 Given the following enthalpy changes:
S(s) + O2(g) →
SO2(g); ΔH = x kJ.
S(s) + 1½O2(g) →
SO3(g); ΔH = y kJ.
In terms of x and y, what is the enthalpy change for the reaction below?
2SO3(g) → 2SO2(g) + O2(g)
|
To construct the equation required you need 2 SO3(g) on the left hand side. Of the equations given, SO3(g) appears on the RHS of equation y. Hence this must be multiplied by 2 and reversed = -2y SO2(g) appear on the RHS of the first equation (energy change x). In the required equation there are 2SO2(g) on the RHS, hence the value of x must be doubled. Finally, they are added together = 2x - 2y |
Q431-10 Given the following enthalpy changes:
CaO(s) + 2HCl(aq) →
CaCl2(aq) + H2O(l)ΔH = x kJ.
CaCO3(s) + 2HCl(aq) →
CaCl2(aq) + H2O(l) + CO2(g) ΔH = y kJ.
In terms of x and y, what is the enthalpy change for the reaction below?
CaO(s) + CO2(g) → CaCO3(s)
Answer|
Constructing the required equation: On the LHS: CaO(s) appears in equation 1 (energy change = x). CO2(g) appears on the RHS in equation 2, which must be reversed so that it appear on the LHS. This also places CaCO3(s) on the RHS as required. Hence the energy change = -y kJ. On adding the two equations together the energy change = (x - y) kJ |