Hess’s Law of Constant Heat Summation

Using solution calorimetry and bomb calorimetry, we can measure the enthalpies of many simple reactions, but there are many more reactions that we can’t measure. However, we can use the additive nature of chemical reactions to find unknown enthalpies.

Consider the following example:

a) C (s) + ½ O₂(g) → CO(g) ΔH = ?

b) CO(g) + ½ O₂(g) → CO₂ (g) ΔH = -283.0 kJ

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

Note that a and b sum up to c.

We can measure the ΔH of b and c in a bomb calorimeter, but not that of a since reaction b would also take place at the same time. However, we can put in CO and react it with more oxygen to measure the ΔH of reaction b.

Just as chemical reactions add, so do their enthalpies:

ΔH_a +ΔH_b = ΔH_c

ΔH_a + -283.0 kJ = -393.5 kJ

ΔH_a = -110.5 kJ

We can diagram this on an energy diagram: figure 6.19

Since enthalpy is a state function, it doesn’t matter what path we take to get from the initial point (C(s) + O₂(g)) to the final point (CO₂(g)). We can stop at as many points along the way, all of the enthalpies will still add up to the overall enthalpy of the overall equation.

Hess’s law says that the heat of a reaction is constant whether it is carried out directly in one step or indirectly in a series of steps.

Some things to remember:

If you reverse a chemical reaction, the sign of ΔH must be changed.

If you multiply a chemical reaction by a number, then ΔH must also be multiplied by that number.

example: exercise 6.14 A

Calculate the enthalpy change for the reaction:

2 CH₄(g) + 3 O₂(g) → 2 CO(g) + 4 H₂O(l)

using:

CO(g) + ½ O₂(g) → CO₂ (g) ΔH = -283.0 kJ

CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2 H₂O(l) ΔH = -890.3 kJ

first arrange the equations so that the chemical species cancels to form the desired reaction:

CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2 H₂O(l) ΔH = -890.3 kJ

CO₂ (g) → CO(g) + ½ O₂(g) ΔH = -(-283.0 kJ) = 283.0 kJ

Then multiply by appropriate values so that sum of reactions form the proper balanced equation:

2(CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2 H₂O(l)) ΔH = 2(-890.3 kJ)

2(CO₂ (g) → CO(g) + ½ O₂(g)) ΔH = 2(283.0 kJ)

2 CH₄ (g) + 4 O₂ (g) → 2 CO₂ (g) + 4 H₂O(l) ΔH = -1780.6 kJ

2 CO₂ (g) → 2 CO(g) + O₂(g) ΔH = 566.0 kJ

Then sum up equations and enthalpies:

2 CH₄(g) + 3 O₂(g) → 2 CO(g) + 4 H₂O(l) ΔH = -1780.6 kJ + 566.0 kJ

ΔH = -1214.6 kJ

Standard Enthalpies of Formation:

Each substance could potentially be made from its elemental components. For example:

C(graphite) + O₂ (g) → CO₂ (g)

Carbon and oxygen are in their most stable elemental forms (what we call their standard state). If we define the enthalpy of each element found in their standard state forms as being zero, then the heats of reaction for the above chemical equation would also be the standard enthalpy of reaction (ΔHº). In this case, we also get the standard enthalpy of formation of CO₂ (ΔHº_f). This is the enthalpy change for the formation of 1 mole of the substance when both products and reactants are in their standard states. (Their physical state at STP)

We can use these ΔHº_f to calculate the enthalpies of reactions:

For 2 NO₂ (g) → N₂O₄ (g)

first we look up in the appendix C the heats of formation of each reactant and product and write their equations:

½ N₂ (g) + O₂(g) → NO₂ (g) ΔHº_f [NO₂] = 33.18 kJ

but we need to form 2 NO₂ so:

N₂ (g) + 2 O₂(g) → 2NO₂ 2xΔHº_f[NO₂]

and also:

N₂ (g) + 2O₂ (g) → N₂O₄(g) ΔHº_f[N₂O₄] = 9.16 kJ

If we flip the first around, we get:

2NO₂ →N₂ (g) + 2 O₂(g) -2xΔHº_f[NO₂]

and now add the above two equations:

2 NO₂ (g) → N₂O₄ (g) ΔHº = ΔHº_f[N₂O₄] + -2xΔHº_f[NO₂]

= 9.16 kJ - (2 x 33.18 kJ)

= -57.20 kJ

So the overall enthalpy ends up being:

ΔHº = S nΔHº_f of products - S nΔHº_f reactants

where n = the stoichiometric coefficient of found in the balanced chemical equations