Stoichiometry of Thermochemical Equations

Consider the following reaction:

1/8 S₈ (s) + O₂ (g) → SO₂ (g)     ΔH = -296.8 kJ

The reaction is exothermic because ΔH is negative. When 1/8 of a mole of S₈ is reacted, 296.8 kJ of heat is released. If 1 mole of oxygen is reacted, 296.8 kJ of heat is released. And, if 1 mole of SO₂ is produced, then 296.8 kJ of heat is released. How much heat would be released if 1 mole of S₈ is reacted? We can see ΔH is an extensive property — it is dependent on the amount of a substance being reacted or undergoing a process.

\(\displaystyle 1.00\;mol\;S_8\;\times\frac{-296.8\;kJ}{1/8\;mol\;S_8}\;=\;2.37\;\times\;10^3\;kJ\)
 
We see if 1 mole of S₈ reacts, eight times more heat is released, 2.37 x 10³ kJ.

Hess’s Law

According to Hess’s Law, the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps.

Recall, enthalpy is a state function, therefore, the enthalpy change, ΔH, associated with a process depends only on the amount of matter that undergoes a change, and on the nature of the initial and final states of the system. If a reaction is carried out in one step or a series of steps, the sum of the enthalpy changes associated with the individual steps must be the same as the enthalpy change associated with a one step process. For example, the combustion of methane can be thought of as occurring in one step,

or in two steps:

The two equations are manipulated and then added to get the overall equation. If an equation is reversed, change the sign of ΔH. If the equation coefficients are divided or multiplied by a number, then divide or multiply ΔH by the same number. The ΔH values are then added together. In the example, when summing the equations, H₂O (g) cancels out. Once the overall equation is determined, the values of ΔH are added.

Hess’s law provides a way to calculate enthalpy changes that are difficult to measure in the lab. The final ΔH is independent of the number of steps, because ΔH is a state function. All we need to do is manipulate the equations and their ΔH values to add to the overall equation and calculate the final ΔH. Below we do an example.

In summary, we can determine ΔH for a reaction that is difficult to measure directly by using Hess’s Law. The value of ΔH is the same whether the reaction takes place in one step or many steps. If presented with a series of reactions, manipulate them so they will add to the overall equation. The equations are manipulated by reversing them or by multiplying or dividing the reaction coefficients by a number. If a reaction is reversed, then change the sign of ΔH. If the coefficients of an equation are multiplied or divided, multiply or divide ΔH of the reaction by the same number. Once the overall equation is determined, then add the values of ΔH.

Worksheet: Hess’s Law
Worksheet: Enthalpy and Stoichiometry

Please watch the following videos before attempting the exercises.

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Exercises

Exercise 1. Consider the following reaction:

2 KClO₃ (s) → 2 KCl (s) + 3 O₂ (g)     ΔH = -89.3 kJ