The enthalpy change for a given reaction is a function of the stoichiometry of the chemical reaction. For example, the Example Problem in Section 5.4a showed that the reaction of carbon monoxide with nitric oxide releases 373 kJ of energy per mole of CO.
|2 CO(g) + 2 NO(g) → 2 CO2(g) + N2(g)||ΔHrxn = -746 kJ|
Consider, however, the reaction of 0.250 mol carbon monoxide. The amount of energy released can be calculated from the enthalpy of reaction.
This calculation is valid because enthalpy is an extensive variable, which means that it is dependent on the amount of substance present; therefore, a change in enthalpy is also dependent on the amount of substance that reacts. The reaction of 250. mol of CO with NO, for example, results in the release of 9.33 × 104 kJ of energy.
Knowing that enthalpy is an extensive variable allows us to look at the relationship between the enthalpy change for a reaction and the chemical equation. Consider, for example, the production of ammonia from elemental nitrogen and hydrogen.
|N2(g) + 3 H2(g) → 2 NH3(g)||ΔHrxn = -23.1 kJ|
In this reaction, 23.1 kJ of energy is released when 2 mol of NH3 is formed. The formation of 1 mol of NH3,
||ΔHrxn = -11.6 kJ|
releases half as much energy, 11.6 kJ. Thus, when a chemical equation is multiplied by a constant, the enthalpy change is also multiplied by that constant.
Consider the decomposition of ammonia to form elemental nitrogen and hydrogen.
|2 NH3(g) → N2(g) + 3 H2(g)||ΔHrxn = +23.1 kJ|
This is the reverse of the equation for the formation of ammonia from elemental nitrogen and hydrogen. Because the original equation represented an exothermic reaction (heat is evolved to the surroundings), the reverse reaction is endothermic (heat is absorbed from the surroundings). When the reactants and products in a chemical equation are reversed, the magnitude of ΔH is identical but the sign of ΔH is reversed.
The reaction of HCl with O2 is exothermic.
|4 HCl(g) + O2(g) → 2 H2O(ℓ) + 2 Cl2(g)||ΔH(1) = -202.4 kJ|
Calculate the enthalpy change for the reaction of water with elemental chlorine to produce HCl and O2.
|H2O(ℓ) + Cl2(g) → 2 HCl(g) + ½ O2(g)||ΔH(2) = ?|
You are asked to calculate the enthalpy change for a given reaction.
You are given the enthalpy change for the reaction under a different set of stoichiometric conditions.
The second reaction is related to the first reaction by (1) reversing reactants and products and (2) multiplication by a constant (× ½). Thus, the enthalpy change for the second reaction is equal to -ΔH(1) × ½.
Is your answer reasonable? The second reaction is the reverse of the exothermic first reaction, so it is endothermic (+ΔH). It also involves only half as many reactants and products, so the amount of energy absorbed is half as much as the amount of energy released in the first reaction.
A chemical reaction shows the relative amounts of reactants and products involved in a chemical reaction. While a balanced equation shows the mole-to-mole ration of reactants and products in the reaction, a typical reaction might involve more or fewer moles of reactants and products. Here we see that not only do the amounts of reactants and products change when the reaction stoichiometry is changed, but also the amount of heat absorbed or released in the reaction also changes.