Reversing the equation give: H2O(l) H2(g) + 1/2O2(g); = +286 kJ → ∆H Note that the sign for is reversed. ∆H Applying Stoichiometry to Heats of Reaction As you have just seen, the quantity of heat obtained from a reaction depends on the amount of reactants. Therefore, we can apply stoichiometry to problems involving the quantity of heat. Lets look at an example. Consider the reaction of methane, CH4 (the principal constituent of natural gas), burning in oxygen at constant pressure. How much heat could you obtain from 10.0 g of methane, assuming you have an excess of oxygen? You can answer this question if you know the enthalpy change for the reaction of 1 mole of methane. The thermochemical equation is CH4(g) + 2O2(g) CO2(g) + H2O(l); = -890.3 kJ → ∆H This calculation involves the following conversions: grams CH4 mole CH4 kJ of heat → → That is, 10,0 g of methane burns in excess oxygen to evolve 556 kJ of heat. If you wanted the result in kilocalories, you would have to concert using the relationship 1 cal = 4.184 J. Lets look at one last example. Calculating the Heat of Reaction Using Stoichiometry How much heat is evolved when 907 kg of ammonia is produced according to the following equation? (Assume that the reaction occurs at constant pressure) N2(g) + 3H2(g) 2NH3(g); = -890.3 kJ → ∆H
Reversing the equation give: H2O(l) → H2(g) + 1/2O2(g); ∆H = +286 kJ Note that the sign for ∆H is reversed. Applying Stoichiometry to Heats of Reaction As you have just seen, the quantity of heat obtained from a reaction depends on the amount of reactants. Therefore, we can apply stoichiometry to problems involving the quantity of heat. Lets look at an example. Consider the reaction of methane, CH4 (the principal constituent of natural gas), burning in oxygen at constant pressure. How much heat could you obtain from 10.0 g of methane, assuming you have an excess of oxygen? You can answer this question if you know the enthalpy change for the reaction of 1 mole of methane. The thermochemical equation is CH4(g) + 2O2(g) → CO2(g) + H2O(l); ∆H = -890.3 kJ This calculation involves the following conversions: grams CH4 → mole CH4 → kJ of heat That is, 10,0 g of methane burns in excess oxygen to evolve 556 kJ of heat. If you wanted the result in kilocalories, you would have to concert using the relationship 1 cal = 4.184 J. Lets look at one last example. Calculating the Heat of Reaction Using Stoichiometry How much heat is evolved when 907 kg of ammonia is produced according to the following equation? (Assume that the reaction occurs at constant pressure) N2(g) + 3H2(g) → 2NH3(g); ∆H = -890.3 kJ