tion:tefan V.Feb 27, 2018H2(G)+12O2(g)→H2O(l) ΔH=−276 kJExplanation:You know that when 2 moles of hydrogen gas react with 1 mole of oxygen gas, you GET 2 moles of water and 572 kJ of HEAT are evolved.2H2(g)+O2(g)→2H2O(l) ΔH=−572 kJDon't forget that the enthalpy change of reaction MUST be negative here to illustrate the fact that heat if being given off by the reaction.Now, in order for this reaction to produce 1 mole of water, all the coefficients of the chemical equation must be halved.(12⋅2)H2(g)+12O2(g)→(12⋅2)H2O(l)This will get youH2(g)+12O2(g)→H2O(l) Now, the enthalpy change for this reaction will be half the value of the enthalpy change for the reaction that produced 2 moles of water.ΔH1 mole H2O=12⋅ΔH2 moles H2O ΔH1 mole H−2O=−572 kJ2=−286 kJ This means that the thermochemical equation that describes the formation of 1 mole of water looks LIKE thisH2(g)+12O2(g)→H2O(l) ΔH=−276 kJTo write the thermochemical equation that describes the decomposition of 1 mole of water into hydrogen gas and oxygen gas, you need to reverse the chemical equationH2O(l)→H2(g)+12O2(g) and change the sign of the enthalpy change of reaction.ΔHreverse=−ΔHforward This means that the thermochemical equation will look like thisH2O(l)→H2(g)+12O2(g) ΔH=+276 kJThis means that when 1 mole of water undergoes decomposition, 276 kJ of heat are being absorbed