![]() ![]() ![]() H 2(g) + 2C(s) + N 2(g) -> 2HCN(g) ΔH = +270.3 kJĪ) first eq ⇒ flip and divide by 2 (puts one NH 3 on the reactant side)ī) second eq ⇒ flip (puts one CH 4 on the reactant side)Ĭ) third eq ⇒ divide by 2 (puts one HCN on the product side)Ģ) Rewite all equations with the changes: In addition, a and c give 5⁄ 2O 2 on the right to cancel out the 5⁄ 2O 2 on the left.ģNO 2(g) -> 3NO(g) + 3⁄ 2O 2(g) ΔH = +174 kJ This is the formation reaction for CO 2 and its value can be looked up, either in your textbook or online.Įxample #4: Given the following information:ĢNO(g) + O 2(g) -> 2NO 2(g) ΔH = −116 kJĢN 2(g) + 5O 2(g) + 2H 2O(ℓ) -> 4HNO 3(aq) ΔH = −256 kJĬalculate the enthalpy change for the reaction below:ģNO 2(g) + H 2O(ℓ) -> 2HNO 3(aq) + NO(g) ΔH = ?Ī) first eq -> flip multiply by 3⁄ 2 (this gives 3NO 2 as well as the 3NO which will be necessary to get one NO in the final answer)ī) second eq -> divide by 2 (gives two nitric acid in the final answer)Ĭ) third eq -> flip (cancels 2NO as well as nitrogen)Ī) step 1a above puts 3⁄ 2O 2 on the right The last one comes from 3⁄ 2O 2 on the left in the third equation and 1⁄ 2O 2 on the right in the second equation.Ĭ(s, gr) + O 2(g) -> CO 2(g) ΔH f o = −394 kJ Sr(s) + C(s, gr) + 3⁄ 2O 2(g) -> SrCO 3(s) ΔH = −1220 kJģ) Here is a list of what is eliminated when everything is added: If everything is right, the oxygen will take care of itself.Ģ) Apply all the above changes (notice what happens to the ΔH values): Why not also multiply first equation by two (to get 2SrO for canceling)? Because we only want one CO 2 in the final answer, not two. Notice that what we did to the third equation also sets up the Sr to be cancelled. ![]() Multiply second eq by 2 (want to cancel 2S, also want 2SO 2 on product side)įlip 3rd equation (want CS 2 as a reactant)ĢS(s) + 2O 2(g) -> 2SO 2(g) ΔH = −593.6 kJ/mol C(s) + 2S(s) ΔH = −87.9 kJ/mol SrCO 3(s) ΔH = −234 kJĢSrO(s) -> 2Sr(s) + O 2(g) ΔH = +1184 kJĢSrCO 3(s) -> 2Sr(s) + 2C(s, gr) + 3O 2(g) ΔH = +2440 kJġ) Analyze what must happen to each equation:Ī) first eq -> flip it (this put the CO 2 on the right-hand side, where we want it)ī) second eq -> do not flip it, divide through by two (no flip because we need to cancel the SrO, divide by two because we only need to cancel one SrO)Ĭ) third equation -> flip it (to put the SrCO 3 on the other side so we can cancel it), divide by two (since we need to cancel only one SrCO 3) Leave eq 1 untouched (want CO 2 as a product) Notice that the ΔH values changed as well.ĥ⁄ 2O 2 ⇒ first & sum of second and third equation We need one H 2 on the reactant side and that's what we have.Ģ) Rewrite all three equations with changes applied:ĢCO 2(g) + H 2O(ℓ) -> C 2H 2(g) + 5⁄ 2O 2(g) ΔH° = +1299.5 kJĢC(s) + 2O 2(g) -> 2CO 2(g) ΔH° = −787 kJ H 2(g) + 1⁄ 2O 2(g) -> H 2O(ℓ) ΔH° = −285.8 kJġ) Determine what we must do to the three given equations to get our target equation:Ī) first eq: flip it so as to put C 2H 2 on the product sideī) second eq: multiply it by two to get 2CĬ) third eq: do nothing. If a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation equals the sum of the enthalpy changes of the other chemical equations.Įxample #1: Calculate the enthalpy for this reaction:ĢC(s) + H 2(g) -> C 2H 2(g) ΔH° = ? kJ The enthalpy of a given chemical reaction is constant, regardless of the reaction happening in one step or many steps. Germain Henri Hess, in 1840, discovered a very useful principle which is named for him: Using three equations and their enthalpies Hess' Law: three equations and their enthalpies - Problems 1 - 10 Hess' Law: two equations and their enthalpies Hess' Law: standard enthalpies of formation Hess' Law: three equations and their enthalpies - Problems 11 - 25 Hess' Law: four or more equations and their enthalpies Hess' Law: bond enthalpies Thermochemistry menu ChemTeam: Hess' Law - using three equations and their enthalpies ![]()
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