![]() ![]() 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 Since air density varies with temperature, the specific weight of air will be function of temperature.ChemTeam: Hess' Law - using three equations and their enthalpies Department of the Interior, Bureau of Reclamation, 1977, Ground Water Manual, from The Water Encyclopedia, Third Edition, Hydrologic Data and Internet Resources, Edited by Pedro Fierro, Jr. Since water density varies with temperature, the specific weight of water will be function of temperature. Specific weight, unlike density, is not absolute, it depends on the value of the gravitational acceleration (g), which varies depending on altitude and latitude. ![]() Specific weight can also be defined as the product between density and gravitational acceleration. Specific weight is defined as the ratio between weight and volume. Example on how to calculate specific weight.
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