CHEM 150 1nd Edition Lecture 16Outline of Last Lecture • Cont. Calorimeter • Measuring ΔH (or ΔE)Outline of Current Lecture • Hess's Law• Enthalpy Change• Standard condition of pressure. Current Lecture-If a thermochemical equation is manipulated mathematically, its ÄH value undergoes thesame manipulation. (See section 5.8)Hess’ Law: The enthalpy change of a process is equal to the sum of the enthalpy changes of the steps.You want to know the enthalpy change for :2 NaBr(s) + 4 H2SO4(aq) -----> Br2(l) + SO2(g) + 4 H2O(l) + 2 NaSO4(aq)You know the following:SO3(g) + H2O(l) -----> H2SO (aq) ΔH= -2227.7KJSO2(g) + 1/2O2(g) -----> SO3(g) ΔH= -98.89KJ1/2Br2(l) + Na(s) -----> NaBr(s) ΔH= -361.06KJ2Na(s) + SO2(g) + O2(g) -----> Na2SO4(aq) ΔH= -1092.68KJQ: Find ΔH for CIF(g) + F2(g) -----> CIF3(l)A: 1/2 x (2CIF + O2 ----> CI2O+ OF2) ΔH= 1/2 x167.5J+ 1/2 x(2F2 + O2 -----> 2O5) ΔH= 1/2 x-43.5J1/2 CI2) + 3OF2 -----> 2CIF3 + 2O2CIF + O2 + F2 ----> 1/2 CI2O + 1/2 OF2+ 1/2 CI2O + 1/2 OF2 ----> CIF3 + O2* CIF(g) + F2(g) -----> CIF3(l) ΔH= 62.04T-1/2 x394.147= -135.056T• Standard Molar Enthalpies of Formation, ΔH °f• The enthalpy change when 1 mole of a substance is formed from its elements in their “standard” state.• ΔH °f of an element in it's standard state is defined as Zero. What is with the ° ? It indicates that we are operating under standard conditions (see page 220). A form of Hess'Law using the standard heats of formation: ΔHrxn =Σ ΔH°f (products) minus Σ ΔH°f (reactants)-ΔH°f for C2H5OH(l) is -277.6KJ/molQ: What is the chemical equation that corresponds to this ΔH?A: 2C(s) + 3H2(g) + 1/2 O2(g) -----> C2H5OH(l) -The enthalpy change of combustion forC2H5OH(l) (ΔH°combustion) is -1234.8 KJ/mol. why is this number different? A: C2H5OH + 3O2(l) + 3O2 (g) -----> 2CO2(g) +
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