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table and the provided Q) using equation below, determine heat of formation of Kdo, 2 Kdo, 13) - 2 kelo, 18) +₂18) DH = 296.2 KJ/mol. son This is the question from from Thermo precisely from . Ah of more chemistry: formation it the Standerd enthalpy of formation (DHg) heat excha hang happens when of a compound gets formed in their one mole elements from its standerd state
Now, AH Reaction equals to EPHg) Re Reactant E(Any) DHI product comes to this question The Reaction given is 2 Kdo 18) -> 2 KAOS) + + O219) given deta's are AH Reaction 296.2 KJ/mol (Angers - 391.2 KJ/mol o kJ/mol. DHg of O2 lets take x kJ/mol AHg of Keo2 =
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Now, Дң (АН) Reactant "; кого, in Reactant there is a moles of kelo, are present дн; { коtоg Now, (AHf) Reactant Анх | Kelo, in there is is 2 mole Reactant kolog reacted.
е • 39 1 •2 yor Imol дн дні: for 2 то (391-2) х2 - 78 2.4 к3/mo . 12 (AHL) Reactant
NOW (AH) Reaction (Aty product (0Hz) Rec Rectant but all the deta's in this equation 296.2 2x - (-782-04) 296.2 - 782. 4 2 x x = .243.1 ks/mal so, SHf of - 243.1 kJ/mol Kelo, is
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