Hey there!:
Given that;
SO2 (g ) = (-296.8 kJ/mol) and SO3 (g )= (-395.7 Kj/mol)
2 SO2(g) + O2(g) --------------> 2 SO3(g)
ΔH reaction = ΔH of product - ΔH of reactant
We know that ΔH O2 = 0 kj
ΔH reaction = [2 x - 395.7] - [2 x - 296.8 +0.0 KJ]
= - 791.4 - (- 593.6)
= - 791.4 + 593.6
= - 197.8 kJ
Because there are 2 mole of SO2 coverted into 2 mole of SO3 :
=- 197.8 kJ/2 mol
ΔH°rxn = - 98.9 kJ/ mol
Hope that helps!
The statement of ΔH°rxn for the conversion of SO2 to SO3 is " -197.8 kJ."
The amount of heat absorbed (+ΔH value) or emitted (-ΔH value) as a result of a chemical reaction is referred to as the Standard Enthalpy of Reaction (ΔH°rxn). The standard enthalpy of production for each component or molecule in the reaction is used to compute ΔH°rxn.
Mathematically,
ΔH°rxn = sum of the product's enthalpies – sum of the reactants' enthalpies
ΔH°rxn = ΣΔH°f (reactants) - ΣΔH°f (products)
ΔH°rxn = [(2*-395.7 kJ/mol) - {(2*-296.8 kJ/mol) + (1*0 kJ/mol)}]
ΔH°rxn = [-791.4 - (-593.6 + 0)] kJ
ΔH°rxn = [-791.4 + 593.6] kJ
ΔH°rxn = -197.8 kJ
Hence the correct option is -197.8 kJ.
Learn more about ΔH°rxn here
https://brainly.com/question/17163919
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