Calculate H°f for benzene, C6H6, from the following data.?
2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(l) H° = -6534 kJ
H°f (H2O) = -285.8 kJ/mol
H°f (CO2) = -393.5 kJ/mol
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2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(l) H° = -6534 kJ
H°f (H2O) = -285.8 kJ/mol
H°f (CO2) = -393.5 kJ/mol
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delta H for the reaction = [sum of heat of formation for products] - [sum of heat of formation for reactants]
(-6534) = [(12)(-393.5) + (6)(-285.8)] - (2X)
(-6534) = -6436.8 - 2X
2X = 97.2
X = +48.6 kJ, which is the heat of formation for benzene.