Question 2 (1 point) Saved The standard heat of combustion for naphthalene, C10Hg(s), is -5156.8 kJ...

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Answer 1

Answer #1

C10H8(s) +12 O2(g) -------------> 10CO2 + 4H2O(l)

$\Delta$ H⁰rxn = $\Delta$ H⁰f products - $\Delta$ H⁰f reactants

$\Delta$ H⁰rxn =10 $\Delta$ H⁰fCO2 + 4 $\Delta$ H⁰f H2O-[ $\Delta$ H⁰f C10H8 + 12 $\Delta$ H⁰f O2]

-5156.8 = 10*-393.5 + 4*-285.9 -[ $\Delta$ H⁰f C10H8 + 12*0]

-5156.8 = -5078.6- $\Delta$ H⁰f C10H8

- $\Delta$ H⁰f C10H8 = -5156.8+5078.6

- $\Delta$ H⁰f C10H8 =- 78.2

$\Delta$ H⁰f C10H8 = 78.2

The standard enthalpy of formation of C10H8 is 78.2 KJ/mole

Answer 2

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