The standard enthalpy of combustion of naphthalene is –5157 kj mol-1. calculate its standard enthalpy of formation. (use data in tables 2c.4 and 2c.5 in the resource section)
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Ответ:
The combustion of naphthalene is given as:
C10H8 (s) + 12 O2 (g) --> 10CO2 (g) + 4 H2O (l)
Enthalpy of combustion ΔH = -5157 kJ/mol
Now, the enthalpy change of a reaction is given
ΔH = ∑nΔHf (products) - ∑nΔHf (reactants)
where n = number of moles
ΔHf = enthalpy of formation
Therefore,
ΔH = [10*ΔHf(CO2) + 4*ΔHf(H2O)] - [1*ΔHf(C10H8) + 12*O2]
-5157 = [10*(-393.5) + 4*(285.83)] - [ΔHf(C10H8) + 12*(0)]
ΔHf(C10H8) = 78.68 kJ/mol
Ответ:
A- crystal
the repeating pattern of a mineral's particles forms a solid called a crystal. A crystal has flat sides, called faces, that meet at sharp edges and corners.