Use the data below to calculate the standard enthalpy change of formation of ketene, C2H2O (CH2=C=O)
standard enthalpy change of formation of CO2 => -395 kJ/mol
standard enthalpy change of combustion of H2 => -286 kJ/mol
standard enthalpy change of combustion of CH2=C=O => -1028 kJ/mol
Please help me solve the question above.. an explanation will be appreciated.
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The equations of the three reactions are:
C (s) + O2 (g) = CO2 (g) ∆H = -395 kJ/mol --- i)
H2(g) + ½ O2(g) = H2O (l) ∆H = -286 kJ/mol --- ii)
CH2=C=O (g) + 2 O2 (g) = 2 CO2 (g) + H2O (l) ∆H= -1028 kJ/mol --- iii)
By a simple mathematical operation: 2 × i) + ii) – iii), one gets the resultant equation as:
2C (s) +2O2 (g) + H2(g) + ½ O2(g) + 2CO2 (g) + H2O (l) = 2CO2 (g) + H2O (l) + CH2=C=O (g) + 2O2 (g)
Or, 2C (s) + H2(g) + ½ O2(g) = CH2=C=O (g)
This is the formation reaction of the ketene indeed.
Keeping the Hess' law in view, the heat of reaction values will also be obtained by the same mathematical operation.
So, ∆H(f) of ketene =[ 2 × -395 + -286 – (-1028) ] kJ/mol = -48kJ/mol.
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