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Use the data below to calculate the standard enthalpy change of formation of ketene,...

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jspake | Student, Undergraduate | eNoter

Posted April 7, 2013 at 7:47 AM via web

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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.
Thanks.

Tagged with chemistry, enthalpy, science

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llltkl | College Teacher | Valedictorian

Posted April 7, 2013 at 2:28 PM (Answer #1)

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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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