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Thermodynamics

Question
CBSEENCH11006213

Calculate the enthalpy of formation of acetic acid if its enthalpy of combustion to CO2(g) and H2O(l) is –867.0 kJ mol–1 and the enthalpies of formation of CO2(g) and H2O(l) are respectively –393.5 and –285.9 kJ mol–1.

Solution

We have the equation
          2 straight C left parenthesis straight s right parenthesis space plus space 2 straight H subscript 2 left parenthesis straight g right parenthesis space plus space straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space CH subscript 3 COOH left parenthesis straight l right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment subscript straight r straight H to the power of 0 space equals space ?
Given that:
left parenthesis straight i right parenthesis space CH subscript 3 COOH left parenthesis straight l right parenthesis space plus space 2 straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space 2 CO subscript 2 left parenthesis straight g right parenthesis space plus space 2 straight H subscript 2 straight O left parenthesis straight l right parenthesis semicolon
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment straight H space equals space minus 867 space kJ
left parenthesis ii right parenthesis space straight C left parenthesis straight s right parenthesis space plus space straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space CO subscript 2 left parenthesis straight g right parenthesis semicolon
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment subscript straight r straight H to the power of 0 space equals space minus 393.5 space kJ
left parenthesis iii right parenthesis space straight H subscript 2 left parenthesis straight g right parenthesis space plus space 1 half straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space space straight H subscript 2 straight O left parenthesis straight l right parenthesis semicolon
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment subscript straight r straight H to the power of 0 space equals space minus 285.9 space kJ
Multiplying equation (ii) by 2 and also equation (iii) by 2, we have,
left parenthesis iv right parenthesis space 2 straight C left parenthesis straight s right parenthesis space plus space 2 straight O subscript 2 left parenthesis straight g right parenthesis space rightwards arrow space space space 2 CO subscript 2 left parenthesis straight g right parenthesis semicolon
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment subscript straight r straight H to the power of 0 space equals space minus 787 space kJ
left parenthesis straight v right parenthesis space space space 2 straight H subscript 2 left parenthesis straight s right parenthesis space plus space straight O subscript 2 left parenthesis straight g right parenthesis space space space rightwards arrow space space space 2 straight H subscript 2 straight O left parenthesis straight l right parenthesis semicolon
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment subscript straight r straight H to the power of 0 space equals space minus 571.8 space kJ space space space space space space space space
Adding equations (iv) and (v), we have,
left parenthesis vi right parenthesis space 2 straight C left parenthesis straight s right parenthesis space plus space 2 straight H subscript 2 left parenthesis straight g right parenthesis space plus space 3 straight O subscript 2 left parenthesis straight g right parenthesis semicolon space rightwards arrow space 2 CO subscript 2 left parenthesis straight g right parenthesis space plus space 2 straight H subscript 2 straight O left parenthesis straight l right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space