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Thermodynamics

Question
CBSEENCH11006210

Calculate the enthalpy of combustion of methane, it is given that heat of formation of CO2(g), H2O(l) and CH4(g) are -393 5 kJ mol–1, –285 .9 kJ mol–1 and –748 kJ mol–1 respectively.

Solution

We have the equation,
CH subscript 4 left parenthesis straight g right parenthesis space plus space 2 straight O subscript 2 left parenthesis straight g right parenthesis space rightwards arrow space space CO subscript 2 left parenthesis straight g right parenthesis space plus space 2 straight H subscript 2 straight O left parenthesis 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 space space space increment subscript straight r straight H to the power of 0 space equals space ? space space space space space space space space space space space space space space space space space space
Given that:
 left parenthesis straight i 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 space rightwards arrow space 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 increment subscript straight r straight H to the power of 0 space equals space minus 393.5 space kJ
left parenthesis ii 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 space space rightwards arrow space space space straight H subscript 2 straight O left parenthesis 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 increment subscript straight r straight H to the power of 0 space equals space minus 285.9 space kJ
left parenthesis iii right parenthesis space straight C left parenthesis straight g right parenthesis space plus space 2 straight H subscript 2 left parenthesis straight g right parenthesis space rightwards arrow space space space space CH subscript 4 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 increment subscript straight r straight H to the power of 0 space equals space minus 748 space kJ
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
Multiplying equation (ii) by 2, we have,
left parenthesis iv right parenthesis 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 space rightwards arrow 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 increment subscript straight r straight H to the power of 0 space equals space minus 571.8 space kJ
Adding equations (i) and (iv), we have,
left parenthesis straight v right parenthesis space space 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 space 2 straight O subscript 2 left parenthesis straight g right parenthesis space space space rightwards arrow space space space 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
                                       increment subscript straight r straight H to the power of 0 space equals space minus 965.3 space kJ
                                                         
Subtracting equation (iii) from equation (v), we have,
         2 straight O subscript 2 left parenthesis straight g right parenthesis space space space rightwards arrow space space CO subscript 2 left parenthesis straight g right parenthesis space plus space 2 straight H subscript 2 straight O left parenthesis l right parenthesis space minus space CH subscript 4 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 space space space increment subscript straight r straight H to the power of 0 space equals space minus 965.3 space plus space 748
or space space CH subscript 4 left parenthesis straight g right parenthesis space plus space 2 straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space space 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 increment subscript straight r straight H to the power of 0 space equals space minus 217.3 space kJ
therefore Enthalpy of combustion of methane is -217.3 kJ mol-1.