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Thermodynamics

Question
CBSEENCH11006205

How is Hess’s law helpful in calculating the enthalpy of formation of carbon monoxide?

Solution
Enthalpy of formation of carbon monoxide cannot be determined experimentally since the combustion of carbon will never stop at CO(g) stage. A small amount of CO2(g) will be always formed. The enthalpy of formation of carbon monoxide can be calculated by Hess’s law.
The thermochemical equation.
straight C left parenthesis straight s right parenthesis space plus space 1 half space straight O subscript 2 left parenthesis straight g right parenthesis space rightwards arrow space space space CO left parenthesis straight g right parenthesis semicolon space space space increment subscript straight r straight H to the power of 0 space equals ?
Experimentally it has been found 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 space rightwards arrow 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 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 CO left parenthesis straight g right parenthesis space plus space 1 half straight O subscript 2 left parenthesis straight g right parenthesis space rightwards arrow 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 negative 283.0 space kJ
Subtracting equation (ii) from (i), we have,
   straight C left parenthesis straight s right parenthesis space plus space 1 half straight O subscript 2 left parenthesis straight g right parenthesis space minus space CO left parenthesis straight g right parenthesis space rightwards arrow space space 0
space space space space space space space space 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 plus 283.0
or space space straight C left parenthesis straight s right parenthesis plus space 1 half straight O subscript 2 left parenthesis straight g right parenthesis space rightwards arrow space space CO 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 increment subscript straight r straight H to the power of 0 space equals space minus 110.5 space kJ