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Thermodynamics

Question
CBSEENCH11006220

Compute the enthalpy of hydrogenation of ethylene if the bond enthalpies of H – H, C – H, C – C and C = C bonds are 436, 485, 347 and 619 kJ mol–1 respectively.

Solution

The required thermochemical equation is

increment subscript straight r straight H to the power of 0 space equals space sum from blank to blank of Bond space enthalpies subscript left parenthesis Reactants right parenthesis end subscript space minus space sum from blank to blank of Bond space enthalpies subscript left parenthesis Products right parenthesis end subscript
space space space space space space space space equals space open square brackets straight H subscript straight C space minus straight C end subscript superscript 0 space plus space 4 space cross times space straight H subscript straight C space minus straight H end subscript superscript 0 space plus space straight H subscript straight H space minus straight H end subscript superscript 0 close square brackets space minus space open square brackets straight H subscript straight C space minus space straight C end subscript superscript 0 space plus space 6 space cross times space straight H subscript straight C space minus straight H end subscript superscript 0 close square brackets
space space space space space space space space equals open square brackets 619 plus 4 left parenthesis 485 right parenthesis plus 436 close square brackets space minus space open square brackets 347 plus 6 left parenthesis 485 right parenthesis close square brackets
space space space space space space space space equals 2995 minus 3257
space space space space space space space space equals negative 262 space kJ
Thus enthalpy of hydrogenation of ethylene is -262 kJ.