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  4. Calculate the enthalpy change of the following reaction H2C=CH2(.g.)+H2(.g.) arrow H3C-CH3(.g.) The bond energy of C-H,C-C,C=C,H-H are 414, 347, 615 and 453 KJ mol- 1 respectively.

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Q. Calculate the enthalpy change of the following reaction

$H_{2}C=CH_{2}\left(\right.g\left.\right)+H_{2}\left(\right.g\left.\right) \rightarrow H_{3}C-CH_{3}\left(\right.g\left.\right)$

The bond energy of $C-H,C-C,C=C,H-H$ are 414, 347, 615 and 453 KJ $mol^{- 1}$ respectively.

NTA AbhyasNTA Abhyas 2020Thermodynamics

Solution:

$\Delta H=\Sigma $ bond energy of reactant $-\Sigma $ bond energy of product

to write bond energy use proper symbols

$\text{E}_{\text{C} \, \text{=} \, \text{C}} \, \text{or} \, \Delta _{\text{C} \, \text{=} \, \text{C}} \text{H}$ $=\left[\right.1\left(\right.C=C\left.\right)+4\left(\right.C-H\left.\right)+1\left(\right.H-H\left.\right)\left]\right.-\left[\right.1\left(\right.C-C\left.\right)+6\left(\right.C-H\left.\right)\left]\right.$

$=-1\left(\right.C-C\left.\right)-2\left(\right.C-H\left.\right)+1\left(\right.C=C\left.\right)+1\left(\right.H-H\left.\right)$

$=-347-2\left(\right.414\left.\right)+1\left(\right.615\left.\right)+1\left(\right.435\left.\right)$

$=-125$ kJ.