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CH2=CH2+H2O->CH3-CH2-OH What's the change of enthalpy in kj/mol c=c is 612 c-o 358 c-c 348 c-h 414 o-h 463 all in kj/mol no H-H
Accepted Answer
The change in enthalpy for the reaction CH2=CH2 + H2O → CH3-CH2-OH can be calculated using the following steps:
1. Identify the bonds broken and formed:
Bonds broken: 1 C=C bond, 1 O-H bond
Bonds formed: 1 C-C bond, 1 C-O bond, 1 C-H bond
2. Calculate the total energy required to break the bonds:
C=C: 612 kJ/mol
O-H: 463 kJ/mol
Total: 612 + 463 = 1075 kJ/mol
3. Calculate the total energy released when forming the bonds:
C-C: 348 kJ/mol
C-O: 358 kJ/mol
C-H: 414 kJ/mol
Total: 348 + 358 + 414 = 1120 kJ/mol
4. Calculate the change in enthalpy (ΔH):
ΔH = Energy released - Energy required
ΔH = 1120 kJ/mol - 1075 kJ/mol = 45 kJ/mol
Therefore, the change in enthalpy for the reaction is 45 kJ/mol.
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