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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
Accepted Answer
The change in enthalpy (ΔH) for the reaction can be calculated using the bond enthalpy values provided. Here's the breakdown:
1. Break bonds in reactants:
C=C (612 kJ/mol) + 1
H-H (436 kJ/mol) = 1048 kJ/mol
2. Form bonds in products:
C-C (348 kJ/mol) + 1
C-O (358 kJ/mol) + 1
O-H (463 kJ/mol) + 5
C-H (414 kJ/mol) = 3077 kJ/mol
3. Calculate ΔH:
ΔH = Energy required to break bonds - Energy released by forming bonds
ΔH = 1048 kJ/mol - 3077 kJ/mol = -2029 kJ/mol
Therefore, the change in enthalpy for the reaction is -2029 kJ/mol. This negative value indicates that the reaction is exothermic, releasing heat to the surroundings.
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