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Using standard heats of formation, calculate the standard enthalpy change for the following reaction.

C2H4(g) + H2O(g) ------> CH3CH2OH(g)

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The enthalpy change of the reaction is -44.79 kJ/mol.

What is the enthalpy change?

We can define the enthalpy change as the energy that is given out or taken in in a reaction. it can be calculated from the enthalpies of formation of species by the use of the relation;

ΔH = ∑Enthalpy of formation of products -  ∑Enthalpy of formation of reactants

ΔHfC2H4(g) = +52kJ/mol

ΔHf H2O(g) = - 241.82 kJ/mol

ΔHf CH3CH2OH(g)= -234.61 kJ/mol

ΔHrxn  = (-234.61) - [52 + (- 241.82 )]

ΔHrxn  = (-234.61) - (-189.82)

ΔHrxn  = -44.79 kJ/mol

Learn more about heat of reaction:https://brainly.com/question/2192784

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