Ethanol (c 2h 6o(l), δh o f = -277.69 kj/mol) can be made by reaction of ethylene (c 2h 4(g) δh o f = 52.26 kj/mol) with water (δh o f = -285.83 kj/mol). what is the enthalpy of reaction for this process?

Enthalpy change = -44.12 kJ

Explanation:

Given:  

ΔH°f(C2H6O(l)) = -277.69 kj/mol  

ΔH°f(C2H4(g)) = 52.26 kj/mol  

ΔH°f(H2O) = -285.83 kj/mol  

To determine:

Enthalpy change for the formation of C2H6O

Calculation:

The given reaction is:

The enthalpy change for the reaction is given as;

where n(products) and n(reactants) are the moles of products and reactants

Substituting the appropriate values for n and ΔH°f:

ΔH = -277.69-(52.26-285.83) = -44.12 KJ