Question

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?

Answer 1

Answer:

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:

$C2H4(g) + H2O(g)\rightarrow C2H5OH(g)$

The enthalpy change for the reaction is given as;

$\Delta H = \sum n(products)\Delta H^{0}f(products)-\sum n(reactants)\Delta H^{0}f(reactants)$

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

Substituting the appropriate values for n and ΔH°f:

$\Delta H = 1\Delta H^{0}f(C2H6O)-[1\Delta H^{0}f(C2H4)+1\Delta H^{0}f(H2O)]$

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