Question

Use the standard enthalpies of formation for the reactants and products to solve for the ΔHrxn for the following reaction. (The ΔHf of C2H4 is 52.26 kJ/mol, CO2 is -393.509 kJ/mol, and H2O is -241.818 kJ.)

C2H4 (g) + 3O2(g) 2CO2 (g) + 2H2O(g)

ΔHrxn = ______ (-345.64 kJ, -583.07 kJ, or -1,322.91 kJ).

The reaction is: _____ (Endothermic or Exothermic).

Answer 1

We can construct an enthalpy diagram for it:

(please take a look at the picture).

-write the equation, indicate the elements that form the reactants/products. The enthalpy change for formation of elements itself is always 0, so nothing needs to be indicated for O2(g).

*make sure you multiply the enthalpy with the no. of moles for reactants/products with coefficients more than 1 in the equation.

Clockwise arrows = anti-clockwise arrows.

Therefore, by Hess's law,

ΔHrxn + ΔHf [C2H4(g)] = ΔHf [CO2(g)] + ΔHf [H2O(g)]

ΔHrxn + 52.26 = 2 (-393.509) + 2 ( -241.818)

ΔHrxn = -1,322.91 kJ/mol

Since the enthalpy change is negative, the reaction is exothermic.