Question

Using the following info and the standard enthalpy of formation for CO2(g) (-393.5KJ/mol), determine the standard enthalpy of formation of tungsten carbide, WC(s):

WC(s)+ 5/2 O2(g) --> WO3(s) + CO2(g) ∆H° = -285.80Kcal

W(s) + 3/2 O2(g) --> WO3(s) ∆H° = -200.16Kcal

Answer 1

This equation must be written for one mole of WC(g). In this case, the reference forms of the constituent elements are O₂(g) and tungsten (W).

WC(s)+ 5/2 O₂(g) --> WO₃(s) + CO₂(g)   ∆H° = -285.80Kcal

The general equation for the standard enthalpy change of formation is given below:

ΔH^oreaction=∑ΔH^of(products)−∑ΔH^of(Reactants)

Plugging in the equation for the formation of WO_{3 and} CO₂ gives the following:

ΔH_reaction^o=( ΔH_f^o[WO₃(s) + CO₂(g)]) - (ΔH_f^o[O₂(g)] + ΔH_f^o[WC(s)])….(1)

Because O₂(g) is in the most elementally stable forms, they each have a standard enthalpy of formation equal to 0

ΔH_f = -393.5 kJ/mol CO₂(g)

ΔH_reaction^o = -285.80Kcal = -1196.587 kJ

Putting these values in (1)

-1196.587 = ( ΔH_f^o[WO₃(s)] -393.5)-(0+ ΔH_f^o[WC(s)]

-803.087= ( ΔH_f^o[WO₃(s)])- (ΔH_f^o[WC(s)])………..(2)

Second Equation

W(s) + 3/2 O2(g) --> WO3(s)   ∆H° = -200.16 Kcal

Plugging in the equation for the formation of WO₃

ΔH_reaction^o= ΔH_f^o[WO₃(s)] - (ΔH_f^o[O₂(g)] + ΔH_f^o[W(s)])

Because O₂(g) and W(s) are in their most elementally stable forms, they each have a standard enthalpy of formation equal to 0

ΔH_reaction^o = -200.16 Kcal = -838.03 kJ

So, ΔH_f^o[WO₃(s)]= -838.03 KJ

Putting this value in (2)

-803.087= ( -838.03)- (ΔH_f^o[WC(s)])

(ΔH_f^o[WC(s)])= ( -838.03) + 803.087

                    ΔH_f^o[WC(s)] = -34.94 KJ