WO3(s) + 3 H2(g) W(s) + 3 H2O(g)?

Question

Tungsten is usually produced by the reduction of WO3 with hydrogen.
WO3(s) + 3 H2(g) W(s) + 3 H2O(g)

Consider the following data.

WO3(s) H2O(g)
H°f (kJ/mol) -839.9 -241.8
G°f (kJ/mol) -763.1 -228.6


(a) Is K > 1 or <1 at 25°C?
K < 1
K > 1
Explain your answer.

(b) What is the value of S° at 25°C?
J/K
(c) What is the temperature at which G° equals zero for this reaction at 1 atm pressure?
K
(d) What is the driving force of this reaction?

Answer 1

K<1

234

324

Increased entropy.

Answer 2

Actually, the above answer is incorrect. K is greater than one because the heat of entropy is more than 50% random (G). As the temperature increases, however, K will approach one (at roughly 50 degrees celsius) and become negative around this same value.

As for the second part of your question, the value of S at 25 is derived from this equation:

S = (deltaTf - delta Ti)/G of reactants.

Hope that helps!!