WO3 (s) + 3 H2 (g) -> W (s) + 3 H20 (g)

Question

Tungsten is obtained commercially by the reduction of WO3 with hydrogen according to the equation above. The following data related to this reaction are available.

DeltaH(kilojoules/mole) for WO3 is
-839.5. DeltaH for H20(g) is -241.6.

Delta G for WO3 is -762.7. Delta G for H20 i -228.2

A) What is the value of the equilibrium constant for the system represented above?

B) Calculate Delta S at 25C for the reaction indicated above

Answer 1

A) First, we need to calculate the standard Gibbs free energy change (ΔG°) for the reaction.
ΔG° = ΣΔG°(products) - ΣΔG°(reactants)
ΔG° = [1 × ΔG(W) + 3 × ΔG(H2O)] - [ΔG(WO3) + 3 × ΔG(H2)]
Using the given values:
ΔG° = [1 × 0 + 3 × (-228.2)] - [(-762.7) + 3 × 0] (Note: ΔG° of W or its Grubbs free energy change, since it is a pure element, is zero)
ΔG° = -684.6 J/mol

Now, we can calculate the equilibrium constant (K) using ΔG° and the following equation:
ΔG° = -RT ln K
where R is the gas constant (8.314 J/mol K) and T is the temperature in Kelvin (25°C = 298.15 K)

Rearrange the equation and solve for K:
ln K = -ΔG° / RT
K = exp(-ΔG° / RT)
K = exp(-(-684.6) / (8.314 × 298.15))
K ≈ 2355.6

The value of the equilibrium constant for the system is approximately 2355.6.

B) We can calculate the standard entropy change (ΔS°) using ΔH° and ΔG° with the following equation:
ΔG° = ΔH° - TΔS°

First, we need to calculate the standard enthalpy change (ΔH°) for the reaction.
ΔH° = ΣΔH°(products) - ΣΔH°(reactants)
ΔH° = [1 × ΔH(W) + 3 × ΔH(H2O)] - [ΔH(WO3) + 3 × ΔH(H2)]
Using the given values:
ΔH° = [1 × 0 + 3 × (-241.6)] - [(-839.5) + 3 × 0] (Note: ΔH° of W, since it is apure element, is zero)
ΔH° = -88.3 kJ/mol

Now, we can solve for ΔS° using ΔG° and ΔH°:
ΔS° = (ΔH° - ΔG°) / T
ΔS° = ((-88.3 × 1000) - (-684.6)) / 298.15 (Note: Convert ΔH° to J/mol by multiplying by 1000)
ΔS° ≈ -294.9 J/mol K

The change in entropy ΔS at 25°C for the reaction is approximately -294.9 J/mol K.