Consider the following reaction shown below and the data in the table for the re

ID: 1003108 • Letter: C

Question

Consider the following reaction shown below and the data in the table for the reduction of tungsten(VI) oxide with hydrogen.

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

a) Is K greater than 1 or less than 1 at 25°C? Explain.

b) What is the value of S° at 25°C? Explain.

c) What is the temperature at which G° equals zero for this reaction at 1 atm? Show or explain how you got your answer.

d) What is the driving force of this reaction, in terms of thermodynamics? (that is, which thermodynamic term, S or H, makes spontaneity possible?)

WO3 (s) H2O (g) Hf° (kJ/mol) -839.9 -241.8 Gf° (kJ/mol) -763.1 -228.6

(a)

Delta G of reaction = delta G of products - delta G of reactants

=(0 + 3 x (-228.6) ) - (-763.1)

= 77.3 kJ/mol

We know that,

delta G = -RTlnK

77.3 kJ/mol = -8.314 J/mol K x 298 K x ln K

K =2.18 x 10-14

K value is less than 1

(b)

delta H of reaction = (3 x -241.6) -(-839.9) =114.5 kJ/mol

delta G = delta H - Txdelta S

delta S = (delta H -delta G)/T

= (114.5 - 77.3 )/298 K

= 0.124 kJ/mol K or 124 J/mol K

(c)

delta G = delta H - T delta S

0 = 114.5 kJ/mol - T x 0.124 kJ/mol

T = 923.4 K

(D) Spontainity is possible only by large values of delta S x T term

So driving force is delta S