Cliffs AP Chemistry, 3rd Edition

Step 2: Solve for ∆S°.

FeO(s) + CO(g) →Fe(s) + CO 2 (g)

∆S°= Σ∆S°products−Σ∆S°reactants

= (27.0 + 214.0) −(61.0 + 190.0) = −10.0 J ⋅K–1⋅mole–1

Step 3: Substitute ∆S°and ∆H°into the Gibbs-Helmholtz equation.

∆G°= ∆H°−T∆S°

=−16 kJ/mole −298 K(−0.0100 kJ ⋅K–1⋅mole–1)

≈−13 kJ/mole

8. Given the balanced equation

H 2 (g) + F 2 (g) ↔2 HF(g)∆G°= −546 kJ/mole

Calculate ∆Gif the pressures were changed from the standard 1 atm to the following

and the temperature was changed to 500°C.

H 2 (g) = 0.50 atm F 2 (g) = 2.00 atm HF(g) = 1.00 atm

A. −1090 kJ/mole

B. −546 kJ/mole

C. −273 kJ/mole

D. 546 kJ/mole

E. 1090 kJ/mole

Answer: B

Realize that you will need to use the equation

∆G = ∆G°+ RTln Q

Step I: Solve for the reaction quotient, Q.

..

.

Q PP.

P

050 200

100

100

HF

HF^22

22

== =

^^

^

^^

^

hh

h

hh

h

ln 1.00 = 0

Step 2:Substitute into the equation.

∆G = ∆G°+ RTlnQ

= −546,000 J + (8.3148 J ⋅K–1⋅mole–1) ⋅773 K(0)

=−546 kJ/mole

Part II: Specific Topics