Cliffs AP Chemistry, 3rd Edition

Answer: B

∆G°= Σ∆G°products−Σ∆G°reactants

=[2(−237.00) + (−394.00)] −(–51.00)

= −817.00 kJ/mole

4. Calculate the approximate standard free energy change for the ionization of hydrofluoric
   acid, HF (Ka= 1.0 × 10 –3), at 25°C.

A. −9.0 kJ

B. −4.0 kJ

C. 0.050 kJ

D. 4.0 kJ

E. 17 kJ

Answer: E

At equilibrium, ∆G= 0 = ∆G°+ 2.303 RT log K(at equilibrium Q = K).

∆G°= −2.303(8.314 J ⋅K–1)(298 K)(log 1.0 × 10 –3)

Rounding,

~ −2.3(8.3)(300)(−3.0) = 17,181 J ≈17 kJ

5. Arrange the following reactions according to increasing ∆S°rxnvalues.
   
   
   1. H 2 O(g) →H 2 O(,)
   
   
   2. 2 HCl(g) →H 2 (g) + Cl 2 (g)
   
   
   3. SiO 2 (s) →Si(s) + O 2 (g)
   
   
   
   

lowest → highest

A. ∆S°(1) < ∆S°(2) < ∆S°(3)

B. ∆S°(2) < ∆S°(3) < ∆S°(1)

C. ∆S°(3) < ∆S°(1) < ∆S°(2)

D. ∆S°(1) < ∆S°(3) < ∆S°(2)

E. ∆S°(3) < ∆S°(2) < ∆S°(1)

Energy and Spontaneity