Cliffs AP Chemistry, 3rd Edition

Part A: Question 3

3. Given: CH 3 OH(aq) + O 2 (g) →HCOOH(aq) + H 2 O(,). Data shown in the table
   (a) Restatement: ∆H°for the oxidation of methyl alcohol.
   ∆H°fHCOOH(aq) = −409 kJ/mole
   ∆H°fH 2 O(l) = −285.84 kJ/mole
   ∆H°fCH 3 OH(aq) = −238.6 kJ/mole
   ∆H°= Σ∆H°fproducts−Σ∆H°freactants
   = (∆H°fHCOOH + ∆H°fH 2 O) −(∆H°fCH 3 OH)
   = [(−409 kJ/mole) + (−285.84 kJ/mole)] – (238.6 kJ/mole)
   = −456 kJ/mole

(b) Restatement: ∆S°for the oxidation of methyl alcohol.

S°HCOOH(aq) = 129 J/mole ⋅K

S°H 2 O(l) = 69.94 J/mole ⋅K

S°CH 3 OH(aq) = 127.0 J/mole ⋅K

S°O 2 (g) = 205.0 J/mole ⋅K

∆S°=ΣS°products−ΣS°reactants

= (S°HCOOH + S°H 2 O) −(S°CH 3 OH + S°O 2 )

= (129 J/K ⋅mole + 69.94 J/K ⋅mole) −(127.0 J/K ⋅mole + 205.0 J/K ⋅mole)

= −133 J/K ⋅mole

(c) (1) Restatement: Is the reaction spontaneous at 25°C?

Explain.

∆G°= ∆H°−T∆S

= −456 kJ/mole −298 K(−0.133 kJ/mole ⋅K)

= −416 kJ/mole

Reaction is spontaneous because ∆G° is negative.

(2) Restatement: If the temperature were increased to 100°C, would the reaction be

spontaneous? Explain.

∆G = ∆H−T∆S

= −456 kJ/mole −373 K(−0.133 kJ/mole ⋅K)

= −406 kJ/mole

Reaction is spontaneous because ∆G is still negative.

Answers and Explanations for the Practice Test