Cliffs AP Chemistry, 3rd Edition

(d) Given: ∆HfusHCOOH = 12.71 kJ/mole at 8.3°C

Restatement: Calculate ∆Sfor the reaction

HCOOH(,) →HCOOH(s)

The temperature at which liquid HCOOH converts to solid HCOOH is known as the

freezing point; it is also the melting point. Because at this particular temperature a state

of equilibrium exists — that is, HCOOH(,) ↔HCOOH(s) — you can set ∆G= 0.

Substituting into the Gibbs-Helmholtz equation yields

∆G = ∆H −T∆S

0= −12.71 kJ / mole −281.3 K(∆S)

(Did you remember to make 12.71 negative, because you want ∆H for freezing, which is

an exothermic process?)

.

S ./

K

∆ kJ mole

281 3

=-12 71

= −0.04518 kJ / mole ⋅K = −45.18 J / mole ⋅K

(e) (1) Given: S°HCOOH(,) = 109.1 J / mole ⋅K

Restatement: What is the standard molar entropy of HCOOH(s)?

HCOOH(,) →HCOOH(s)

∆S°= Σ∆S°products−Σ∆S°reactants

= S°HCOOH(s) −S°HCOOH(l)

−45.18 J/mole ⋅K = S°HCOOH(s) −109.1 J/mole ⋅K

S°HCOOH(s) = 63.9 J/mole⋅K

(2) Restatement: Is magnitude of S°HCOOH(s) in agreement with magnitude of S°

HCOOH(,)?

The magnitude of S° HCOOH(s) is in agreement with the magnitude of S°HCOOH(,) be-

cause the greater the value of S°, the greater the disorder; the liquid phase has higher en-

tropy than the solid phase.

(f) Given: (Ka= 1.9 × 10 –4)

Restatement: ∆G°for the ionization of methanoic (formic) acid at 25°C.

∆G°= −2.303R ⋅T logKa R= 8.314 J/K

= −2.303(8.314 J/K)⋅298 K(−3.72)

= 2.1 × 104 J = 21 kJ

Part IV: AP Chemistry Practice Test