(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