Consider the following reaction 2SO2 (g) + O2 (g) = 2SO3 (g) the free energies o
ID: 818683 • Letter: C
Question
Consider the following reaction 2SO2 (g) + O2 (g) = 2SO3 (g)
the free energies of formation at 25 C are given below
SO2 = -300
SO3 = -371
a) Is entropy change for the reaction negative, positive, or 0? explain.
b) Is the enthalpy change for the reaction negative, positive, or zero? explain.
c) Is the reaction spontaneous at 25 degree. Explain why or why not.
d) Is the reaction spontaneous at all temperatures? Explain your answer.
e) Is the value of the equilibrium constant for the reaction at 298 K less than one, zero, or greater than one. Show your resoning.
Explanation / Answer
2 SO2(g) + O2(g) => 2 SO3 (g)
(a) Since moles of gas decreases => entropy decreases
=> final entropy < initial entropy
=> entropy change DS = final - initial entropy < 0
=> entropy change is negative
(b) Free energy change DG = DGf(products) - DGf(reactants)
= 2 x DGf(SO3) - 2 x DGf(SO2) - DGf(O2)
= 2 x (-371) - 2 x (-300) - 0
= -142 kJ
DG = DH - T x DS = -142 kJ
Since DS < 0 and T > 0 => DH < 0
=> enthalpy change is negative
(c) Since DG = -142 kJ is negative at 25 deg C => reaction is spontaneous at 25 deg C
(d) For spontaneous reaction: DG = DH - T x DS < 0
=> T < DH/DS
=> reaction is NOT spontaneous at all temperatures (only spontaneous when T is less than DH/DS)
(e) DG = -RT ln K
K = exp(-DG/RT) = exp(142 x 1000/(8.314 x 298) = 7.78 x 10^24 > 1
Thus equilibrium constant K is greater than 1