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88 3 THE SECOND AND THIRD LAWS
the entropy change of the system plus surroundings to be evaluated. For exam-
ple, at constant volume and temperature the change in the Helmholtz energy
is expressed in terms of the internal energy change and the entropy change
of the system: dA = dU − TdS. If this expression is divided by −T to give
−dA/T = −dU/T + dS the two terms on the right can bothe be identi�ed as
entropy changes.
�e �rst term, −dU/T , is equal to the entropy change of the surroundings
because dqsur = −dq, and at constant volume dq = dU . �e second term is
the entropy change of the system. �us the sum of the two is the total en-
tropy change, which the Second Law shows must be positive in a spontaneous
process. �erefore, the change in the Helmholtz energy is an indicator of the
total entropy change, even though the former refers only to the system. Similar
considerations can be applied to the Gibbs energy.
It is also possible to express the criterion for spontaneity in terms of the change
in H, U or S for the system. For example, as shown in Topic 3D, dSU ,V ≥ 0.
However, the variables which are being held constant (here U and V ) do not
correspond to such easily realizable conditions such as constant temperature
and volume (or pressure) so such criteria are less applicable to chemical sys-
tems.
Solutions to exercises
E3D.1(b) �e standard reaction Gibbs energy is given by [3D.9–100], ∆rG−○ = ∆rH−○ −
T∆rS−○ . �e standard reaction enthalpy is given in terms of the enthalpies
of formation by [2C.5b–55], ∆rH−○ = ∑J νJ∆fH−○(J), where νJ are the signed
stoichiometric numbers.
(i)
∆rH−○ = ∆fH−○(Zn2+ ,(aq)) + ∆fH−○(Cu ,(s))
− ∆fH−○(Zn ,(s)) − ∆fH−○(Cu2+ ,(aq))
= (−153.89 kJmol−1) + 0 − 0 − (+64.77 kJmol−1)
= −218.66 kJmol−1 .
Given the result for the previous execise, ∆rS−○ = −21.0 JK−1mol−1.
∆rG−○ = (−218.66 kJmol−1) − (298.15 K) × (−0.0210 kJK−1mol−1)
= −212.40 kJmol−1 .
(ii)
∆rH−○ = 12∆fH−○(CO2 ,(g)) + 11∆fH−○(H2O ,(l))
− ∆fH−○(sucrose ,(s)) − 12∆fH−○(O2 ,(g))
= 12 × (−393.51 kJmol−1) + 11 × (−285.83 kJmol−1)
− (−2222 kJmol−1) − 12 × 0
= −5644.25 kJmol−1 = −5644 kJmol−1 .