5 Steps to a 5 AP Chemistry 2019

Thermodynamics ❮ 129

[-750.2 + (-237.2) + 2(-16.6)] - [2(-203.9) + (-604.2)]

= - 8.6 kJ/mole

b. C 2 H 5 OH(l) C 2 H 4 (g) H 2 O(g)

• 174.18 68.12 -228.6


• 174.18 - [68.12 + (-228.6)] = - 13.7 kJ/mol

c. CaSO 4 (s) SO 2 (g) H 2 O(l) Ca(s) H 2 SO 4 (l)

• 1,320.3 -300.4 -237.2 0.0 -689.9 kJ/mol
  [(-1,320.3) + (-300.4) + 2(-237.2)] - [(0.0) + 2(-689.9)]
  = - 715.3 kJ/mol

Thermodynamics and Equilibrium

Thus far, we have considered only situations under standard conditions. But how do we

cope with nonstandard conditions? The change in Gibbs free energy under nonstandard

conditions is:

DG = DG° + RT ln Q = DG° + 2.303 log Q

Q is the activity quotient, products over reactants. This equation allows the calculation

of DG in those situations in which the concentrations or pressures are not 1.

Using the previous concept, calculate DG for the following at 500.K:

2 NO(g) + O 2 (g) → 2 NO 2 (g)

2.00 M 0.500 M 1.00 M

(Assume DGf° = DGf^500 )

DGf° (86.71 0.000 51.84) kJ/mol

DGrxn = 2(51.84) - [2(86.71) + 0.000] = - 69.74 kJ/mol

DG^500 = DGrxn + RT ln Q

=

=− +













=− ×

Q

[NO]

[NO] [O ]

( 69.74kJ)(1,000 J/kJ) 8.314

J

molK

(500.K)ln

(1.00)

(2.00)(0.500)

7.262 10 J/mol

2

2 2 2 2 2 4

Note that Q, when at equilibrium, becomes K. This equation gives us a way to calculate the

equilibrium constant, K, from a knowledge of the standard Gibbs free energy of the reaction

and the temperature.

If the system is at equilibrium, then DG = 0 and the equation above becomes:

DG° = - RT ln K = - 2.303 RT log K

For example, calculate DG° for:

2O (g)3 32  O(g) Kp=×4.17 1014