5 Steps to a 5 AP Chemistry

[−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)

−1320.3 −300.4 −237.2 0.0 −689.9 kJ/mol

[(−1320.3) +(−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:

ΔG=ΔG°+RTln Q=ΔG°+2.303 log Q

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

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

Using the previous concept, calculate ΔGfor 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 ΔGf°=ΔGf^500 )

ΔGf° (86.71 0.000 51.84) kJ/mol

ΔGrxn=2(51.84) −[2(86.71) +0.000] =−69.74 kJ/mol

ΔG^500 =ΔGrxn+RTln Q

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 ΔG=0 and the equation above becomes:

ΔG°=−RTln K=−2.303 RTlog K

For example, calculate ΔG°for:

23 Og32p() Og()K =× 41710.^14

Q=

=− +

[]

[[]

(.

NO

NO] O

kJ)(1000J/kJ) 8.314 J

mo

2

2

2

2

69 74

ll K

(500.K )ln

⎛ (1.00)^2

⎝

⎜

⎞

⎠

⎟

=−

(. )(. )

.

2 00 0 500

7 262

2

×× 104 J/mol

130  Step 4. Review the Knowledge You Need to Score High