18.5 Gibbs Free Energy and the Equilibrium Constant
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Gibbs Free Energy Under Nonstandard Conditions
ΔG = ΔG° + RTlnQ (Free energy under non-standard conditions)
Calculate ΔG for the following reaction at 298K when
PN2 = 0.1atm, PH2 = 0.1atm, and PNH3 = 10.0atm.
N2(g) + 3H2(g) → 2NH3(g) ΔG° = -33.3kJ
| ΔG and Q | |
|---|---|
| ΔG < 0 | Q < K |
| ΔG > 0 | Q > K |
| ΔG = 0 | Q = K |
Gibbs Free Energy and the Equilibrium Constant
ΔG° = -RTlnKeq
Calculate the equilibrium constant for the following reaction at 298K.
N2(g) + 3H2(g) → 2NH3(g) ΔG° = -33.3kJ
| ΔG° and Keq | |
|---|---|
| ΔG° < 0 | Keq > 1 |
| ΔG° > 0 | Keq < 1 |
| ΔG° = 0 | Keq = 1 |