Chem 300 - Ch 26/#2 Today’s To Do List

1. Chem 300 - Ch 26/#2 Today’s To Do List • Calculating KP from ΔrGo • ΔrG vs ΔrGo • Reaction Direction • van’t Hoff Equation • Temperature & rHo

2. Calculating KP • rGo = - RT ln KP(T) • N2(g) + 3H2(g)  2NH3(g) @ 298K • rGo = 2 fGo(NH3) – [fGo(N2) + 3 fGo(H2)] • rGo = 2 fGo(NH3) – [ 0 + 0 ] • rGo = 2 ( - 16.5 kJ/mol) = -33.0 kJ/mol • RT = (8.31 J/mol-K)(298 K) = 2.48 kJ/mol • ln KP = - (-33.0kJ/mol)/(2.48kJ/mol) = 13.3 • KP = 6.0 x 105Thermodynamically exact

3. Degree of Dissociation (α) • H2O  H2 + ½ O2 @ 2300 K • rGo(2300K) = 118.1 kJ/mol • First Calculate KP-- • ln KP = - rGo/RT = - 118.1/(8.31x2300) • ln KP = - 6.17  KP = 2.08 x 10-3 • Calculate α

4. H2O  H2 + ½ O2

5. If ideal gases • KP = P(H2) P½(O2)/P(H2O) • Subst from prev. slide: • KP = α3/2 P1/2/(1- α)(2 + α)1/2 = 2.08 x 10-3 • Thus α is small: α << 1 • KP  α3/2 P1/2/ 21/2 Since P = 1 • α  0.0205 small dissociation

6. Reaction Direction • rG(T) = rGo(T) + RT ln Q • rGo(T) = - RT ln KP • rG(T) = - RT ln KP + RT ln Q = RT ln (Q/KP) • If (Q/KP) < 1  rG(T) < 0 Rxn spontan. • If (Q/KP) > 1  rG(T) > 0 Not spontan. • If (Q/KP) = 1  rG(T) = 0 At equilibr.

7. Example • 2 SO2 + O2 2 SO3 KP = 10 @ 900K • Initial Conditions: • P(SO2) = 0.20 bar • P(O2) = 0.20 bar • P(SO3) = 0.10 bar • Q = P2(SO3) / P2(SO2) P(O2) =(0.01)/(0.04)(0.20) = 1.25 < KP Spontan. • rG(T) = RT ln (Q/KP) = (8.31)(900) ln (1.25/10) • rG(T) = - 15.6 kJ

8. KP, T, @ van’t Hoff • Recall Gibbs-Helmholtz: • [(rGo/T)/ T]P = - rHo/T2 • Subst rGo(T) = - RT ln KP(T) • [(ln KP/ T]P = rHo/RT2 • If rHo> 0  KP(T) increases with T • If rHo< 0  KP(T) decreases with T • Integrate: • ln (K2/K1) = - rHo(1/T2 – 1/T1)/R

9. H2 + CO2 = CO + H2OEndothermic Rxn

10. When rHo not constant3H2 + N2= 2NH3

11. Le Chatelier’s Principle • If a chemical reaction at equilibrium is subjected to a disturbance, it responds in a way that tends to minimize the effect.

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