MQ-2 on Monday, Feb. 21 at 6:30 pm Covering Chapters 13 (osmosis…), 14, and 15

MQ-2 on Monday, Feb. 21 at 6:30 pm

Covering Chapters 13 (osmosis…), 14, and 15

Review session on Sunday, Feb 20at 1:00 pm – 3:00 pm

in Stillman 100

Alternate Exam time for studentswith scheduled class conflicts:

5:00 – 6:20 pm in 159 MacQuigg Labssign up by email to mathews

Second MQ exam – Chem 122Monday, 21 February

6:30 pmLAB INSTRUCTOR LOCATION

Christopher Beekman* 250 Knowlton Hall

Chitanya Patwardhan “

Ramesh Sharma “

Mark Lobas 180 Hagerty Hall

Edwin Motari* “

Roxana Sierra “

Lin Sun “

All Others 131 Hitchcock Hall

Knowlton Hall - 2073 Neil AvenueHitchcock Hall - 2070 Neil AvenueHagerty Hall - 1775 College Rd

Chapter 15 Chemical Equilibrium

15.1 The Concept of Equilibrium

15.2 The Equilibrium ConstantThe Magnitude of Equilibrium ConstantsThe Direction of the Chemical Equation and Keq

Other ways to Manipulate Chemical Equations and Keq

Units of Equilibrium Constants– Kp and Kc

15.3 Heterogeneous Equilibria

15.4 Calculating Equilibrium Constants

15.5 Applications of Equilibrium ConstantsPredicting the Direction of Reaction

Calculation of Equilibrium Concentrations

15.6 Le Châtelier’s PrincipleChange in Reactant or Product ConcentrationsEffects of Volume and Pressure ChangesEffect of Temperature ChangesThe Effect of Catalysts

Completion of Chapter 15:

15.6 Le Châtelier’s PrincipleChange in Reactant or Product ConcentrationsEffects of Volume and Pressure ChangesEffect of Temperature ChangesThe Effect of Catalysts

Recall from last week:

CO(g) + Cl2(g) = COCl2(g)

2 H2S(g) = 2 H2(g) + S2(g)

C(s) + S2(g) = CS2(g)

C(s) + CO2(g) = 2 CO(g)

What happens if we increase PT ???

Consider effect of Change in Temperature

Exothermic reaction:

CO + 3 H2 = CH4 + H2O ΔH = -206 kJ/mol

or

CO + 3 H2 = CH4 + H2O + 206 kJ

therefore incr temp shifts eq to left!

ifi

f

TTR

H

K

K 11ln

Where Kf is the equilibrium constant at temperature Tf andKi is the equilibrium constant at temp. Ti .

Example: Consider a reaction which has Keq = 1.38 x 105 at 800 Kand Keq = 4.9 x 1027 at 298 K.

What is ΔH for the reaction?

molkJmolJxx

H

KKKmolJ

H

x

x

/214/1014.21011.2

84.450

298

1

800

1

)314.8(109.4

1038.1ln

53

1127

5