Chemical Equilibrium Chapter 15

Factors that Affect Chemical Equilibrium Changes in Concentration Changes in Pressure/Volume Changes in Temperature Effect of Catalysts

Le Châtelier’s Principle ChangeShift Equilibrium Change Equilibrium Constant Concentrationyesno Pressure/volume yesno Temperature yes Catalystno

At 1000 K, the equilibrium constant (K c ) for the reaction is 3.8 x Suppose you start with moles I 2 in a 2.30 L flask at 1000 K. What are the concentrations of these gases at equilibrium? I 2 (g) 2 I (g) Let x be the change in concentration of I 2 Initial (M) Change (M) Equilibrium (M) x-x+2x ( – x)(2x) [I]2[I]2 [I2][I2] K c = (2x) 2 ( – x) = 3.8 x Solve for x #15.34 p 504

4x x x x = 0 ax 2 + bx + c = 0 -b ± b 2 – 4ac  2a2a x = x = 4.29 x ; x At equilibrium, [ I ] = 2x = 8.58 x M At equilibrium, [ I 2 ] = – x = 1.94 x M I 2 (g) 2 I (g) Initial (M) Change (M) Equilibrium (M) x-x+2x ( – x)(2x) K c = (2x) 2 ( – x) = 3.8 x x x x – 1.88 x = 0

Exam #3 Covers Chapters 14 and Multiple choice 60% qualitative 40% quantitative All equations provided Constants & Periodic Table provided

Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- (aq) + 3I - (aq) 2SO 4 2- (aq) + I 3 - (aq) Experiment [S 2 O 8 2- ][I - ] Initial Rate (M/s) x x x rate = k [S 2 O 8 2- ] x [I - ] y Double [I - ], rate doubles (experiment 1 & 2) y = 1 Double [S 2 O 8 2- ], rate doubles (experiment 2 & 3) x = 1 k = rate [S 2 O 8 2- ][I - ] = 2.2 x M/s (0.08 M)(0.034 M) = 0.08/M s rate = k [S 2 O 8 2- ][I - ]

Sample Exercise Determining the Rate Law for a Multistep Mechanism The decomposition of nitrous oxide, N 2 O, is believed to occur by a two-step mechanism: (a) Write the equation for the overall reaction. (b) Is the proposed mechanism consistent with the equation for the overall reaction? (c) Write the rate law for the overall reaction. (d) What is the molecularity of each step of the mechanism? (e) Identify the intermediate(s).

Practice Exercise A 2.00 L flask is filled with 1.00 mol of H 2 and 4.00 mol of I 2 at 720 K. At this temperature, K c for this reaction is H 2 (g) + I 2 (g) ⇌ 2 HI (g) Calculate the equilibrium concentrations of all species at this temperature. Ans: [H 2 ] = M [I 2 ] = M [HI] = 1.87 M