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Chemical Equilibrium Chapter 15. Factors that Affect Chemical Equilibrium Changes in Concentration Changes in Pressure/Volume Changes in Temperature Effect.

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Presentation on theme: "Chemical Equilibrium Chapter 15. Factors that Affect Chemical Equilibrium Changes in Concentration Changes in Pressure/Volume Changes in Temperature Effect."— Presentation transcript:

1 Chemical Equilibrium Chapter 15

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

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

4 At 1000 K, the equilibrium constant (K c ) for the reaction is 3.8 x 10 -5. Suppose you start with 0.0456 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) 0.01980 -x-x+2x (0.0198 – x)(2x) [I]2[I]2 [I2][I2] K c = (2x) 2 (0.0198 – x) = 3.8 x 10 -5 Solve for x #15.34 p 504

5 4x 2 + 3.8 x 10 -5 x - 7.52 x 10 -7 = 0 ax 2 + bx + c = 0 -b ± b 2 – 4ac  2a2a x = x = 4.29 x 10 -4 ; -4.38 x 10 -4 At equilibrium, [ I ] = 2x = 8.58 x 10 -4 M At equilibrium, [ I 2 ] = 0.0198 – x = 1.94 x 10 -2 M I 2 (g) 2 I (g) Initial (M) Change (M) Equilibrium (M) 0.01980 -x-x+2x (0.0198 – x)(2x) K c = (2x) 2 (0.0198 – x) = 3.8 x 10 -5 x 2 + 9.50 x 10 -6 x – 1.88 x 10 -7 = 0

6 Exam #3 Covers Chapters 14 and 15 34 Multiple choice 60% qualitative 40% quantitative All equations provided Constants & Periodic Table provided

7 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) 10.080.0342.2 x 10 -4 20.080.0171.1 x 10 -4 30.160.0172.2 x 10 -4 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 10 -4 M/s (0.08 M)(0.034 M) = 0.08/M s rate = k [S 2 O 8 2- ][I - ]

8 Sample Exercise 14.14 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).

9 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 50.5. H 2 (g) + I 2 (g) ⇌ 2 HI (g) Calculate the equilibrium concentrations of all species at this temperature. Ans: [H 2 ] = 0.065 M [I 2 ] = 1.065 M [HI] = 1.87 M


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