StudentLecture_11 - Housekeeping Details Reading Chapter 16 Reading for next time Sect 10.1-10.4 CH 17 3.7 4.6-4.7 Lab Chapter 7 Bromocresol Green

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Housekeeping Details Reading: Chapter 16 Reading for next time: Sect. 10.1-10.4; CH 17; 3.7, 4.6-4.7 Lab: Chapter 7, Bromocresol Green Equilibrium Systems HW6: Due as usual. Final Exam: Tues 05/03, 7:00-9:00 pm, Lambert
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Summary of Steps Step 1: Determine the direction the reaction shifts to come to equilibrium. Step 2: Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations or pressures in terms of one change, . Step 3: Solve for and the equilibrium concentrations or pressures. Step 4: Check the assumptions and arithmetic.
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Example 2 Determine the concentrations when a mixture of F 2 , and H 2 with [F 2 ] i = 0.050 M, [H 2 ] i = 0.100 M come to equilibrium at an elevated temperature. H 2 (g) + F 2 (g) = 2 HF(g) K = 115 Which direction will the reaction shift? Q = [HF] 2 [H 2 ][F 2 ] = ________________________ K > Q but _____________________________ The change will ____________________ What happens if we assume that is small? Look out for this!
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Assuming is small H 2 (g) + F 2 (g) = 2 HF(g) K = 115 [F 2 ] i = 0.050 M, [H 2 ] i = 0.100 M, and [HF] i = 0 M H 2 (g) + F 2 (g) = 2 HF(g) [ ] i 0.100 0.050 0 - - 2 [ ] eq 0.100 - 0.050 - 2 Assume is small. K = [HF] 2 [H 2 ][F 2 ] = (2 ) 2 (0.100)(0.050) = 115 __________ Shift ________________________
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A Simplifying Technique If K < 1 and Q > 1 or K > 1 and Q < 1 ASSUME the reaction gives a ______________ as it _______________ and then it comes to _________________________.
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After 100% shift [ ] 100% ______ ____ _____ K ___ Q ____ K___Q H 2 (g) + F 2 (g) = 2 HF(g) K = 115 [F 2 ] i = 0.050 M, [H 2 ] i = 0.100 M, and [HF] i = 0 M H 2 (g) + F 2 (g) = 2 HF(g) Initial concs. 0.100 0.050 0 K > 1 Q < 1 A Simplifying Technique ______ ____ _____ [ ] eq ______ ____ _____
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= 1.74 x 10 -3 A Simplifying Technique (0.100 - 2 ) 2 (0.050 + )( ) = 115 Assume is small. [H 2 (g)] = ________________ = 0.050 + 0.00174 = 0.048 [F 2 (g)] = = 1.74 x 10 -3 [HF(g)] = ________________ = 0.100 - 0.00348 = 0.097 __________________ x 100 = 3.5 % Check #1: ____________ [ ] 100%
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Check #2 H 2 (g) + F 2 (g) = 2 HF(g) K = 115 K = [HF] 2 [H 2 ][F 2 ] = 115 (0.097) 2 (0.048)(1.74 x 10 -3 ) = 113 Agrees ± 2 in last significant figure
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This note was uploaded on 02/27/2012 for the course CHEM 116 taught by Professor Stevenson during the Spring '08 term at Purdue University-West Lafayette.

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StudentLecture_11 - Housekeeping Details Reading Chapter 16 Reading for next time Sect 10.1-10.4 CH 17 3.7 4.6-4.7 Lab Chapter 7 Bromocresol Green

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