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BIOCH499A (12)

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School
Department
Biochemistry
Course
BIOCH499A
Professor
Gail Amort- Larson
Semester
Winter

Description
I hope to finish reviewing your exams and watch the Super Bowl on Sunday … Bizarro by Dan Piraro Read 18.1 – 18.5 Practice PS #3 is posted. 1 2 Quantitative chemical equilibrium The equilibrium constant 3.00 mol H 2 (g)nd 6.00 mol F 2 (g)re mixed in a 3.00 L  Determination of K flask. They react to equilibrium to form HF (g) K =  Determination of the position of equilibrium (i.e. all 1.15 x 10 2. concentrations at equilibrium) Find the position of equilibrium (i.e. concentrations).  Simplifying calculations H 2 (g) + F 2 (g)  2 HF (g) Initial1.0002.0.000M Change – x – x + 2x Equil. 1.000 – x 2.000 – x + 2x [HF] 2 (2x)2 2 K = [H 2[F2] = (1.000 – x)(2.000 – x)= 1.15 x 10 3 4 Rearrange and solve: Simplifying calculations 2 111x – 345x + 230 = 0  K is small  K is large This has the form of a quadratic equation K is very small 2 –b  b – 4ac If you add (or subtract) two numbers, one very large ax + bx + c = 0 x = 2a compared with the other one, the result can be a = 111 b = – 345 c = 230 approximated by the larger number. x1= 2.14 M impossible 1–x1is –ve in E (ICE) Large number  small number  large number x2 = 0.968 M Say 100 + 1  100 error is 1% Or 100 – 1  100 error is 1% The position of equilibrium is: Anything below 5% is acceptable When K is very small, the change x is also very small [H 2 = 1.000 – 0.968 = 0.032 M Large/small is relative. Generally K values with [F2] = 2.000 – 0.968 = 1.032 M negative powers of 10 are considered small, while those [HF] = 2 x 0.968 = 1.936 M with positive powers are considered large. But always look for the relative magnitude to other data. 5 6 K is very small K is small  x is small –5 K = 1.6 x 10 for the gas phase rxn of NOCl Approximate 0.50 – 2x  0.50 decomposing to nitric oxide and chlorine. Calculate the –5 4x 3 1.6 x 10 M  0.502 equilibrium concentrations when 0.5 M nitrosyl –2 chloride are introduced into the flask. x = 1 x 10 M N2OCl  2 NO + Cl Position of equilibrium: (g) (g) 2 (g) [NOCl] = 0.48 M I 0.5 0 0 M C – 2x + 2x + x [NO] = 0.02 M [Cl2] = 0.01 M E 0.5 – 2x + 2x + x 2 calc – approx –5 PNO P C2 (2x) x check  x 100% < 5% K = 1.6 x 10 = P 2 = (0.50 – 2x)2  calc  NOCl 0.48 – 0.5 Cubic equation which is hard to solve !!!  0.48  x 100% = 4.2% < 5% valid Remember: under 5% is acceptable 7 8 4 Consider N 2O5 (g) 4 NO 2 (g) O2 (g) –5 (4x) 1.25 K = 1.0 x 10  3.52 K = 1.0 x 10 at a certain temperature. If 2 O5at 3.5 Solve x = 0.025 atm atm is introduced to a container filled wit2 O at 1.25 atm, and the rxn is allowed to reach equilibrium. What is the position of equilibrium? Position of equilibrium: P = 3.5 – 2 (0.025) = 3.45 atm 2 N 2 5 (g) 4 NO 2 (g) O 2 (g) N 2O 5 I 3.5 0 1.25 atm PNO = 4x = 0.10 atm 2 C – 2x + 4x + x PO 2 = 1.25 + x = 1.275 atm E 3.5 – 2x + 4x 1.25 + x P 4P 4 K = 1.0 x 10 = NO 2 22 = (4x)(1.25 +2x) PN 2 5 (3.5 – 2x) 3.45 – 3.50
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