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BIOCH499A
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Gail Amort- Larson
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Biochemistry

BIOCH499A

Gail Amort- Larson

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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