CHM 2046C Module Ten-Chapter 16 Sample Test
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Module Ten: Chemical Equilibria Chapter 16 |
Possible |
Actual |
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A.
Equilibrium Constant Derivation from Reaction Rates Lecture |
10 |
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B.
Writing Equilibrium Constant Expressions Section 16.2 p760-1 |
10 |
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C.
Meaning of the Equilibrium
Constant: K; Section 16.2 p765-7 |
10 |
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D.
Meaning of the Equilibrium
Quotient: Q; Section 16.2 p767-9 |
10 |
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E.
Determination of Equilibrium Constants from Lab Data Section 16.3-16.4 |
10 |
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10 |
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G.
Determination of Equilibrium Conc from Kc Problems Section 16.4-16.5 |
10 |
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10 |
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10 |
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J.
Key Terms - Chapter 16 & More about Equations & Eq Coefficients 16.5 |
10 |
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25 |
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Module Ten Total: |
125 |
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Part A: Equilibrium Constant Derivation 10 points
Derive the equilibrium constant expression
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[C]c [D]d |
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Kc = --------------- |
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[A]a [B]b |
from the rate expressions of the following reversible reaction:
aA + bB ç è cC +
dD
Part B: Equilibrium Constant Expressions 10 points
Write equilibrium constant expressions, Kc, for the following reactions they represent:
a. PCl3 (g) + Cl2 (g) ç è PCl5 (g)
Kc =
b. 2 NOCl (g) ç è 2 NO (g) + Cl2 (g)
Kc =
c. 4 HCl (g) +
O2 (g) ç è 2 H2O (g) +
2 Cl2 (g)
Kc =
d. CS2 (g) + H2 (g) ç è CH4 (g) + H2S (g)
Kc =
e. CaCO3 (s) ç è CaO (s) + CO2 (g)
Kc =
f. NH3
(g) + H2O (l) ç è NH4
1+(aq) + OH1- (aq)
Kc =