Northern Arizona University
CHM 152, General Chemistry II
Sections 2 and 3, Spring 2004
Quiz #4
Dr. Brandon Cruickshank
February 18, 2004
Name _____________________________
1.
True/False Statements.
a)
[2 pts each]
At equilibrium, the total concentration of products equals the total concentration of reactants, that is
[products] = [reactants]
False. At equilibrium, the concentrations of the reactants and products are no longer changing with
time.
b) Equilibrium is the result of the termination of all chemical change.
False. Equilibrium is dynamic. Reactants are going to products and products are going to reactants.
The rates of the forward and reverse reactions are equal.
c)
When the value of Kc for a reaction is very large, it indicates that the equilibrium concentration of the
products are large compared to the equilibrium concentrations of reactants.
TRUE
d) In many cases, a catalyst increases the rate of a reaction by decreasing the activation energy.
TRUE
2.
For the following equation, write the expressions for both the equilibrium constants Kc and Kp.
[4 pts]
C (s) + 2 N2O (g) U CO2 (g) + 2 N2 (g)
Kc =
3.
[CO2 ][N 2 ]2
[N 2 O]2
Kp =
PCO2 PN2
PN2
2
2O
For the reaction, A + 2B → products, the experimentally determined rate law is: Rate = k[A][B].
If I indicates an intermediate and P indicates a product, which of the following is the most likely reaction
mechanism? [4 pts]
A + B → I (slow)
I → P (fast)
b) B + B → I (slow)
A + I → P (fast)
c) A + B → I (fast)
I + B → P (slow)
d) A + B → I (slow)
I + B → P (fast)
e) B + B → I (fast)
A + I → P (slow)
a)
4.
Consider the following reaction:
2 NO (g) + Cl2 (g) U 2 NOCl (g)
a)
4
The value of Kp for this reaction is 2.5 × 10 at 37°C. What is the value of Kc for this reaction at 37°C?
∆n
Kp = Kc(RT)
4
−1
2.5 × 10 = Kc[(0.0821)(310 K)]
5
Kc = 6.4 × 10
b) For the following reaction at 37°C, what is the value of Kp?
[2 pts]
2 NOCl (g) U 2 NO (g) + Cl2 (g)
1
Kp
K p' =
1
K p' =
5.
2.5 × 104
= 4.0 × 10−5
Consider the following equilibrium process at 700°C:
2 H2 (g) + S2 (g) U 2 H2S (g)
−5
Analysis shows that there are 2.50 moles of H2 and 1.35 × 10 mole of S2 present in a 9.00 L flask at
6
equilibrium. If the equilibrium constant Kc for the reaction is 5.58 × 10 , what is the concentration of H2S at
equilibrium?
[5 pts]
2.50 mol
= 0.278 M
9.00 L
[H 2 ] =
1.35 × 10−5 mol
= 1.50 × 10−6 M
9.00 L
[S2 ] =
K =
[H 2S]2
[H 2 ]2 [S2 ]
5.58 × 106 =
[H 2S]2
(0.278 M ) 2 (1.50 × 10−6 M )
[H 2S]2 = 0.647
[H 2S] = 0.804 M
Potentially Useful Information
R = 0.0821 L⋅atm/mol⋅K
Kp = Kc(RT)
∆n
K = °C + 273
molarity( M ) =
moles of solute
L of solution
[3 pts]
PV = nRT
K=
[products]x
[reactants]y