2 AB(g) Kc = 16
3.00 mol A2, 3.00 mol B2, and 3.00 mol AB added to a 1.00 L container. (8)
Calculate [A2], [B2], and [AB] at equilibrium.
1. Run [A]
[B]
[C]
Rate
1
1.00 M 1.00 M 1.00
M 0.150 M/s
2
2.00 2.00
2.00 4.80
3
1.50 1.00
1.00 0.338
4
1.50 3.00
1.00 9.13
(a) Write rate equation and describe how orders determined.
(6)
(b) Calculate rate constant.
(3)
(c) Determine rate when [A] = [B] = [C] = 3.00M
(3)
2. Reaction first-order in A.
(15)
(a) Calculate k if 30.0% A remains after 10.0 min.
(b) Calculate time needed for 95.0% of A to react.
(c) Calculate % A remaining after 17.5 min.
3. At 300K the rate constant is 0.100 s-1 and at 350K
the rate constant is 0.800 s-1. (10)
(a) Determine activation energy.
(b) Determine rate constant at 400 K.
4. A2(g) + B2(g)
2 AB(g) Kc = 16
3.00 mol A2, 3.00 mol
B2, and 3.00 mol AB added to a 1.00 L container.
(8)
Calculate [A2], [B2],
and [AB] at equilibrium.
5. A(g) + B(g)
AB(g) T = 300K
(a) 2.00 mol each A, B, and AB added
to a 1.00 L container. Calculate [A], [B], and [AB]
when equilibrium reached (Kc = 1.0).
(8)
(b) Calculate Kp.
(4)
6. 2 A(g) + 2 B(g)
A2B2(g) + heat (Ea = 10 kJ
DH = -90 kJ)
(a) State and account for shift when temperature
decreased.
(3)
(b) State and account for shift when everything
doubled.
(3)
(c) State and account for shift when volume increased.
(3)
(d) State and account for shift when [B] doubled
and [A2B2] tripled. (3)
(e) Draw energy profile to scale labeling values
for Ea and DH.
(3)
(f) Explain why transition state resembles reactant.
(3)
(g) Explain why K>1.
(3)
7. Provide brief explanation: (16)
(a) A small increase in temperature results
in large increase in rate of reaction.
(b) The concentration of solids & liquids
are omitted from equilibrium expression.
(c) For a two step reaction, the rate depends
on the slow step.
(d) The initial concentration has no influence
on the half-life of a first order reaction.
Answers
1. (a) rate = k[A]2[B]3
runs 1&2
establish 5th order overala
runs 3&4
establish 3rd order in B (if triple [B], rate 27x)
runs 1&3
establish 2nd order in A (increase [A] by 1.5x, rate 2.25x)
(b) rate = k[A]2[B]3
0.150 M/s = k [M5]
k = 0.150 M-4s-1
(c) rate = [0.150][3]5 = 36.5 M/s
2. (a) ln (A0/At) = kt
ln (100/30)
= k(10 min) k =
0.12 min-1
(b) ln (100/5) = (0.12)t
t = 25 min
(c) ln (100/At) = (.12)((17.5)
At
= 12.2 or 12.2%
3. (a) Ea = (R)[(T1T2)/(T2-T1)]ln (k2/k1) = (0.00831)[(350)(300)/(50)]ln 8 = 36.3 kJ/mol
(b) ln (k2/k1) =
(Ea
/R)[(T2 - T1)/(T1T2)]
= (36.3/.00831)[(400-300)/(300)(400)] = 3.64
(k2/0.10
s-1) = 38 k2
= 3.8 s-1
4. A2(g) +
B2(g)
2 AB(g)
Eq: 3-x
3-x 3+2x
(3+2x)/(3-x) = 4 x = 1.50 [A2] = [B2] = 1.50M [AB] = 6.00M
5. A + B
AB T = 300K
2-x
2-x 2+x
(a) (2+x)/(2-x)2
= 1
x2 - 5x +2 = 0 x
= 0.44 [A]
= [B] = 1.56M [AB] = 2.44M
(b) Kp = Kc(RT)Dn
= (1)[(0.0821)(300)]-1 = 0.041
6. (a) Reaction exothermic, SHIFTS
RIGHT when Temperature decreased
(b) K =[A2B2]/[A]2[B]2
= 2/(22)(22) = 1/8 SHIFTS
RIGHT
(c) If pressure
decreased, favors greater # gas molecules, SHIFTS LEFT
(d) Q = 3/(12)(22)
= 3/4 SHIFTS RIGHT
(e)
(f) Reactant
closer to transition state in terms of energy.
(g) When
equilibrium reached, there will be much more product than reactant
(see energy profile).
7. (a) Increasing the temperature dramatically
increases the number of effective collisions.
(b) These concentrations
are constant and can be represented in K.
(c) The slow step has the
higher energy barrier and it is assumed the slow step takes about the same
time as the slow and fast steps combined.
(d) Half-life defined as
the time required for half the initial concentration to react. Therefore,
it is
independent of initial concentration.