CHEM 1250 Study Guide - Quiz Guide: Rate Equation, Rice Chart, Chloroform
Dr. Zellmer Chemistry 1250 Wednesday
Time: 30 mins Spring Semester 2016 April 20, 2016
Quiz III
Name KEY Rec. TA/time
1. (8 pts) The following mechanism has been proposed for the gas-phase reaction of chloroform, CHCl
3
,
and chlorine. k
1
Cl
2
(g)
W
2 Cl(g) (fast, equilibrium)
k
-1
Cl(g) + CHCl
3
(g)
6
HCl(g) + CCl
3
(g) (slow)
Cl(g) + CCl
3
(g)
6
CCl
4
(g) (fast)
------------------------------------------------------------------
Cl
2
(g) + 2Cl(g) + CHCl
3
(g) + CCl
3
(g)
6
2 Cl(g) + HCl(g) + CCl
3
(g) + CCl
4
(g)
(a) What is the overall reaction? (cancel things that appear on both sides)
Cl
2
(g) + CHCl
3
(g)
6
HCl(g) + CCl
4
(g)
(b) What are the intermediates in the mechanism?
Cl(g) & CCl
3
(g) Things that don’t appear in overall eqn., but appear in mechanism are either
intermediates or catalysts. Intermediates appear for first time as a product
and used in following step(s) as a reactant. A catalyst appears first as a
reactant and then reappears later as a product to be used again.
(c) What is the rate-determining step (explain why)?
Step 2 - slowest step rate for entire rxn can NOT be faster than SLOWEST step in mech.)
(d) What is the rate law predicted by this mechanism?
r = k
2
[Cl] [CHCl
3
]
based on reactants in step 2 (slowest, rate-determining step)
Can’t have intermediate, Cl, in the rate law - use step 1 (fast equilibrium step) to find [Cl]
For step 1
rate
forward 1
= rate
reverse 1
r = k
2
[Cl] [CHCl
3
]
Write rate law for each rate r = k
2
{ (k
1
/k
-1
) [Cl
2
] }
1/2
[CHCl
3
]
k
1
[Cl
2
] = k
-1
[Cl]
2
r = k
2
(k
1
/k
-1
)
1/2
[Cl
2
]
1/2
[CHCl
3
]
solve for [Cl] r = k [Cl
2
]
1/2
[CHCl
3
]
[Cl] = { (k
1
/k
-1
) [Cl
2
] }
1/2
where, k = k
2
(k
1
/k
-1
)
1/2
Substitute this into rate
eqn base on step 2
Copyright R. J. Zellmer, April 20, 2016
2. (10 pts) For the following reaction K
p
equals 5.42 x 10
!3
, at 80EC.
AB (s)
W
A (g) + B (g)
a) (5 pts) What are the equilibrium partial pressures of A and B if solid AB is placed in a closed
container and decomposes until equilibrium is reached at 80EC? (Show the ICE table. State any
assumptions made and check your percent error.)
Use ICE (equil.) table (in Molarity)
AB (s)
W
A (g) + B (g)
initial C 0 atm 0 atm
change -x + x + x
-----------------------------------------------------------------
equil constant + x + x
K
p
= P
A
C P
B
= 5.42 x 10
!3
AB (s) doesn’t appear in K
(pure solids and liquids do not
x
2
= 5.42 x 10
!3
appear in K)
x = (5.42 x 10
!3
)
1/2
= 7.362 x 10
!2
atm
P
A
=
P
B
= 7.36 x 10
!2
atm
b) (3 pts) What is the value of K
c
for this reaction?
K
p
= K
c
(RT)
Δ
n
(Δn = change in moles of gas, products - reactants)
Δn = 2 ! 0 = 2 Do NOT include the solid (if you do Δn = 1)
K
c
= K
p
(RT)
!
Δ
n
R = 0.08206 LCatm/molCK NOT 8.314 J/molCK
T = 80EC + 273 = 353 K (must be in kelvin)
K
c
= 5.42 x 10
!2
{(0.08206)(353 K)}
!2
= 6.459 x 10
!5
= 6.46 x 10
!5
***** cont. on next page *****
Copyright R. J. Zellmer, April 20, 2016
