Name:____________
Chemistry 114
Third Hour Exam
1. (13 points) I am studying the reaction A+B6C using the method of initial rates. In my
kinetic experiments I have the following results
rate (mole/min)
1.36x10-5
3.60x10-5
14.37x10-5
[A](M)
.1
.1
.2
[B](M)
.1
.2
.2
Determine the order of this reaction with respect to A, B and the overall k of this reaction.
2A. (6 points)Assume the reaction A6B is second order with respect to A. If the reaction
has a half life of 23 minutes when the concentration of A is .05 M, what is the k for the
reaction?
2B. (6 points) Using the rate constant you found for 2A, determine how long it will take for
the concentration of A to go from 0.5M to 0.3M. (If you didn’t get a rate constant in 2A, just
write one down and finish the problem anyway, I’ll try to figure it out.)
Note: this is NOT a ½ time!
2
3. (13 points) For the reaction 2H2(g) + 2NO(g) 6 N2(g) + 2H2O(g) the observed rate law is:
rate = k[NO]2[H2]
Which of the following proposed mechanisms are consistent with the experimental data?
WHY?
I
2H2(g) + 2NO(g) 6 N2(g) + 2H2O(g)
Kinetic doesn’t match
III
H2(g) + 2NO(g) 6N2O(g) + H2O(g) Slow
N2O(g) + H2(g) 6 N2(g) + H2O(g) Fast
All matches, consistent
II
H2(g) + NO(g) 6H2O(g) + N(g)
slow
N(g) + NO(g) 6 N2(g) + O(g)
fast
H2(g) + O(g)6 H2O(g)
fast
Kinetics of slow step doesn’t match
IV
H2(g) + 2NO(g) 62 HNO(g)
Slow
HNO(g) + H2(g)6NH3(g) + O(g) Fast
Sum doesn’t match
4. (12 points) The energy profile of a certain 2 step reaction is shown below.
On the above diagram please indicate
A. The position of reactants and products
B. The activation energy for the overall reaction
C. The )E for the overall reaction
I threw out the D through F
D. Which point indicates the energy of the first intermediate in the reaction
pathway?
First and only intermediate is in the valley between the two peaks
E. Which point indicates the energy of the second intermediate in the
reaction pathway?
There is no second intermediate
F. Which step in the reaction pathway (step 1 or 2) is the rate limiting step in
the overall reaction?
The Ea for both reactions is about the same so the rates are about the same
3
5. (12 points) When the reaction
N2(g) + 3Cl2(g) 62NCl3(g)
is at equilibrium at a certain temperature, the concentration of the component gases is
found to be [N2]=.0014M, [Cl2]=.00043M, and [NCl3] = .19M
What is the Kc for this reaction?
What is the KP for this reaction? (Assume T=25oC)
What is the KC for the reaction NCl3(g)6½ N2(g) + 3/2 Cl2(g)?
Reverse reaction x ½ so:
6. (13 points) Define the following
A. Heterogenous equilibrium
An equilibrium involving products or reactants that are not in the same phase.
B. Reaction Quotient (Q)
The number you get when you plug your current set of conditions into the
equilibrium expression.
C. Catalyst
A substance that speed up a reaction but is not used up in the reaction.
D. Elementary Step
A reaction whose rate can be written directly from it’s molecularity because it
describes a molecular collision that is part of a reaction mechanism.
E. Order of a reaction
The exponent associated with concentration in a rate equation.
4
7. (13 points) Acetic acid undergoes the reaction
CH3COOH(aq) +H2O(l) 6CH3COO-(aq) + H3O+(aq)
The K for this reaction is 1.75x10-4
A. Write the equilibrium expression for this reaction
1.75x10-4= [CH3COO-][H3O+]/[CH3COOH]
B. If my initial concentration of CH3COOH is 1M, but CH3COO- and H3O+ are both
0, will the reaction move to the right (products) or left (reactants)?
Will move to fill in the 0 concentration of products, to the right
C. Assume I wanted you to calculate X, the concentration of H3O+ after the solution
reaches equilibrium. What would the concentration of CH3COOH, CH3COO-, and H2O be
at equilibrium? (Write an expression for each concentration using X, don’t try to solve for
X)
[CH3COOH] = 1-X
[CH3COO-]=X
[H3O+] = X
[H2O] = 55.5M-X . 55.5M = constant, that is why we can drop it out of the
equilibrium expression
D. Write the new equilibrium expression using X in the concentration of all the
chemicals
8. (12 points)
Assume I have the following exothermic reaction at equilibrium in the gas phase
N2(g) + 3H2(g)6 2NH3(g)
Which way with the equilibrium shift (right, left, remains unchanged) under the
following conditions
A. H2 gas is added to the system 6
Put in excess reactant, get rid of excess
B. NH3 gas is added to the system 7
Put in excess product, get rid of product
C. The pressure of the system is increased by adding N2 gas6
N2 is not inert, is a reactant, get rid of excess reactant
D. The pressure of the system is decreased by doubling the volume7
Product has smaller volume than reactants, increase volume favors reactants
E. The temperature of the system is increased7
Exothermic, heat is a product, get rid of excess product
F. Water is introduced to the system, so the following reaction occurs:6
NH3(g) + H2O(l) 6NH4OH(aq)
NH3 is a reaction product, this reaction will remove NH3, so system will try to make
more NH3
5
9, (5 points extra credit) - On the back of this page solve the equation you wrote for 7D
Only the + concentration makes sense so:
[H+]=[CH3COO-]=.0131M
[CH3COOH] = 1-.0131 =.9869M