Name:_____________
(4 points)
Chemistry 114
First Hour Exam
Remember- Show all work for partial credit. Each problem is worth 14 points. You
must to all 7 problems.
1. Consider the reaction 2N2O5(g) 64NO2(g) + O2(g)
If you start with .15 moles of N2O5 in a 2 liter container, and it contains .12 moles of
N2O5 after 10 minutes
What is
ÄM? = (.15 moles-.12 mole)/2L = .015mol/L
Ät = 10 minutes; ÄM/Ät = .0015 mol N2O5 /LAmin
What is
?
What is
?
2. Consider the reaction A + B + C6 D
A. Write the rate law equation for this reaction:
rate = [A]x[B]y[C]z
B. Determine the order parameters for A, B, and C given the following data
[A] (mol/L)
[B] (mol/L)
[C](mol/L)
Initial rate (mol/L sec)
.1
.1
.1
1.6e-5
.25
.1
.1
4e-5
.25
.2
.1
1e-4
.25
.2
.2
4e-4
3. You are studying the reaction A + B 6 C.
If the reaction is 1st order with respect to A and 2nd order with respect to B, and
[A]=.001 M while [B] is 1M, predict the appearance of the following plots
Plot
Slope (+, 0, or -)
Linearity (straight or
curved)
Y = [A], X=t
negative
Curved
Y = ln[A], X=t
negative
Straight
Y = 1/[A], X =t
Positive
Curved
Y=[B], X=t
zero
Straight
Now I change the conditions and [A]=1M and [B]= .001 M
Plot
Slope (+, 0, or -)
Linearity (straight or
curved)
Y = [B], X=t
negative
Curved
Y = ln[B], X=t
negative
Curved
Y = 1/[B], X =t
Positive
Straight
Y=[A], X=t
zero
Straight
4. Define the following terms:
Intermediate
A species that is both generated and consumed in a reaction mechanism
Elementary step
A step in a reaction mechanism whose molecularity can be written directly from the
reaction.
Bimolecular step
A step in which only two molecules must collide.
Rate determining step
The slow step in a reaction pathway.
Steric Factor
The factor ñ in the equation rate = Zñe-Ea/RT that refers to the orientation of a molecule.
Inhibitor
A substance that slow a reaction down.
Equilibrium Constant
When the reaction aA + bBWcC + dD reaches equilibrium; K = [C]c[D]d/[A]a[B]b
5. Give the Arrhenius plot below, determine the activation energy for this reaction.
Ea = ?
Slope = -Ea/R; -1804=-Ea/8.314J/KAmol
Ea=1804×8.314=15000J/mol = 15kJ/mol
Given that one point on this line is the point k=.02 T=25oC, what is the k of this reaction
when T = 200oC?
6. Write the reaction quotient for the following chemicals reactions:
H2(g) + F2(g) W2HF(g)
Q=[HF]2 / [H2][F2]
N2(g) + 3H2(g) W2NH3(g)
Q=[NH3] / [N2][H2]3
Cu(NH3)62+ (aq) + 6Cl-(aq) WCuCl62+(aq) + 6NH3(aq)
Q=[CuCl62+][NH3]6 / [Cu(NH3)62+][Cl-]6
2C4H10(g) + 13O2(g) W8CO2(g) + 10H2O(g)
[CO2]8[H2O]10 / [C4H10]2[O2]13
7. The reaction 2SO2(g) + O2(g)W2SO3(g) has an equilibrium constant (in terms of
concentration) of 3.57
What is the equilibrium constant (in terms of concentration) for the reaction
2SO3(g)W2SO2(g) + O2(g)
This is reverse of original equation so Knew = 1/Kold
=1/3.57 = .280
What is the equilibrium constant (in terms of concentration) for the reaction
4SO3(g)W4SO2(g) + 2O2(g)
This is the previous equation multiplied by 2 so Knew = (Kold)2
=.2802 = .0785
If T = 25oC,what is the equilibrium constant (in terms of pressure) for the reaction
2SO2(g) + O2(g) W 2SO3(g)
This is the original KC now being expressed in terms of pressure
KP = KC (RT)Än; Än=2-(2+1) = -1
3.57(RT)-1
=3.57 / (.08206)(25+273)
=.146
What is the equilibrium constant (in terms of pressure) for the reaction
2SO3(g)W2SO2(g) + O2(g)
This is the reverse of the above equation so Knew = 1/Kold
=1/.146 = 6.85