Chem 103 Final Portage Questions and
Accurate Detailed Answers
In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below, the following data
table is obtained:
2 N2O5 (g) → 4 NO2 (g) + O2 (g)
1. Using the [O2] data from the table, show the calculation of the average rateover the measured
time interval from 0 to 3000 secs.
2. Using the [O2] data from the table, show the calculation of the instantaneous rate early in the
reaction (0 secs to 300 sec).
3. Explain the relative values of the average rate and the early instantaneous rate. - Answer: 1. The
average rate over the measured time interval from 0 to 3000 secs is:
rate = ∆[O2] / ∆t = (0.046 - 0) / 3000 - 0 = 1.53 x 10-5mol/L•s
2. The early instantaneous rate over the measured time interval from 0 to 300 secs is:
rate = ∆[O2] / ∆t = (0.009 - 0) / 300 - 0 = 3.00 x 10-5mol/L•s
3. The early instantaneous rate is larger since the concentrations of reactants is higher during the
earliest stages of the reaction.
The following rate data was obtained for the hypothetical reaction:
A+D→X+Y
Experiment # [A] [D] rate
1 0.50 0.50 2.1
2 1.00 0.50 8.4
3 1.00 1.00 33.6
1. Determine the reaction order with respect to [A].
2. Determine the reaction order with respect to [D].
3. Write the rate law in the form rate = k [A]n [D]m (filling in the correct exponents).
4. Show the calculation of the rate constant, k. - Answer: 1. 2nd order with respect to [A] : 8.4/2.1
=k (1.00)n (0.50)m / k (0.50)n(0.50)m : n = 2
2. 2nd order with respect to [D] : 33.6/8.4 =k (1.00)n (1.00)m / k (1.00)n(0.50)m : m = 2
3. rate = k [A]2 [D]2
,4. k = 2.1 / (0.50)2 (0.50)2 = 33.6
ln [A] - ln [A]0 = - k t 0.693 = k t1/2
An ancient sample of paper was found to contain 19.8 % 14C content as compared to a present-day
sample. The t1/2 for 14C is 5720 yrs. Show the calculation of the decay constant (k) and the age of
the paper. - Answer: 0.693 = k (5720)
k = 0.693/5720 = 1.2115 x 10 -4
ln [A] - ln [A]0 = - k t
ln 19.8 - ln 100 = -(1.2115 x 10-4) t
t = -1.6195 / -(1.2115 x 10-4) = 13,368 years
Using the potential energy diagram below, state whether the reaction described by the diagram is
endothermic or exothermic and spontaneous or nonspontaneous, being sure to explain your
answer. - Answer: Large Eact = nonspontaneous ∆H- = exothermic
Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.900 mole of CO
and 2.70 mole of H2 in a 9.00 liter container forms an equilibrium mixture containing 0.346 mole of
H2O and corresponding amounts of CO, H2, and CH4.
CO (g) + 3 H2 (g) CH4 (g) + H2O (g) - Answer: At equilibrium
H2O = 0.346 mole (as stated)
CH4 = 0.346 mole (1 mole of CH4 forms for every mole of H2O that is formed)
CO = 0.900 - 0.346 mole (1 mole of CO reacts for every mole of H2O that is formed)
H2 = 2.70 - 3 x 0.346 mole (3 mole of H2 reacts for every mole of H2O that is formed)
Change all amounts to moles/L before entering in Kc expression:
H2O = 0.346 mole / 9.00 L = 0.0384 M
CH4 = 0.346 mole / 9.00 L = 0.0384 M
CO = 0.554 mole / 9.00 L = 0.0616 M
H2 = 1.662 mole / 9.00 L = 0.185 M
Kc = [CH4] [H2O] = [0.0384] [0.0384] = 3.78
[CO] [H2]3 [0.0616] [0.185]3
, Explain the terms substrate and active site in regard to an enzyme. - Answer: The substrate is the
substance whose reaction rate is increased by an enzyme. The active site is the group of atoms on
the surface of an enzyme where the substrate binds to undergo the reaction catalyzed by the
enzyme.
The reaction below has the indicated equilibrium constant. Is the equilibrium mixture made up of
predominately reactants, predominately products or significant amounts of both products and
reactants. Be sure to explain your answer.
2 H2 (g) + S2 (g) 2 H2S (g) Kc = 9.39 x 10-5 - Answer: The very small Kc indicates that this equilibrium
mixture will be composed of mostly reactants.
The equilibrium reaction below has the following equilibrium mixture concentrations:
H2O = [0.0380], CH4 = [0.0380], CO = [0.0620] and H2 = [0.186] with Kc= 3.62.
If the concentration of H2O at equilibrium is increased to [0.100], how and for what reason will the
equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use this to confirm
your answer.
CO (g) + 3 H2 (g) CH4 (g) + H2O (g) - Answer: Kc = 3.62 when H2O = [0.0380], CH4 = [0.0380], CO =
[0.0620] and H2 = [0.186 M]
When H2O = [0.100], Q = [0.100] [0.0380] = 9.52[0.0620] [0.186]3
The reaction must shift briefly in the reverse direction to decrease the [H2O] to come back to
equilibrium.This is in agreement with Qc > Kcwhich also predicts the reaction will proceed to the left.
The equilibrium reaction below has the Kc = 3.93. If the volume of the system at equilibrium is
decreased from 6.00 liters to 2.00 liters, how and for what reason will the equilibrium shift? Be sure
to calculate the value of the reaction quotient, Q, and use this to confirm your answer.
CO (g) + 3 H2 (g) CH4 (g) + H2O (g) - Answer: When volume decreases from 6.00 to 2.00, the
pressure triples and the concentration of all gases (CO, H2, CH4, and H2O) triples so:
(at equilibrium) Qc =Kc = [CH4] [H2O] = 3.93[CO] [H2]3
(volume 1/3 = pressure tripled = conc tripled) Qc = [3 CH4] [3 H2O] = Kc [3 CO] [3 H2]3 9
The reaction must shift briefly in the direction that decreases the pressure by going toward the side
with the lesser moles of gas (forward direction : 4 moles of gas yields 2 moles of gas) to come back
to equilibrium. This is in agreement with Qc < Kc: the reaction will proceed to the right (in the
direction of the products).
The equilibrium reaction below has the Kc = 0.254 at 25oC. If the temperature of the system at
equilibrium is decreased to 0oC, how and for what reason will the equilibrium shift? Also show and
explain how and why the Kc value will change.
Accurate Detailed Answers
In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below, the following data
table is obtained:
2 N2O5 (g) → 4 NO2 (g) + O2 (g)
1. Using the [O2] data from the table, show the calculation of the average rateover the measured
time interval from 0 to 3000 secs.
2. Using the [O2] data from the table, show the calculation of the instantaneous rate early in the
reaction (0 secs to 300 sec).
3. Explain the relative values of the average rate and the early instantaneous rate. - Answer: 1. The
average rate over the measured time interval from 0 to 3000 secs is:
rate = ∆[O2] / ∆t = (0.046 - 0) / 3000 - 0 = 1.53 x 10-5mol/L•s
2. The early instantaneous rate over the measured time interval from 0 to 300 secs is:
rate = ∆[O2] / ∆t = (0.009 - 0) / 300 - 0 = 3.00 x 10-5mol/L•s
3. The early instantaneous rate is larger since the concentrations of reactants is higher during the
earliest stages of the reaction.
The following rate data was obtained for the hypothetical reaction:
A+D→X+Y
Experiment # [A] [D] rate
1 0.50 0.50 2.1
2 1.00 0.50 8.4
3 1.00 1.00 33.6
1. Determine the reaction order with respect to [A].
2. Determine the reaction order with respect to [D].
3. Write the rate law in the form rate = k [A]n [D]m (filling in the correct exponents).
4. Show the calculation of the rate constant, k. - Answer: 1. 2nd order with respect to [A] : 8.4/2.1
=k (1.00)n (0.50)m / k (0.50)n(0.50)m : n = 2
2. 2nd order with respect to [D] : 33.6/8.4 =k (1.00)n (1.00)m / k (1.00)n(0.50)m : m = 2
3. rate = k [A]2 [D]2
,4. k = 2.1 / (0.50)2 (0.50)2 = 33.6
ln [A] - ln [A]0 = - k t 0.693 = k t1/2
An ancient sample of paper was found to contain 19.8 % 14C content as compared to a present-day
sample. The t1/2 for 14C is 5720 yrs. Show the calculation of the decay constant (k) and the age of
the paper. - Answer: 0.693 = k (5720)
k = 0.693/5720 = 1.2115 x 10 -4
ln [A] - ln [A]0 = - k t
ln 19.8 - ln 100 = -(1.2115 x 10-4) t
t = -1.6195 / -(1.2115 x 10-4) = 13,368 years
Using the potential energy diagram below, state whether the reaction described by the diagram is
endothermic or exothermic and spontaneous or nonspontaneous, being sure to explain your
answer. - Answer: Large Eact = nonspontaneous ∆H- = exothermic
Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.900 mole of CO
and 2.70 mole of H2 in a 9.00 liter container forms an equilibrium mixture containing 0.346 mole of
H2O and corresponding amounts of CO, H2, and CH4.
CO (g) + 3 H2 (g) CH4 (g) + H2O (g) - Answer: At equilibrium
H2O = 0.346 mole (as stated)
CH4 = 0.346 mole (1 mole of CH4 forms for every mole of H2O that is formed)
CO = 0.900 - 0.346 mole (1 mole of CO reacts for every mole of H2O that is formed)
H2 = 2.70 - 3 x 0.346 mole (3 mole of H2 reacts for every mole of H2O that is formed)
Change all amounts to moles/L before entering in Kc expression:
H2O = 0.346 mole / 9.00 L = 0.0384 M
CH4 = 0.346 mole / 9.00 L = 0.0384 M
CO = 0.554 mole / 9.00 L = 0.0616 M
H2 = 1.662 mole / 9.00 L = 0.185 M
Kc = [CH4] [H2O] = [0.0384] [0.0384] = 3.78
[CO] [H2]3 [0.0616] [0.185]3
, Explain the terms substrate and active site in regard to an enzyme. - Answer: The substrate is the
substance whose reaction rate is increased by an enzyme. The active site is the group of atoms on
the surface of an enzyme where the substrate binds to undergo the reaction catalyzed by the
enzyme.
The reaction below has the indicated equilibrium constant. Is the equilibrium mixture made up of
predominately reactants, predominately products or significant amounts of both products and
reactants. Be sure to explain your answer.
2 H2 (g) + S2 (g) 2 H2S (g) Kc = 9.39 x 10-5 - Answer: The very small Kc indicates that this equilibrium
mixture will be composed of mostly reactants.
The equilibrium reaction below has the following equilibrium mixture concentrations:
H2O = [0.0380], CH4 = [0.0380], CO = [0.0620] and H2 = [0.186] with Kc= 3.62.
If the concentration of H2O at equilibrium is increased to [0.100], how and for what reason will the
equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use this to confirm
your answer.
CO (g) + 3 H2 (g) CH4 (g) + H2O (g) - Answer: Kc = 3.62 when H2O = [0.0380], CH4 = [0.0380], CO =
[0.0620] and H2 = [0.186 M]
When H2O = [0.100], Q = [0.100] [0.0380] = 9.52[0.0620] [0.186]3
The reaction must shift briefly in the reverse direction to decrease the [H2O] to come back to
equilibrium.This is in agreement with Qc > Kcwhich also predicts the reaction will proceed to the left.
The equilibrium reaction below has the Kc = 3.93. If the volume of the system at equilibrium is
decreased from 6.00 liters to 2.00 liters, how and for what reason will the equilibrium shift? Be sure
to calculate the value of the reaction quotient, Q, and use this to confirm your answer.
CO (g) + 3 H2 (g) CH4 (g) + H2O (g) - Answer: When volume decreases from 6.00 to 2.00, the
pressure triples and the concentration of all gases (CO, H2, CH4, and H2O) triples so:
(at equilibrium) Qc =Kc = [CH4] [H2O] = 3.93[CO] [H2]3
(volume 1/3 = pressure tripled = conc tripled) Qc = [3 CH4] [3 H2O] = Kc [3 CO] [3 H2]3 9
The reaction must shift briefly in the direction that decreases the pressure by going toward the side
with the lesser moles of gas (forward direction : 4 moles of gas yields 2 moles of gas) to come back
to equilibrium. This is in agreement with Qc < Kc: the reaction will proceed to the right (in the
direction of the products).
The equilibrium reaction below has the Kc = 0.254 at 25oC. If the temperature of the system at
equilibrium is decreased to 0oC, how and for what reason will the equilibrium shift? Also show and
explain how and why the Kc value will change.
