Download Kinetics II - Principles of Chemistry I - Homework 10 | CH 302 and more Quizzes Chemistry in PDF only on Docsity!
This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 10.0 points A and B react to form C according to the single step reaction
A + 2 B → C.
Which of the following is the correct rate equation for [B] and the correct units for the rate constant of this reaction?
∆[B]
∆t
= −k [A] [B]^2 ; 1/M^2
∆[B]
∆t
= − 2 k [A] [B] ; 1/(M · s)
∆[B]^2
∆t
= − 2 k [A] [B]^2 ; 1/(M · s)
∆[B]
∆t
= − 2 k [A] [B]^2 ; 1/(M^2 ·s) cor-
rect
∆[B]
∆t
− 2 k [A] [B] [C]
; 1/(M · s)
Explanation: By definition, Rate = −
∆[A]
∆t
∆[B]
∆t
∆[C]
∆t Since this is a simple-step reaction the rate law can be written from the balanced equation and is third order, where the units of k are M − (^2) · s− (^1).
Rate = −
∆[B]
∆t
= k [A] [B]^2
or
∆[B]
∆t
= − 2 k[A][B]^2
002 10.0 points The reaction
NO 2 + CO 2 → CO + NO 3
has a rate law that is second order in NO 2. Which of these statements describes the mechanism that explains this unexpected rate law?
- A single-step reaction mechanism in
which a bimolecular collision between NO 2 and CO 2 is the rate determining step.
- A multi-step reaction mechanism in which a first bimolecular collision between NO 2 molecules is the rate determining step. cor- rect
- A single-step reaction mechanism in which a bimolecular collision between NO 2 molecules is the rate determining step.
- A multi-step reaction mechanism in which a first unimolecular decomposition of NO 2 is the rate determining step.
- A single-step reaction mechanism in which a first unimolecular decomposition of NO 2 is the rate determining step.
Explanation:
003 10.0 points Consider the mechanism NO 2 + F 2 → NO 2 F + F k 1 , slow F + NO 2 → NO 2 F k 2 , fast What is the rate law?
- rate = k 2 [NO 2 ]^2
- rate = k 1 [NO 2 ] [F 2 ] correct
- rate = k 1 k 2 [NO 2 ]^2
- rate = k 2 [NO 2 ] [F]
- rate = k 1 [NO 2 F] [F]
Explanation:
004 10.0 points Determine the overall balanced equation for a reaction having the following proposed mech- anism Step 1: B 2 + B 2 −→ E 3 + D slow Step 2: E 3 + A −→ B 2 + C 2 fast and write an acceptable rate law.
- E 3 + A −→ B 2 + C 2 ; R = k [E 3 ] [A]
- A + B 2 −→ C 2 + D; R = k [B 2 ]^2 correct
- B 2 + B 2 −→ E 3 + D; R = k [B 2 ]^2
- A + B 2 −→ C 2 + D; R = k [A][B 2 ]
Explanation: Step 1: B 2 + B 2 −→ E 3 + D slow Step 2: E 3 + A −→ B 2 + C 2 fast balanced equation, rate law =?
A + B 2 −→ C 2 + D
(from the 2 molecules of B 2 in the rate- determining step)
005 10.0 points A reaction rate increases by a factor of 710 in the presence of a catalyst at 37◦C. The activation energy of the original pathway is 106 kJ/mol. What is the activation energy of the new pathway, all other factors being equal?
Correct answer: 89.0791 kJ/mol.
Explanation: T = 37◦C = 310 K Ea = 106 kJ/mol Using the form
rate = A e−Ea/R T^ ,
A a constant, of the Arrhenius equation, the ratio of the two conditions gives
ratecat rateuncat
A e−Ea,^ cat/(R T^ ) A e−Ea/(R T^ ) Taking logarithm of both sides gives
ln(710) =
−Ea, cat R T
Ea R T Ea, cat = Ea − R T ln(710) = 106 kJ/mol
−
kJ K · mol
× (310 K) ln(710) = 89.0791 kJ/mol
In practice, the new pathway also has a different pre-exponential factor.
006 10.0 points A given reaction has an activation energy of 24.52 kJ/mol. At 25◦C the half-life is 4 min- utes. At what temperature will the half-life be reduced to 20 seconds?
- 100.◦C
- ≥ 150 ◦C
- − 59. 9 ◦C
- 115 ◦C
- 25.5◦C
- 57.9◦C
- − 1. 19 ◦C
- 125 ◦C correct
- 75.0◦C
Explanation: Use the Arrhenius equation. The 20 second reaction is running 12 times that of the 240 reaction. Put that ratio
in for
the ln
ka kb
term. Remember to use Kelvin for the temperatures.
007 10.0 points For the reaction
HO(g) + H 2 (g) → H 2 O(g) + H(g)
a plot of ln k vs
T
gives a straight line with a slope equal to − 5. 1 × 103 K. What is the activation energy for the reaction?
- 5 .1 kJ/mol
- 0 .61 kJ/mol
- 42 kJ/mol correct
have the necessary energy and the slower the reaction.
013 10.0 points Consider the following potential energy dia- gram.
a
b c (^) d A+B e
X+Y
Reaction progress
Potential energy
If a catalyst were added, which arrow would change, and how?
- the length of arrow b would be smaller. correct
- the length of arrow a would be larger.
- the length of arrow b would be larger.
- the length of arrow c would be larger.
- the length of arrow d would be larger.
- the length of arrow e would be smaller.
- the length of arrow a would be smaller.
Explanation:
014 (part 1 of 2) 10.0 points Consider potential energy diagram
A
B
Reaction progress
Energy (kJ)
What is the change in enthalpy (∆H) for
the reaction A → B?
- −350 kJ
- +100 kJ
- +350 kJ
- −100 kJ correct
- −50 kJ Explanation:
Hi = HA = 450 kJ Hf = HB = 350 kJ ∆H = Hf − Hi = HB − HA = 350 kJ − 450 kJ = −100 kJ
Notice that ∆H is negative; the reaction is exothermic.
015 (part 2 of 2) 10.0 points What is the activation energy Ea for the reac- tion in the previous question?
- +100 kJ correct
- +550 kJ
- +200 kJ
- +450 kJ
- +350 kJ Explanation: Activation energy Ea is the additional en- ergy that must be absorbed by the reactants in their ground states to allow them to reach the transition state:
Ea = Etransition state − Ereactant = 550 kJ − 450 kJ = 100 kJ
016 10.0 points A catalyst facilitates a reaction by
- shifting the position of the equilibrium of the reaction.
- increasing the activation energy for the reverse reaction.
- lowering the activation energy of the re- action. correct
- making the reaction more exothermic.
- decreasing the temperature at which the reaction will proceed spontaneously.
Explanation:
017 10.0 points Compound A reacts with compound B and forms products C and D according to the equation A + B → C + D.
This reaction is found to proceed very slowly at first, then to proceed very quickly until virtually all A and B have been consumed. Suggest an explanation for this.
- The reaction has a small activation en- ergy.
- The reaction is irreversible.
- One of the products is a catalyst for the reaction. correct
- Compound B is really a very reactive metal ion.
- Reaction rates increase as reactant con- centrations decrease as a general rule.
Explanation:
018 10.0 points Which of the following does NOT affect the rate of a reaction?
- the value of Ea
- the presence of a catalyst
- the temperature of the reactants
- the value of ∆H correct
Explanation: The presence of a catalyst increases reac- tion rate, as does increasing the temperature of the reactants. A large Ea slows down a reaction.
019 10.0 points Which of the following statements is true?
- The exponents in the rate-law must match the coefficients in the balanced chemical equa- tion for the reaction.
- The rate-law for a reaction can be pre- dicted from the balanced chemical equation.
- If the exponents in the rate-law do not match the coefficients in the balanced equa- tion, then we know that the reaction does not take place in one step. correct
- If the exponents in the rate-law match the coefficients in the balanced chemical equation, then we know that the reaction takes place in one step.
Explanation: Rate laws are written based on experimen- tal data, not the balanced equation. The coefficients in the balanced equation are used in writing the rate law only in the case where the reaction is known to take place in only one step.
020 10.0 points “Reaction mechanisms usually involve only unimolecular or bimolecular steps.” Is this statement true or false?
- True, because a rate-determining step might be of higher molecularity.
- False
- True, because collisions of higher molecu-
reactants
products
Reaction coordinate
Energy (kJ) 100
reactants products
Reaction coordinate
Energy (kJ) 100
reactants
products
Reaction coordinate
Energy (kJ) 100
correct
- reactants products
Reaction coordinate
Energy (kJ) 100
reactants
products
Reaction coordinate
Energy (kJ) 100
Explanation:
