LT 33: multi-step energy profiles, catalysts

LT 33: multi-step energy profiles, catalysts

10th Grade

41 Qs

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LT 33: multi-step energy profiles, catalysts

LT 33: multi-step energy profiles, catalysts

Assessment

Quiz

Chemistry

10th Grade

Medium

NGSS
HS-PS1-4, HS-PS1-5, HS-PS3-2

+2

Standards-aligned

Used 5+ times

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41 questions

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1.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Media Image

The slowest step in a reaction mechanism requires the collision represented above to occur.  Which of the following most likely indicates how the addition of a solid catalyst could increase the rate of the reaction?

The catalyst could change the reaction from second order to third order

The catalyst could increase the particles’ speed, thereby increasing the value of the rate constant, k

The catalyst could decrease the particles’ speed, making it less likely that the particles will rebound without reacting when they collide

The catalyst could adsorb (hold) one of the particles, making a successful (reaction-producing) collision with the other particle more likely

Tags

NGSS.HS-PS1-5

2.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Media Image

The energy diagram for the reaction X + Y  --> Z is shown.  The addition of a catalyst to this reaction would cause a change in which of the indicated energy differences?

I only

II only

III only

I and II only

Tags

NGSS.HS-PS1-4

3.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Media Image

The energy diagram for the reaction X + Y  --> Z is shown.  At what point on the graph is the activated complex present?

A

B

C

D

Tags

NGSS.HS-PS1-4

4.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Media Image

The energy diagram for the reaction X + Y  --> Z is shown.  What would happen if the kinetic energy of the reactants was not enough to provide the needed activation energy?

The rate of the reaction would tend to increase

The activated complex would convert into products

The reactants would continue to exist in their present state

The products would form at an unstable energy state

Tags

NGSS.HS-PS1-4

5.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Media Image
Media Image
Media Image
Media Image

Tags

NGSS.HS-PS1-4

6.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Media Image

O3(g) + O(g)  →  2O2(g)

The decomposition of O3(g) in the upper atmosphere is represented by the equation above.  The potential energy diagram for the decomposition of O3(g) in the presence and absence of NO(g) is given below. Which of the following mechanisms for the catalyzed reaction is consistent with the equation and diagram above?

2O3(g)  +  2NO(g) →  4O2(g)  +  N2(g) (slow) 

O3(g) + NO(g)→  NO2(g) + O2(g) (slow)

NO2(g) + O(g) → NO(g) + O2(g) (fast)

NO2(g) + O3(g) → NO(g) + 2O2(g) (slow)

NO(g)  +  O(g)  → NO2(g)  (fast)

NO2(g) + O(g) → NO3(g)  (slow)

NO3(g) + O3(g)→ NO2(g) + 2O2(g) (fast)

Tags

NGSS.HS-PS1-4

7.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Media Image

The diagram above illustrates how the reaction N2(g) + 3H2(g)--> 2NH3(g) occurs on the surface of Ru(s).  The rate of this reaction is determined by the amount of energy required to break the bond in N2. This bond is weakened when N2 is adsorbed on Ru(s). Based on this information, which of the following provides the best reason for the use of Ru(s) for the synthesis of NH3?

It promotes the proper orientation of N and H atoms to form new H-H bonds

It provides a reaction path with a lower activation energy

It forms a bond between Ru and N that is stronger than the bond in N2.

It decreases the frequency of collisions between N2(g) and H2(g)

Tags

NGSS.HS-PS1-4

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