Unit 5 - Kinetics

Question 1

Define the collision theory

Answer 1

For an elementary reaction to successfully produce products, reactants must collide with the correct orientation and sufficient amount of energy (activation energy) to break reactant bonds and rearrange them into product bonds

Question 2

How do some factors increase the rate of reaction? (general)

Answer 2

Factors affect the rate of reaction by affecting the frequency of particle collisions and/or the proportion of successful collisions that result in products

Question 3

Define the rate law and its components.

Answer 3

Rate law for mX + nY → Z + W rate = k [X] m [Y] n

k = rate constant (temperature dependent); don’t forget UNITS
m = order of X
n = order of Y
m + n = overall order

Question 4

How do you write the rate law if given a set of reactions?

Answer 4

Write the rate law based on the slowest elementary step (since it is the rate determining step)

Molar coefficients in balanced equation act as exponents for reactant concentrations

Cannot write reaction intermediates in rate law – substitute for another reactant from one of the steps

Question 5

Use steady state approximation to write the rate law of the following reaction.

STEP 1 – NO2 (g) + F2 (g) ⇌ NO2F2 (g) fast

STEP 2 – NO2F2 (g) → NO2F (g) + F (g) slow

STEP 3 – F(g) + NO2 → NO2F (g) fast

Answer 5

1. Write the rate law based on the slowest elementary step aka the rate determining step.

rate = k2 [NO2F2]

2. Replace the reaction intermediate with reactants

rate forward = rate reverse
k forward [NO2] [F2] = k reverse [NO2F2]
[NO2F2] = (k forward/k reverse) [NO2] [F2]

3. Substitute your new expression into the initial rate law

rate = k [NO2] [F2]
k = k2 (k forward/k reverse)

Question 6

For first order reactions, half life _____

Answer 6

Remains constant

Decreasing from 100% to 50% will take the SAME amount of time as decreasing from 50% to 25%) - not the case for zero and second order reactants

Question 7

Catalyst

Answer 7

Consumed in the first step (slowest) and regenerated in the final step (amount remains the same)

Increases the # of effective collisions by providing a new alternative reaction pathway with a lower activation energy; they DO NOT add energy

Question 8

Reaction intermediate

Answer 8

Produced in the first step and consumed in the following step

Question 9

Endothermic vs Exothermic (temperature)

Answer 9

Endothermic reactions are cold because energy is absorbed from the surroundings (work is done on the system)

Exothermic reactions are hot because energy is released into the surroundings (work is done on the surroundings)

Question 10

Describe the components of a potential energy graph.

Answer 10

Transition state = highest peak 
# of elementary steps in a reaction = # of transition states or peaks in potential energy graph

Activation energy = energy of transition state - energy of reactants
The slowest elementary step has the highest activation energy to overcome

ΔH (change in energy) = energy of products - energy of reactants

Question 11

Compare the potential energy graphs for exothermic vs endothermic reactions

Answer 11

In exothermic reactions, energy is released, so the energy of the products is lower than the energy of the reactants

In endothermic reactions, energy is absorbed, so the energy of the products is higher than the energy of the reactants

Question 12

Compare the potential energy graphs for catalyzed vs uncatalyzed reactions

Answer 12

The activation energy is lower in catalyzed reactions (transition state peak is smaller) compared to uncatalyzed reactions.

However, ΔH (change in energy) between the reactants and products stays the same in both cases