22. Reaction Kinetics

Question 1

Define ‘rate of reaction’.

Answer 1

Change in concentration / time taken.

• units: mol/dm3 (min or h in slow reactions).
• this method gives an average rate, as the rate changes as [R] decreases.

Question 2

How can you obtain a more accurate value for the rate of reaction?

Answer 2

Use shorter time intervals (Δt = 0) and draw tangents at different [R] values.

Question 3

How can you monitor the rate of a very fast reaction?

Answer 3

Stopped-flow spectrophotometry.

• Small volume of reactant into a mixing chamber at very high speed.
• Reacting mixture moved into observation cell where the rate is monitored by the transmission of UV radiation through the sample.
• Graph automatically generated (rate against time).

Question 4

Define ‘rate equation’.

Answer 4

Rate = k[A]m [B]n

• equation relating the initial rate of reaction to the concentrations of reactants involved.
• it can only be found experimentally.

Question 5

Define ‘k’.

Answer 5

Rate constant - a proportionality constant relating the rate of a chemical reaction at a given temperature, to the concentration of the reactants involved.

Question 6

Define ‘order of reaction’.

Answer 6

The power to which a reactant’s concentration is raised in the rate equation. It can be 0, 1, 2 or 3, but is rarely higher. A fractional order indicates the presence of a free radical.

Question 7

How would you find a rate equation experimentally?

Answer 7

Vary the concentrations of one reactant whilst keeping the other(s) constant. Do this for each respective reactant and deduce the effect of each on the rate of reaction.

Question 8

How would you calculate the units of ‘k’?

Answer 8

• Write the rate equation.
• Rearrange in terms of k.
• Substitute the units. Don’t forget s-1 for the rate.
• Solve to find the units of k.

Question 9

Name three methods to deduce the order of reaction.

Answer 9

• Graphing rate against [R]
• Graphing [R] against time
• Successive half-lives (from [R] against time).

Question 10

Describe the method of ‘graphing rate against [R]’.

Answer 10

• Zero-order = horizontal line
• First-order = linear increase (directly proportional)
• Second-order = upward curve (proportional to the square, cube etc).

Question 11

Describe the method of ‘graphing [R] against time’.

Answer 11

• Zero-order = linear decrease (rate = gradient, constant rate of decline).
• First-order = shallow curve
• Second-order = deeper curve with longer tail.

Question 12

Define ‘half-life’.

Answer 12

The time taken for [R] to decrease to half its original value. Independent of the original [R].

Question 13

Describe the ‘successive half-lives (from [R] against time)’ method.

Answer 13

• Zero-order = successive half-lives decrease with time
• First-order = relatively constant
• Second-order = increase with time.

Question 14

How would you find ‘k’ from the initial [R] and rate?

Answer 14

• Rearrange rate equation in terms of k.

- Substitute the values and units, solve.

Question 15

How would you find ‘k’ from half-life data?

Answer 15

For a FIRST-ORDER reaction:

k = 0.693 / half-life (s)

Question 16

How would you find the order of a reaction from monitoring the course of the reaction?

Answer 16

• Plot a graph of [R] or [P] against time.
• Draw tangents at different concentrations to find the rate.
• Plot a graph of rate against [R] or [P].
• Determine the overall order from the shape of the curve.

Question 17

How would you find the order of a reaction from initial rate data?

Answer 17

• Conduct several experiments, varying initial [R].
• Find the initial rates by drawing a tangent at the start of each curve OR by measuring [R] after the reaction starts.
• For each [R], plot a graph of initial rate against [R].

Question 18

Define ‘rate-determining step’.

Answer 18

The slowest step in a reaction, on which the overall rate depends. Reactants in this step will appear in the rate equation. They can be either pure reactants, or intermediates formed from reactants.

Question 19

What is the reaction called based on the number of species in the rate equation?

Answer 19

One = unimolecular, two = bimolecular, three = tri.
- trimolecular is rare because it is highly unlikely for three particles to collide simultaneously, and in the correct orientation.

Question 20

Define ‘homogeneous catalyst’.

Answer 20

A catalyst in the same phase as the reaction mixture.

Question 21

Define ‘heterogeneous catalyst’.

Answer 21

A catalyst in a different phase to the reaction mixture.

Question 22

How does homogeneous catalysis work?

Answer 22

Involves a change in oxidation states of the ions involved in catalysis. Ions of transition elements are useful due to their variable oxidation states.

Question 23

Describe the iodine-peroxodiosulfate reaction (homogeneous).

Answer 23

S2O8 2- + 2I - -> 2SO4 2- + I2

• Both ions are negatively charged so there is repulsion.
• Fe3+ catalyst is attracted to the negative charge.
• 2Fe 3+ + 2I - -> 2Fe 2+ + I2 (reduction)
• 2Fe 2+ + S2O8 2- -> 2Fe 3+ + 2S2O4 2- (oxidation)
• The standard electrode potentials for each of these reactions must be between those involving the reactants.

Question 24

Describe the formation of acid rain (homogeneous).

Answer 24

SO2 + 1/2 O2 -> SO3 (oxidation)

• SO2 + NO2 -> SO3 + NO
• NO + 1/2 O2 -> NO2

Question 25

How does heterogeneous catalysis work?

Answer 25

Gaseous molecules react at the surface of a solid catalyst.

• Diffusion: molecules diffuse to surface.
• Adsorption: weak bonds (eg. VDWs) form between molecules and solid’s atoms (PHYSICAL).
• Weakening: bonds within the molecules weaken and break (CHEMICAL).
• Reaction: molecules are chemically adsorbed now - they are held in a favourable orientation for reaction with another species.
• Desorption: weak bonds between product and catalyst break, product diffuses away.

Question 26

Describe the Haber process (heterogeneous).

Answer 26

• Diffusion of nitrogen and hydrogen
• Adsorption (bonds strong enough to weaken covalent bonds, weak enough to be breakable)
• Reaction to form ammonia
• Ammonia desorbed, diffuses away.

Question 27

Describe the use of transition elements in catalytic converters (heterogeneous).

Answer 27

Honeycomb structure with small beads coated with Pt, Rh, Pd.

• NOx and CO adsorbed
• Covalent bonds weakened
• New bonds form (nitrogen, carbon dioxide)
• Desorption and diffusion.

Question 28

Give examples of different strengths of adsorption.

Answer 28

• Too weak: Ag (not enough d orbitals)
• Too strong: W
• Ideal: Ni/Pt (weak enough to break, strong enough to weaken covalent bonds).

Question 29

How does temperature affect the rate constant?

Answer 29

Increase in temperature increases k, increases overall rate.

Question 30

Describe the graph of rate of reaction against [substrate].

Answer 30

Initially first-order (directly proportional), then zero-order as graph plateaus.

Question 31

What does Km say about the rate?

Answer 31

High Km = low rate, low Km = high rate.

Question 32

Describe the structure of an active site.

Answer 32

R-groups (COO-, NH3+, OH-, HC chains) are in the active site, allowing for the binding of a substrate. Formed from tertiary structure.

Question 33

How does temperature affect enzyme activity?

Answer 33

Above optimum, VDWs and H bonds in tertiary structure break, denaturing the enzyme.

Question 34

How does pH affect enzyme activity?

Answer 34

• COOH and -NH2 groups transfer H+ ions to allow ionic bonds to form between substrate and active site.
• An increase or decrease in H+ disrupts the ionic bond and the substrate cannot bind to the active site.
• In extreme cases, the ionic bonds in the tertiary structure also break, denaturing the enzyme.