Kinetics 2: Experimental derivation of rate laws & constants

Question 1

NB:

Collision theory is useful only for simple reactions.

We propose rate laws, then see whether data is consistent, rather than trying to infer rate laws from data.

Rate laws are differential equations, the simplest of which can be solved manually.

Question 2

• Write the rate law of the following first order reaction, in terms of A, both in terms of the rate constant, and in differential form: A –> products
• Sketch [A] against time

Answer 2

r = k_1st[A]

d[A]/dt = -k_1st[A]

since recall r = -d[A]/dt

Question 3

Solve the rate law below, between t = 0 and t = t

A –> products

r = -d[A]/dt = k_1st[A]

Answer 3

result should be

[A]_t = [A]₀exp(-k_1stt)

exp decay in accordance with graph

Question 4

Prove that, for first order reactions, you can use a quantity proportional to conc in order to determine the rate constant - ie you don’t need to know the absolute rate.

Answer 4

Say the quantity I is proportional to conc. Then I = y[A], where y is unknown.

[A]₀ = I₀/y, [A]_t = I/y

Recall [A]_t = [A]₀exp(-k_1st)

ln[A]_t = -k_istt + ln[A]₀

^straight line, slope -k1st

ln(I/y) = -k_1stt + ln(I_o/y)

lnI - lny = -k_stt - lny + lnI₀

lnI = -k_1stt + lnI₀

^also a straight line, slope k1st, so serves as a relative measure of conc

Question 5

A first order reaction has stoichiometric eq A –> B. Produce and solve the rate law in terms of B.

Answer 5

d[B]/dt = k_1st[A]

quantity of B formed results from loss of same quantity of A, so:

[A]t = [A]₀ - [B]_t

Recall [A]_t = [A]₀exp(-k_1stt)

[A]₀ - [B]_t = [A]₀exp(-k_1stt)

[B]_t = [A]₀(1 - exp(-k_1stt))

as expected, this predicts [B] rises from 0 and tends towards [A]₀

Question 6

Produce and solve the rate law for the second order reaction A –> products.

Show how this gives a straight line plot.

Answer 6

rate law

d[A]/dt = -k_2nd[A]²

∫(from [A]_t to [A]₀) 1/[A] d[A] = k_2nd ∫(from t=t to t=0) dt

1/[A]_t = k_2ndt + 1/[A]₀

straight line

Plot 1/[A]_t against k_2ndt, slope = k_2nd, intercept = 1/[A]₀

whereas [A] against time decreaes hyperbolically from [A]₀, then falls asymptotically towards 0

Question 7

Can relative concentrations be used to find the value of the 2nd order rate constant?

Answer 7

No, only absolute concs

Question 8

The reaction A + B –> C has rate law d[A]/dt = -k_2nd[A][B]. Simplify and solve this law.

Answer 8

1: 1 stoichiometry means [A] = [B] always, so d[A]/dt = -k_2nd[A]²
* As before:*

∫(from [A]_t to [A]₀) 1/[A] d[A] = k_2nd ∫(from t=t to t=0) dt

1/[A]_t = k_2ndt + 1/[A]₀ or 1/[B]_t = k_2ndt + 1/[B]₀

Question 9

Describe the use of the isolation method

Answer 9

Reaction measurements are taken with one reagent in significant excess, so that its conc doesn’t change significantly throughout the reaction. This is repeated for all reagents in order to obtain pseudo-rate laws

Question 10

Describe the limitations of the isolation method.

Answer 10

Putting one reagent in excess may

• make reaction too fast to measure accurately
• alter the mechanism

Question 11

The reaction A + B –> products is thought to have rate law d[B]/dt = -k_3rd[A]²[B].

• Use the isolation method to simplify this law
• Explain how the isolation method is used to determine 3^rd experimentally

Answer 11

law simplification

Say we put A in excess.

d[A]/dt ~ 0, so k_3rd[A]² is effectively constant

Rate law becomes d[B]/dt = -k_eff[B]

where k_eff = k_3rd[A]², and is a pseudo-1st-order rate constant (but isn’t really since it’s conc-dependent)

determining k3rd

• Measure [B] throughout reaction, for several different excess concs of A
• Obtain k_eff, the gradient, from each graph
• Plot k_eff against [A]₀², giving a straight line. k_3rd is the gradient

Question 12

A rate law has the form r = k[A]ⁿ. Explain how the differential method is used to determine the value of n.

Answer 12

r = k[A]ⁿ

lnr = nln[A] + lnk

Plot lnr against ln[A] –> straight-line graph with gradient = n

Question 13

What is the drawback of the differential method?

Answer 13

Requires plotting functions of rates, rather than concs, but rates are much harder to measure. This is since taking the accurate slope of a graph of conc against time is hard, esp if reactions are fast such that rates change quickly.

Question 14

define half-life

Answer 14

time taken for the conc of a given reagent to reach half its initial value

Question 15

Give the expression for the half-life for a first-order reaction.

Answer 15

t_1/2 = ln2/k_1st

Question 16

Prove that the half-life for a first-order reaction is independent of inital concentration.

Answer 16

Begin with integrated rate law

Recall [A]_t = [A]₀exp(-k_1st)

ln[A]_t = -k_1st + ln[A]₀

When t = 1_1/2, [A]_t/2 = [A]₀/2

So ln([A]₀/2) = -k_1stt_1/2 + ln[A]₀

k_1stt_1/2 = ln[A]₀ - ln([A]₀/2) = ln(2[A]₀/[A]₀) = ln2

t_1/2 = ln2/k_1st

Expression doesn’t contain [A]₀ so is independent of it

Question 17

Give the expression for the half-life for a second order reaction.

Answer 17

t_1/2 = 1/(k_2nd[A]₀)

Question 18

Prove that the half-life for a second order reaction is dependent on the initial concentration.

Answer 18

2nd order integrated rate law

1/[A]_t = k_2ndt + 1/[A]₀

At t_1/2, [A]_t/2 = [A]₀/2

So 2/[A]₀ = k_2ndt_1/2 + 1/[A]₀

t_1/2 = 1/(k_2nd[A]₀)

half-life contains initial conc so is dependent on it

Question 19

A reaction has a 1st order rate constant of 2000 s^-1 at 400 K. Calculate the half-life.

Answer 19

t_1/2 = ln2/k_1st

= ln2/2000 = 0.3 ms = 3 x 10^-4 s

Question 20

rates of reactions can’t be compared using rate constants- have to take conc into account

A second-order reaction 2A –> B + C has rate contant 0.8 mol^-1 dm³ s^-1 at 280 K, and the initial conc of A = 4.4 x 10^-3 mol dm^-3. Calculate the half-life.

Answer 20

t_1/2 = 1/(k_2nd[A]₀)

= 1/(0.8 x 4.4 x 10^-3) = 280 s