Kinetics and Sn1 & Sn2 Flashcards

1
Q

What are the two methods for determining rate equations?

A

The continuous method and the initial-rate method

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

What are the two steps of the continuous method?

A

1) Draw a concentration-time graph

2) Find out the half-life of the reaction

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

If the half-life is constant, what does this tell you about the reaction?

A

It is first order with respect to that reactant

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

If the half-life increases over time, what does this tell you about the reaction?

A

It is second order or above with respect to that reactant

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

If the concentration-time graph is a straight line (with a negative gradient), what does this tell you about the reaction?

A

The concentration of the reactant has no impact on the rate, so the reaction is zero order with respect to that reactant

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

How does the initial-rate method work?

A

Several reaction mixtures are set up, and the initial rate of each is recorded.
Then through comparison you can find the relationship between concentration of a reactant and rate, allowing you to work out the order of the reaction

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

What will zero order look like on a rate-concentration graph?

A

A straight horizontal line

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

What will first order look like on a rate-concentration graph?

A

A straight diagonal line passing through the origin

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

What will second order look like on a rate-concentration graph?

A

A curved line, however from that you cannot directly determine that it is second order. You must then plot rate against concentration^2. If that is a straight diagonal line (like first order) then it is second order

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

If a reaction is ‘elementary’, what can you deduce straight from the stoichiometric equation?

A

The rate equation.
e.g. NO + O3 -> NO2 + O2
Rate = k[NO][O3]

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

When is a reaction ‘elementary’?

A

When a reaction takes place from a single collision between two reactant particles

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

How many species are involved in the rate determining step in Sn1?

A

One

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

What is the slow step of Sn1 of alkaline hydrolysis?

A

The slow ionisation of the halogenoalkane to form a carbocation, therefore only one species (the halogenoalkane) is involved

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

How many species are involved in the rate determining step in Sn2?

A

Two

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

What is the slow step of Sn2 of alkaline hydrolysis?

A

Where the halogenoalkane forms a partial bond with the attacking OH, and the halogen, therefore two species (the halogenoalkane and the OH) are involved

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Does Sn1 or Sn2 involved a carbocation?

17
Q

What do tertiary halogenoalkanes undergo?

18
Q

What do primary halogenoalkanes undergo?

19
Q

What two reasons explain why tertiary halogenoalkanes undergo sn1?

A
  • Tertiary carbocations are relatively stable, as they have three electron-donating carbons around it
  • Steric hinderance
20
Q

How does steric hinderance impact tertiary halogenoalkanes?

A

Because of the bulky alkyl groups surrounding the central carbon attached to the halogen atom, there is very little space for the nucleophile to attack. Consequently the weak carbon-halide bond must first be broken to allow space for the nucleophile to attack

21
Q

What is the first step to find the activation energy of a reaction?

A

Plot ln(k) against 1/T

22
Q

What axis should ln(k) be on?

A

The y axis (vertical)

23
Q

What axis should 1/T be on?

A

The x axis (horizontal)

24
Q

What is ln(k)?

A

ln(1/time)

25
What is 1/T?
1/Temperature in kelvin
26
After plotting ln(k) against 1/T, what is the next step to find the activation energy?
Find the gradient of the line plotted
27
After finding the gradient of the line, what is the next step to find the activation energy?
The gradient is equal to Ea/-R | Therefore if we multiply the gradient by -R (which is 8.31) we will find the activation energy in J mol-1