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

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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

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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

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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

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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

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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

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7
Q

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

A

A straight horizontal line

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8
Q

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

A

A straight diagonal line passing through the origin

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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

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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]

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11
Q

When is a reaction ‘elementary’?

A

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

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12
Q

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

A

One

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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

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14
Q

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

A

Two

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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

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16
Q

Does Sn1 or Sn2 involved a carbocation?

A

Sn1

17
Q

What do tertiary halogenoalkanes undergo?

A

Sn1

18
Q

What do primary halogenoalkanes undergo?

A

Sn2

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
Q

What is 1/T?

A

1/Temperature in kelvin

26
Q

After plotting ln(k) against 1/T, what is the next step to find the activation energy?

A

Find the gradient of the line plotted

27
Q

After finding the gradient of the line, what is the next step to find the activation energy?

A

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