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Chemistry A Level AQA > 3.1.9- Rate equations (PAPER 2) > Flashcards

3.1.9- Rate equations (PAPER 2) Flashcards

1
Q

What are the 4 factors that affect rate?

A

Temperature- higher temp means particles have more kinetic energy so vibrate more= more successful collisions above EA= higher ROR

SA- More particles on surface so more exposed particles increasing ROR

Catalyst- Higher ROR by providing alternate pathway with lower activation energy

Conc/pressure- more particles so more frequent successful collisions, increased likelihood that reactant particles collide.

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2
Q

Draw a graph of a product against a reactant- conc on y axis, time on x axis.

A

Product has upwards curve, reactant has downwards curve- they cross once.

Curve inwards/ outwards

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3
Q

Explain the shape of this graph

A

ROR the fastest at the beginning as higher likelihood of successful collisions as more particles at beginning, slope of reaction decreases with time as conc of reactants decrease.

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4
Q

What is rate of reaction?

A

The amount of reactant used/product formed over time.

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5
Q

What does rate of reaction depend on?

A

The slowest step in the reaction mechanism- the rate determining step.

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6
Q

How to find rate of a reaction?

A

Change in conc or moles of product/ time

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7
Q

How to find rate using a gradient?

A

Draw a tangent to the curve, change in y/ change in x

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8
Q

What is the rate expression?

A

Rate (mol dm−³ s−¹) = k [A]m [B]n

[A]= concentration of substance UNITS of moldm-3

m and n= orders of reaction

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9
Q

How to find total order for a reaction?

A

Sum of the separate orders- m+n

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10
Q

How to know which substance affects rate?

A

Species that appear in chemical equation but not rate expression do not affect rate.

A species that is not in chemical equation can be in rate expression eg catalyst

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11
Q

Which factors increase the rate constant k?

A

Increasing the temperature

Adding a catalyst

Larger value of k indicates faster reaction, reaction with large activation energy has small rate constant so slow reaction.

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12
Q

What is an order of reaction?

A

Relationship between the rate of a chemical reaction and the concentration of the species taking part in it

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13
Q

What does it mean if a reactant is in 0 order?

A

A change in concentration of this reactant has no overall effect on the rate.

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14
Q

Draw the graph of a reaction in 0 order: rate on y axis, conc on bottom

A

Conc of this species does not impact rate

Shown as horizontal straight line from middle y axis to right.

Rate= k

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15
Q

What does it mean if a reactant is in 1st order?

A

Changes in concentration of this reactant have a proportional change on the rate- eg if [A] doubles, rate doubles.

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16
Q

Draw the graph of a reaction in 1st order: rate on y axis, conc on bottom

A

Conc of species and rate are directly proportional

Diagonal straight line from origin to right

Rate= k[A]

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17
Q

What does it mean if a reactant is in 2nd order?

A

Changes in concentration of this reactant have a squared proportional change on the rate- eg if [A] doubles, rate quadruples

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18
Q

Draw the graph of a reaction in 2nd order: rate on y axis, conc on bottom

A

Rate is proportional to conc squared

Curved line from origin reaching to left

Rate = k[A]2

19
Q

What are the shapes of a 0, 1st and 2nd order reaction graphs with conc on y axis and time on x axis?

A

0 order- diagonal going down from left to right
1st order- curly line from left to right like half a U
2nd order- steeper and smaller curly line from left to right

20
Q

How to find initial rate using a graph?

A

Take the gradient of the tangent at 0 minutes- at the very start of the reaction.

Change in y/ change in x

21
Q

How do we use initial rates to work out the rate equation for a reaction?

A

Repeat the experiment several times changing the concentrations of A, B and C one at a time in each experiment to find effect of changing conc of each one on rate.

Calculate initial rate for each experiment

Record concentrations of reactants and their initial rates- work out order with respect to each reactant by looking where the other reactant (s) remain constant.

22
Q

Work out the rate equation of this reaction.

2NO (g) + Cl2 (g) —–> 2NOCl (g)

Initial [NO] Initial [Cl2] Initial rate
1 0.20 0.10 0.63
2 0.20 0.30 1.92
3 0.80 0.10 2.58
4 0.50 0.50 ?

A

Where [NO] is changing and [Cl2] is constant:

1&3 [NO] x4 from 0.20 to 0.80, rate has also x4 from 0.63 to 2.58 1ST ORDER

Where [Cl2] is changing and [NO] is constant:

1&2 [Cl2] x3 from 0.10 to 0.30, rate has also x3 from 0.63 to 1.92 ALSO 1ST ORDER

Rate equation= k[NO][Cl2]

23
Q

Calculate the missing data from this table

A

Calculate k using any experiment

experiment 1-
k= rate/[NO][Cl2]
k= 0.63/ (0.20x0.10)
k= 31.5

rate = k [NO][Cl2]
= 31.5 x 0.50 x 0.50= 7.88 moldm-3 s-1

24
Q

What are the units of initial rate, conc and k in this reaction?

A

Initial rate= moldm-3 s-1
Conc= moldm-3

k= moldm-3 s-1/ moldm-3 x moldm-3
k= mol-1dm3 s-1

25
Q

What to do if there if the concentration of the opposite reactant does not remain constant at any point?

A

Write an additional column where you change the rate according to the reactant which order you know, ignore the reactant you cannot work out.

Work out the difference between the new rate and the one affected by only one reactant, this is the order of the other reactant.

26
Q

Describe the initial rates method for recording rate of reaction: iodine clock

A

H202 + 2H+ 2I- —> 2H20 + I2

Add sodium thiosulfate and starch which acts as an indicator- sodium thiosulfate reacts immediately with I2, when there is no more sodium thiosulfate I2 reacts with starch= blue.

Varying conc of I2 and/or H202 and keeping everything else constant will change the time taken for it to turn blue, can work out order of reaction: reaction vessel on paper with cross, time until cannot see cross.

27
Q

3 ways rate can be measured in experiments if no colour change?
Continuous monitoring

A

1- Change in pH of reaction

pH can change if H+ ions are used up or produced, pH meter can measure pH at regular intervals and can calculate H+

2- Volume of gas produced

measure vol of gas produced over given time using gas syringe- use ideal gas equation to find moles of gas produced, use molar ratio to find reactant concentrations.

3-Amount of mass lost

reactants that produce a gas- reaction mixture on balance, measure vol of mass lost as gas is produced

28
Q

How to measure rate of reaction using a colorimeter?

A

Colorimeter measures the absorbance of a coloured sample, more concentrated= more dark so more light absorbed.

Plot a calibration curve- range of known concentrations of iodine, absorption measured for each one.

29
Q

What is the rate determining step?

A

The slowest step in a multi-step reaction.

Substances not in the rate equation will not be in the rate determining step.

30
Q

How to know which substances will be in rate determining step?

A

If the substances are in the rate equation.

31
Q

Use this multi-step reaction to determine rate-determining step?

Step 1 A + B —–> 2C (fast)
Step 2 2C ——-> D (slow)
Step 3 D + E —–> F + G (fast)

Overall equation= A + B + E —–> F + G

C is catalyst so not in overall equation.

A

Step 2 = rate determining step as slowest

C= one of reactants so must appear in rate equation- 2nd order with respect to C (due to big 2), C= intermediate from A + B so A+B must be in rate equation.

1st order with respect to A+ B

Rate equation= k[A][B][C]^2

31
Q

Use this rate equation to find the rate determining step?

Step 1 NO + NO —–> N2O2
Step 2 N2O2 + O2 —–> 2NO2

Rate = k[NO]2[O2]

A

Ratio of moles we are looking for is 2NO and 1 O2

Use the equations and cross off the molecules which match the ratio we need

Step 1- 2 NO molecules can be crossed off
Step 2- One O2 atom can be crossed off

2nd step= rate determining step as this is the one where the last part of the rate equation was found

31
Q

Use this rate determining step to find the reaction mechanism?

This reaction
(CH3)3CBr + OH- —–> (CH3)3COH + Br-

Mechanism 1
(CH3)3CBr + OH- —–> (CH3)3COH + Br- (SLOW)

Mechanism 2
(CH3)3CBr —–> (CH3)3C+ + Br- (SLOW)

(CH3)3C+ + OH- ——-> (CH3)3COH (FAST)

Rate equation
k[(CH3)3CBr]

A

Rate equation shows us that only (CH3)3CBr can appear in rate determining step

Mechanism 1 shows (CH3)3CBr and OH- whereas Mechanism 2 shows only (CH3)3CBr in step 1

Reaction mechanism likely to be mechanism 2 as rate determining step matches ratio in rate equation.

32
Q

What is the Arrhenius equation?

A

k = Ae^-Ea/RT

Rate= Arrhenius constant (e^x on calculator) to power of activation energy (J) / gas constant (8.31JK-1mol-1) x temp (K)

33
Q

How does activation energy affect rate constant?

A

As activation energy (Ea) decreases, the rate constant (k) gets bigger as rate of reaction increases as more particles have enough energy to react when they collide.

34
Q

How does temperature affect rate constant?

A

As temperature increases, the rate constant increases as particles have more kinetic energy and are more likely to collide with at least the activation energy.

35
Q

How to simplify Arrhenius equation?

A

Take natural log of both sides to remove exponential.

ln K= ln A - Ea/RT

36
Q

Rearrange Arrhenius for Ea?

A

ln K= ln A - Ea/RT

Ea/RT = lnA-lnk

Ea= RT (lnA-lnk)

37
Q

Calculate Ea in Kjmol-1 reaction at 330k, rate constant 1.30x10^-4, Arrhenius constant is 4.55x10^13, gas constant is 8.31

A

Ea= RT (lnA-lnk)

(ln 4.55x10^13 - ln 1.30x10^-4) x 330 x 8.31

110780 Jmol-1 or 110.8Kjmol-1

38
Q

How to use an Arrhenius plot to work out activation energy?

A

Use Arrhenius equation ln K= ln A - Ea/RT

Plot graph lnk on y axis, 1/T (x10^-3) on x axis

Gradient represents -Ea/R

39
Q

Use gradient of -11,000 to work out Ea?

A

Gradient = -Ea/ R

-11,000= -Ea/ 8.31
Ea = - (-11000x8.31)
= 91410Jmol-1
=91.4kJmol-1

40
Q

How to use Arrhenius plot to work out Arrhenius equation?

A

Substitute the value of the gradient and the coordinates of any point on the line into the equation lnK= - Ea/RT + lnA

41
Q

Use coordinates 3.1x10^-3, -4 and gradient -1.1/0.1x10^-3 to work out Arrhenius equation?

A

y = mx + c

lnK= - Ea/RT x 1/T (x axis) + lnA

-4 = -1.1/0.1x10^-3 x 3.1x10^-3 + x

-4= -11000 x 3.1x10^-3 + lnA

lnA= -4 + 11000 x 3.1x10^-3

lnA= -4 + 34.4 = 30.1
have to do exponential log (inverse ln)

e^30.1 = 1.18x10^13 dm3mol-1s-1

Chemistry A Level AQA (25 decks)

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