Kinetics - CI

Question 1

What is the rate equation

Answer 1

rate = k[A]ˣ[B]ʸ

k = rate constant
x = order of reaction with respect to A
y = order of reaction with respect to B

Question 2

How to fin the overall order of a reaction

Answer 2

the powers added together in the rate equation
e.g. rate = k[H]¹[Cl]² overall order = 3

Question 3

Using rate = k[A]ˣ, if x=0. What is the order with respect to that reactant. What will the rate equation be and how does it affect the rate

Answer 3

zero order with respect to A. Rate = k[A]⁰ -> rate = k so the rate does not depend on the concentration of that particular reactant

Question 4

Using rate = k[A]ˣ, if x=1. What is the order with respect to that reactant. What will the rate equation be and how does it affect the rate

Answer 4

first order with respect to A. Rate = k[A]¹ -> rate = k[A] so the rate is directly proportional to the concentration of that reactant (if [A] doubles the rate doubles)

Question 5

Using rate = k[A]ˣ[B]ʸ, if x=2. What is the order with respect to that reactant. What will the rate equation be and how does it affect the rate

Answer 5

second order with respect to A. Rate = k[A]² if [A] doubles the rate quadruples

Question 6

How do you find the orders w.r.t reactants and how

Answer 6

experimentally, to find the order w.r.t one particular reactant we need to change the concentration of it and see what happens to the rate. To find the effect on rate we measure the amount of product made (or reactant used)

Question 7

Common variables to measure change in concentration

Answer 7

mass lost, pH, colorimetry, volume of gas, conductivity, quenching titration (take sample, stop reaction, test sample)

Question 8

Name the 4 ways to find the order w.r.t a particular reactant

Answer 8

multiple trials, single trial, quenching, clock reaction

Question 9

How to find the order w.r.t a particular reactant - Multiple trials

Answer 9

we are changing the conc of one reactant (5 trials, 5 concs) but keeping initial conc of other reactants the same, measure chosen variable at set time intervals, plot a conc vs time graph, calculate initial rate by tangent at t=0 as at initial rate the other reactants concs are constant, calculate gradient then draw a rate vs conc graph to find the order

Question 10

How to find the order w.r.t a particular reactant - Single trial

Answer 10

do one trial and look at how the conc of one reactant changes as the reaction progresses, all other reactants must be in large excess so effectively the conc stays constant, plot a conc vs time graph to find order. If possibly first order check for half life but if not clear find the gradient (rate) at 5 different conc on the curve and plot a rate vs conc graph

Question 11

How to find the order w.r.t a particular reactant - Quenching

Answer 11

variation on single trial - small samples of the reaction mixture are removed at regular time intervals, samples are
immediately quenched to stop the reaction, by putting them into an ice bath to lower the temperature, sample is then analysed,
e.g. by titration.

Question 12

How to find the order w.r.t a particular reactant - Clock reaction

Answer 12

do multiple trials (5) and change the conc of one of the reactants, measuee time taken for the reaction to reach a fixed point when a set amount of product is made (usually colour change). The time is converted to rate by doing 1/time. It is important that only a small % of reactants have reacted as we can assume the conc of all the other substances are effectively constant, as for initial rates. This is repeated, changing the initial conc of the reactant but keeping the others the same. This allows for the measurement of the initial rate of reaction by simply recording the time it takes for the observable endpoint to appear.

Question 13

Progress curves - concentration against time graph of zero order

Answer 13

https://chem.libretexts.org/@api/deki/files/8555/Image01.jpg?revision=1&size=bestfit&width=253&height=244

Question 14

Progress curves - concentration against time graph of first order

Answer 14

https://lh5.googleusercontent.com/proxy/KWu_zCFNlD1_Pdx8OyqLXxC0OfDRmsk25K3EpWfTqeBX_q87GXR0pgpMqf8PuypC0PY89tRM-_eoI48UBZae5aljc8Tu6D3h7htLCuoGRusGOUi3Ryqvvpsy13zFoZaqKgZ_7zlTtv4wOl4FwMm8MQ

Question 15

Progress curves - why is the first order graph of conc against time special

Answer 15

it has an exponential shape so for the first order reaction the half life is constant (time taken for half the reactant to get used up).
https://chem.libretexts.org/@api/deki/files/17871/firstorderint_lin.png?revision=1

Question 16

Progress curves - concentration against time graph of second order

Answer 16

https://lh6.googleusercontent.com/proxy/OHxlISNS-IzNblMGXijDu6Z2cVC8sPvFECoEUbv5zyjjKUVZEPYsav20VQsEIl7mxojPuhcvAgYnlhSAMys-Y_QLr-NBVyAqI7_V-LqLjcD_etjBuO32qWAMwE6m3EULLkgWFNwY6IgFlB_tlMBIIUI

Question 17

Progress curves - rate against concentration graph of zero order

Answer 17

https://lh6.googleusercontent.com/proxy/hzXn-lbFNMDUBKdIYua5Sgk88OLlvbhAcIpKhY5_yvJNGee7Wk19xVtAkJRcHnofLPg6muGAq0aQAV8uSF4DWA6hO7Y_yn71f3BqPaOLmRjnyFU8YHXAOwLuTAYglSX6UyV03AiRY2EO2UxLFLUL

Question 18

Progress curves - rate against concentration graph of first order

Answer 18

https://lh4.googleusercontent.com/proxy/PSmxwntK1g08B9UcJ-HYnbTjnQ1rQyDquNKmsPC-VvT_0RpAcXHA7FRlKoA2ApNdwDeb0WeIf0czaPn_ztjL1tii2wuSNqPxrSj0_ERb04xZ-mR6MMW0VLZeNMxL78jRALrvouj1VFH3orQYpLxBQw

Question 19

Progress curves - rate against concentration graph of second order

Answer 19

https://lh5.googleusercontent.com/proxy/EGPfIfZX-VTYeiq1ZEr8f3gCtB_s84yK7qLqfKOgoSixBFurExqgdhtehogzYfCmZrHVPA2QYnzrYhqGAnYbf7DO4lkkrJhzac5U4vWjfyJf4dIbMvBd9IBv_7it9Cml5XVVA67xMWNZIxgzKEstgmQ

all progress graphs = https://s3.eu-west-2.amazonaws.com/elements.cognitoedu.org/7404157d-aa19-451b-b7f0-cf068583d47c/reaction-order-rate-concentration-graphs.png

Question 20

Progress curves - rate against concentration graph of second order with concentration squared

Answer 20

https://o.quizlet.com/0tl-8JpDOvs9Bh-yTihvBg.png

Question 21

Many reactions occur in more than one step so we can propose a mechanism for the reaction which shows the separate steps.
e.g. A + B -> Q
Q + B -> D
What is the overall equation for this reaction

Answer 21

A + 2B -> D

Question 22

Mechanism of a reaction
A + B -> Q
Q + B -> D
What is Q in this reaction

Answer 22

the intermediate

Question 23

When finding the mechanism of a reaction and the separate steps. The steps occur at rates and what is the slowest step called and why is it important

Answer 23

the rate determining step (RDS) and the overall rate of reaction depends on this slow step. Only chemical which are involved in or before the rate determining step affect the overall rate of reaction

Question 24

If the reaction has the separate steps:
A + B -> Q step 1 slow
Q + B -> D step 2 fast
what is the rate equation and why

Answer 24

rate = k[A][B]
because the mechanism shows one lot of A and one lot of B in or before the RDS therefore they are both first order

Question 25

If the reaction has the separate steps: A + A -> Q step 1 slow Q + B -> D step 2 fast what is the rate equation and why

Answer 25

rate = k[A]² because the mechanism shows two lots of A and zero lots of B in or before the RDS therefore A if second order and B is zero order so not in the rate equation

Question 26

If the reaction has the separate steps: A + B -> Q step 1 fast Q + B -> D step 2 slow what is the rate equation and why

Answer 26

rate = k[A][B]² because the mechanism shows one lot of A and two lots of B in or before the RDS therefore A is first order and B is second order. Q is an intermediate so not shown in the rate equation

Question 27

If the conc of a reactant is raised to the power of 1 in the rate expression... (3 things)

Answer 27

- only one lot of that chemical in involved in or before the RDS (rate determining step) - it is first order with respect to that reactant - if we double the conc of the reactant the rate will double

Question 28

If the conc of a reactant is raised to the power of 2 in the rate expression... (3 things)

Answer 28

- two lots of that chemical in involved in or before the RDS - it is second order with respect to that reactant - if we double the conc of the reactant the rate will quadruple

Question 29

If the conc of a reactant is not included in the rate expression... (3 things)

Answer 29

- none of that chemical in involved in or before the RDS (rate determining step) - it is zero order with respect to that reactant - if we double the conc of the reactant the rate will stay the same

Question 30

How can you predict a mechanism of a reaction

Answer 30

use intermediates (can be 'made up' chemicals) steps must add up to overall equation the RDS must be chosen to fit the rate equation highly unlikely that more than 2 molecules will collide simultaneously so only have 2 molecules in each step

Question 31

Why is 2A + B -> Q step 1 slow Q + B -> D step 2 fast not likely

Answer 31

as each step should have a maximum of 2 particles colliding as it is statistically unlikely all 3 particles will collide simultaneously

Question 32

If A + B -> C + Q step 1 slow Q + B -> D + E step 2 fast Why can't the rate expression be rate = k[A][B]²

Answer 32

as the mechanism shows one lot of A and one lot of B in or before the RDS therefore both would be first order. However, the rate equation shows A as first order but B as second order, so there should be two lots of B in or before the RDS