rate equations

Question 1

how can reactants be expressed mathematically
A+ B = C

Answer 1

-double conc of A then rate is directly prop to A so the rate does not change - order is 1
-double A quadruples the rate is value squared and order is 2
-double A had no effect on rate so order is zero

Question 2

what is the rate equation

Answer 2

rate = k[A]^m[B]^n

Question 3

how do scientists find the order of the reactants

Answer 3

through experiments - DO NOT USE BALANCE NUMBERS

Question 4

what is the rate constant dependent on

Answer 4

temp

Question 5

how to find overall order

Answer 5

add all orders together

Question 6

order of a catalyst?

Answer 6

same method as reactants

Question 7

what are initial rate experiments

Answer 7

conc changed one at a time but everything else remains the same
how to measure rate immediately at the start eg use disappearing cross method

Question 8

how to find orders using table

Answer 8

start with reactant A - look for two experiments where the conc of A has changed and everything else is the same- then work out order using principles eg if conc of A doubles and rate doubles = first order
then do same for B and then C
then add total orders

Question 9

what to do if cant see relationship easily

Answer 9

divide rate value for bigger conc by smaller

Question 10

how to rearrange rate equation to find K

Answer 10

k = rate/reactants

Question 11

what are the units for K

Answer 11

mol dm s-1
the mol and dm will change
work out using same method as for kc and kp

Question 12

what does the conc-time graph of a zero order reactant look like

Answer 12

In a zero-order the concentration of the reactant is inversely proportional to time
This means that the concentration of the reactant decreases with increasing time
The graph is a straight line going down

Question 13

what does the graph for a conc-time graph for first order look like

Answer 13

In a first-order reaction the concentration of the reactant decreases with time
The graph is a curve going downwards and eventually plateau

Question 14

what does the graph for a conc-time graph for second order look like

Answer 14

In a second-order reaction the concentration of the reactant decreases more steeply with time
The concentration of reactant decreases more with increasing time compared to in a first-order reaction
The graph is a steeper curve going downwards

Question 15

what is the rate -conc graph for zero order

Answer 15

In a zero-order reaction the rate doesn’t depend on the concentration of the reactant
The rate of the reaction therefore remains constant throughout the reaction
The graph is a horizontal line
The rate equation for this one reactant is rate = k

to find k - to find the gradient - work out from conc-time graph

Question 16

how to plot a rate-conc graph

Answer 16

The progress of the reaction can be followed by measuring the initial rates of the reaction using various initial concentrations of each reactant
These rates can then be plotted against time in a rate-time graph

Question 17

what is the rate -conc graph for first order

Answer 17

In a first-order reaction the rate is directly proportional to the concentration of a reactant
This means that if you doubled the concentration of the reactant, the rate would also double
If you increased the concentration of the reactant by a factor of 3, the rate would increase by this factor as well
The graph is a straight line
The rate equation for this one reactant is rate = k [A]

find k to find the gradient- use rate - conc graph
change in y over change in x

Question 18

what is the rate -conc graph fo

Answer 18

In a second-order reaction, the rate is directly proportional to the square of concentration of a reactant
The graph is a curved line
The rate equation for this one reactant is rate = k [A]2

k = gradient of rate-conc^2

Question 19

how do u draw rate-conc from conc-time for first and second order

Answer 19

draw tangents at specific conc and calculate the gradients
use these to plot rate-conc
with gradients on x-axis

Question 20

what is the rate determining step

Answer 20

the slowest step in the reaction

Question 21

how to determine which reactants are in the rate equation

Answer 21

If a reactant appears in the rate-determining step, then the concentration of that reactant will also appear in the rate equation

Question 22

what is the overall rate of the reaction dependent on

Answer 22

the slowest step thus the RDS

Question 23

how can u predict the RDS

Answer 23

The overall reaction equation and rate equation can be used to predict a possible reaction mechanism of a reaction
This shows the individual reaction steps which are taking place

-deduce which reactants are which order
-then this is how many molecules are in the RDS
-compare to mechanism given
-pick which one is RDS by seeing if it matches rate equation and overall

Question 24

example of RDS prediction

Answer 24

For example, nitrogen dioxide (NO2) and carbon monoxide (CO) react to form nitrogen monoxide (NO) and carbon dioxide (CO2)
The overall reaction equation is:
NO2 (g) + CO (g) → NO (g) + CO2 (g)

The rate equation is:
Rate = k [NO2]2

From the rate equation it can be concluded that the reaction is zero order with respect to CO (g) and second order with respect to NO2 (g)
This means that there are two molecules of NO2 (g) involved in the rate-determining step and zero molecules of CO (g)
A possible reaction mechanism could therefore be:
Step 1:

2NO2 (g) → NO (g) + NO3 (g) slow (rate-determining step)

Step 2:

NO3 (g) + CO (g) → NO2 (g) + CO2 (g) fast

Overall:

2NO2 (g) + NO3 (g) + CO (g) → NO (g) + NO3 (g) + NO2 (g) + CO2 (g)

= NO2 (g) + CO (g) → NO (g) + CO2 (g)

Question 25

are species that appear after the RDS included in the rate equation

Answer 25

no

Question 26

what species do u use in the RDS to compare to the rate equation

Answer 26

the ones BEFORE the RDS

Question 27

how to identify a catalyst

Answer 27

When a rate equation includes a species that is not part of the chemical reaction equation then this species is a catalyst

Question 28

how to identify intermediates

Answer 28

produced in one of the mechanisms of the reaction but are not involved in the overall reaction
they should also cancel out

Question 29

what does the Arrhenius equation show

Answer 29

how changes in temp and Ea affect the value of k

Question 30

what is the Arrhenius equation

Answer 30

k = A x e ^ -(Ea/RT)

Question 31

what are the units for the Arrhenius eq

Answer 31

A= constant - has same units as k - given in the question
e= exponential = constant with a value of 2.7182…
R= gas constant= 8.31 J K-1 mol-1
T= temp in kelvin
Ea= J mol-1
k=rate constant = no. of successful collisions resulting in a reaction per second

Question 32

what factor of collision theory does the Arrhenius constant use

Answer 32

molecules have to be in the correct orientation in order to react

Question 33

example calc using Arrhenius eq

Answer 33

a) chemical reaction is carried out at 20 c and has an Ea of 50 KJ mol-1
calculate the fraction of molecules that are able to react at this temp

20 = 293k and 50KJ = 50000 j
work out value of -(Ea/RT)
then work out fraction of moles able to react
e to power of value of -(Ea/RT)

b) value of A is 4.79x10^9 s-1 calc the value of k

k= 4.79x10^9 x e to the power
then find units

Question 34

where is e on the calculator

Answer 34

shift then natural log button

Question 35

what effect has 10 degrees increase in T had on the value of k

Answer 35

doubled

Question 36

how does the change in the value of k by 10 degrees affect the rate

Answer 36

value of k double so rate doubles

only works with Ea 50Kj

Question 37

what is another form of the Arrhenius eq

Answer 37

ln k = ln A - Ea/RT

this is using the natural log form which enables u to eliminate the exponential to find Ea

Question 38

how to rearrange ln k = ln A -Ea/RT

Answer 38

Ea =RT (lnA - lnk)

Question 39

how can u use the natural log form of Arrhenius eq to find the Ea on a graph

Answer 39

ln k or ln (rate) = lnA - Ea/R (1/T)
l l l l
y= c + m x

then find gradient of line ( big triangle_
change in y /change in x
then should get a negative no.
then as two negative cancel gradient becomes +ve
then x this by the gas constant
and divide by 100 to get into KJ

Question 40

how do u draw a graph using Arrhenius

Answer 40

ln k on y-axis
(1/T) on x -axis