rate equations Flashcards

1
Q

what are in the orders of rate in respect to A

A

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

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2
Q

what is the rate equation

A

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

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3
Q

how do scientists find the order of the reactants

A

through experiments - DO NOT USE BALANCE NUMBERS

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4
Q

what is the rate constant dependent on

A

temp

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5
Q

how to find overall order

A

add all orders together

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6
Q

order of a catalyst?

A

same method as reactants

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7
Q

what are initial rate experiments

A

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

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8
Q

how to find orders using table

A

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

OR
work out K
put the values in formula/ rearrange to find which one u are looking for

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9
Q

what to do if cant see relationship easily

A

divide rate value for bigger conc by smaller

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10
Q

how to rearrange rate equation to find K

A

k = rate/reactants

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11
Q

what are the units for K

A

mol^-1 dm^3 s-1
the mol and dm will change
work out using same method as for kc and kp

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12
Q

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

A

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

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13
Q

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

A

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

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14
Q

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

A

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

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15
Q

what is the rate -conc graph for zero order

A

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

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16
Q

how to plot a rate-conc graph

A

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

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17
Q

what is the rate -conc graph for first order

A

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

18
Q

what is the rate -conc graph for second order

A

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

19
Q

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

A

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

20
Q

what is the rate determining step

A

the slowest step in the reaction

21
Q

how to determine which reactants are in the rate equation

A

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

22
Q

what is the overall rate of the reaction dependent on

A

the slowest step thus the RDS

23
Q

how can u predict the RDS

A

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

24
Q

example of RDS prediction

A

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)

25
Q

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

A

no

26
Q

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

A

the ones BEFORE the RDS

27
Q

how to identify a catalyst

A

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

28
Q

how to identify intermediates

A

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

29
Q

what does the Arrhenius equation show

A

how changes in temp and Ea affect the value of k

30
Q

what is the Arrhenius equation

A

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

31
Q

what are the units for the Arrhenius eq

A

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

32
Q

what factor of collision theory does the Arrhenius constant use

A

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

33
Q

example calc using Arrhenius eq

A

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

34
Q

where is e on the calculator

A

shift then natural log button

35
Q

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

A

doubled

36
Q

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

A

value of k double so rate doubles

only works with Ea 50Kj

37
Q

what is another form of the Arrhenius eq

A

ln k = ln A - Ea/RT

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

38
Q

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

A

Ea =RT (lnA - lnk)

39
Q

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

A

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 1000 to get into KJ

40
Q

how do u draw a graph using Arrhenius

A

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

41
Q

Suggest why chloroethanoic acid is a stronger acid than ethanoic acid.

A

M1 Curve steeper at first & flattens at same point on y axis
M2 Cl is an electron withdrawing group or negative inductive effect
M3 Weakens the O-H bond / increase polarity of O-H bond

42
Q
A