Rate equations (y13) Flashcards
rate = k[A]^m [B]^n
m and n = orders of reaction with respect to reactants A and B
K = rate constant
rate K constant values
varies with temperature
rate equations show –> In rate equations, the mathematical relationship between rate of reaction and concentration gives
information about the mechanism of a reaction that may occur in several steps.
rate constant affected by temperature equation
k = Ae–Ea/RT
where A is a constant, known as the Arrhenius constant,
Ea is the activation energy and T is the temperature in K
key mark scheme points when making a graph
-sensible scales (cover at least half of the paper and not above squares)
-line through points must be smooth
-line of best fit ignores anomalies
-calculate gradient
expectations for a catalyst to be in rate equation
-It must have a measurable and quantifiable effect on the rate of reaction
-The catalyst must be homogeneous
-If a chemical appears in a rate equation but is not one of the reactants, then it is a catalyst
write the rate law for this equation
2NO (g) + O2 (g) –> 2NO2 (g)
K {NO]^2 {H2}^2
rate of reaction and units
Rate of reaction is the change in concentration of a product or reactant per unit time
Units of ROR = mol dm^-3 s^-1
Average
Instant
Average = averaged over the whole time of the experiment
Instant = rate at a particular time
rate expression
-rate expression tells us how the concentration of reactants and catalysts affect the reaction rate. Derived from experimental data
-rate = K{A}^a {B}^b
-rate of reaction depends on some or all the species in the reaction vessel
how can rate data be determined experimentally
-data of concentration and initial rate
-plotting a graph and look at shape
-plotting a graph with logarithmic scale of concentration and rate (will always be straight, gradient is the rate order)
changing temperature and pressure
If temperature increases or a catalyst is used the rate of reaction increases but concentrations remain constant. Rate constant (K) increases.
Rate constant (K) is unaffected by pressure change. When pressure increases, concentration increases and K remains constant
determining rate equation
K = rate / {R}
R = reacts with s to make products
Rate = K {R}
1) Find pair of experiments -> concentrations must be same
2) calculate how many times greater concentration of other reactant was
3) find rate reactant
4) calculate K and find units
order of a reaction
The order of a reactant shows how the concentration of a reactant affects the rate of reaction
It is the power to which the concentration of that reactant is raised in the rate equation
The order can be 0, 1 or 2
When the order of reaction of a reactant is 0, this means that it has no effect on the rate of the reaction and therefore is not included in the rate equation at all (unchanged)
When the order of reaction of a reactant is 1, the rate is directly proportional to the concentration of that reactant (x A)
When the order of reaction of a reactant is 2, the rate is directly proportional to the square of the concentration of that reactant (x A^2)
The overall order of reaction is the sum of the powers of the reactants in a rate equation
E.g Rate = k [NO]2 [H2]
Second-order with respect to NO
First-order with respect to H2
Third-order overall (2 + 1)
half life
The half-life (t1/2) is the time taken for the concentration of a limiting reactant to become half of its initial value
concentration vs time graph
-Graphs of concentration of reactant against time give some clues to the order or the reagent
-straight line = reaction is zero order –> constant with change if conc so conc doesn’t havent rate of reaction
-curve = first or second order
arrhenius equation
k = A e − E a R T
calculating gradient then change in concentration
1) Draw tangent then do change in y / change in X to get gradient
2) change in concentration x (A)2
3) Square root answer to get new {A}
A as pre-expoential factor
pre-exponential factor is related to collision frequency and the orientation of the particles (Same units as K)
rearranged arrhenius question (remove expoential)
Easier way of figuring out Arrhenius equation –> LnK = LnA – (Ea / RT)
rearranged to find activation energy
-Ea = RT (Ink – InA)
rearrnaged to find T
T = Ea / (InA – InK) x R
rearranged to find A
InA = InK + Ea / RT
arrhenius equation in relation to y=mx + c
Y= Ink = y-axis
M = Ea / R
X = 1/T = x axis
C = InA
rearrnged rate constant
K = rate / [C] [D]
units of K = mol-1dm3s-1
calculate Ea at 25 degrees in KJ
InA = 16.9
R = 8.31
K = 0.028
temp = 298
-Ea = RT (InK-InA) / 1000
(8.31) (In0.028 - 16.9) = 51
dont need In for A in equation as it is already provided in the equation
order reaction
relationship between concentration of the reactant and rate of reaction
order of reaction
-straight line –> if you change conc by x2 the rate does nothing
-straight line diagnoal –> if you change conc by 2 sometimes rate will be x2
-curve –> if you change conc by x2 sometimes the rate will be x4
zero order
A + B –> C
-increase A by a factor of X there is no change in rate
1st and 2nd order
A + B –> C
1st –> if we increase A by x and the rate increases by x we describe the order with respect to A as one
2nd –> if we increase A by x and the rate increases by X sqaured the rate is respective to A as two
K [NO]2 [H2]
NO = 2nd
H2 = 1st
which statement is correct
K = [X]2 [Y]
the rate of the reaction is halved if the concentraiton of X is halved and the concentration of Y is doubled
units of rate constant
The units of k for a zero-order reaction are M/s,
the units of k for a first-order reaction are 1/s,
and the units of k for a second-order reaction are 1/(M·s)
how to know if reaction is 2nd order
A reaction is second order overall when it is second order in one reagent, zero order in all others, or first order for two reagents (1 + 1 = 2). For example, the combination reaction A + B → C would be second-order overall if first-order in both A and B.
1, 2, 3 calculating units
rate = moldm-3s-1
reactants = moldm-3
One:
rate = K [A]1 –> moldm-3s-1 /moldm-3 so units are s-1
Two:
rate = K [A]2 = mol-1dm3s-1
Three: mol-2 dm6 s-1
half life
zero order = half life decreases over time + gradient is constant
1st order = half life is constant + gradient decreases exponentially
2nd order = half life increases over time + gradient decreases even faster
explain why the student is able to deduce that iodine is zero order
-volume of sodium is directionally proportional
-propanone was in large excess = so only iodine is varied
calculate the order of experiment with respect to B
B is doubled so rate double = 1st order
determine Ea from table
subtract 2 temperatures close together or at ends of each table
In(1/time) - In(1/time) / 1/t - 1/ T
if activation energy is high
rate constant is low
if a reactant has a zero order, do you include it in the rate expression
no
[A] =
rate / K
calculate missing value in rate table
-find K first
-rearrange K = rate / [A] equation to find missing values
method to measure change in volume of gas
- Measure 50 cm3 of the 1.0 mol dm–3 hydrochloric acid
and add to conical flask. * Set up the gas syringe in the stand - Weigh 0.20 g of magnesium. * Add the magnesium ribbon to the conical flask, place the
bung firmly into the top of the flask and start the timer. * Record the volume of hydrogen gas collected every 15
seconds for 3 minutes.
what is the rate determining step in the 2 equations
1) (CH3)3CBR –> (CH3)3C+ + Br-
2) (CH3)3C+ + OH- –> (CH3)3COH
step 1 = longer to break strong bond in haloalkane
zero order if ….
appears in mechanism after the slow step
if NO2 apperars twice in rate determing step it is in
second order
when drawing mechanisms refer to organic chem e.g if it a haloalkane and OH its
nucleophilic subsitution
rate determing step
-rate determining step is the slowest part of the reaction that is key for the production of products
-reactions often occur in multiple steps as it is very unlikely that all the reactants collide at the same time
equation for
H2O2 (Aq) + 2H+(aq) + 2I- (Aq) —> I2 (aq) + 2H2O
1) H2O2 + I- –> HOI + OH- (Slow = rate detemining step)
2) HOI + I- –> I2 + OH- (fast)
3) 2OH- + 2H3O+ –> 4H2O (l) (fast)
A + B + C + D +E
in this equation C is the rate determining step as it limits the rate at which products are produced
A + 2B –> C + D
Rate equation = K = [A][B]
Step 1 = A + B –> P (slow = RDS)
Step 2 = B + P –> C + D (Fast)
A catalyst reacts with the reactants as is reproduced in a later step. It is not shown in the overall reaction, only the rate equation e.g K = [A] [H+] –> H+ is the catalyst in the overall equation A + 2B –> C + D
Step 1 = A + H+ –> AH (Slow RDS)
Step 2 = AH+ + B –> AB + H+ (fast)
Step 3 = AB + B –> D + E (fast)
intermediate
A + B + C –> Y + Z
Step 1 = A + B –> D (D is an intermediate as it is used in 2nd reaction to make products and is not shown in the overall equation)
Step 2 = D + C –> Y + Z
total order of an equation
Total order of K {CH3] [H+} = 2 as 1 + 1 = 2
true or false –> any reactant involved after rate determing steps will be zero order
true
when will a step no longer affect the rate of reaction
when the step occurs outside the rate detemining step
rate equation for the reaction
slow = A + C –> X
fast = X + B –> D + E
K = [A]1 [C]1
rate equation for the reaction
fast = A + C –> X
slow = X + B –> D + E
K = [A]1 [B]1 [C]1
is step 1 or 2 the rate detemiing step
E + F –> EF = step 1
EF + F –> G = step 2
step 2
step which requires catalyst =
step which creates intemediate =
rate determiining step
y = mx + c
lnK = -Ea/RT + InA
gradient (m) =
-Ea / R
to get A from InA on calculator do
shift In to get exponential
true or false -> if doubling the concentration of S has no effect on rate it is in the zero order
true
so rate equation doesnt include
Show how these data can be used to deduce the rate expression for the
reaction between A and B
-B = experiment 1 and 2 is constant
-A x3 and rate = 3^2 so 2nd order in experiment 1 and 2
-A in experiment 2 and 3 only increases by 2 not 2^2 so B halves = respect to order 1
-rate = k [A]2 [B]
why does doubling the temperature have a much greater effect than doubling conc of E
rate = K [E]
-reaction occurs when molecules have energy greater than or equal to activation energy
-doubling temperature increases kinetic energy
-doubling E only doubles number with E
explain whether these haloalkanes would react with NaOH by the same mechanism
No – the mechanisms are different because with X the reaction is first order with respect to [NaOH] but with Y it
is zero order with respect to [NaOH]
iodine clock
H2O2 + 2I- + 2H+ –> I2 + 2H2O
reactants = colourless
-products = black
react in the prescence of an acid
initial rate -> iodine clock
-take gradient at t = O (initial rate)
-repeat after changing conc of a different reactant
-not time for colour change
-determine order
large excess =
zero order
how to calculate rate when [H+] = 0.35
draw tangent at 0.35
find gradient
explain why the use of a large excess of H2O2 and I- means the rate of reaction at a fixed temperature depends only on the concentration of H+ (aq)
-H2O2 and I- concentrations are constant (negligble change)
-H2O2 and I- have no effect on rate as they are zero order
state and explain what must be done to each sample before it is treated with an alkali
-stop the reaction
use ice
explain how the graph shows the order with respect to H+ is zero
-contstant gradient
-H+ is decreasing but rate is not changing
explain how you could use a series of experiments to determine the order of this reaction with respect to A
A + B + C –> D + E
-reagent X reacts with E
-measure known conc and volume of A and B
-add known conc and volume to container
-measure conc and vol of C in separate container
-add c and set timer
-record time for blue-black colour
-repeat with different conc of A but same for B and C
-keep constant temp using water bath
-plot graph of 1/time vs conc of A
-deduce order from graph
units for 2nd order reaction
mol-1dm3s-1
why is the reaction fastest at the start
highest concentration of reactants
more frequent and successful collisions
what order does the graph show and why
1st order
straight line graph through the origin
Explain why the use of a large excess of H2O2 and I– means that the rate of reaction
at a fixed temperature depends only on the concentration of H
concentration is negligble
so is in zero order
Explain, giving brief experimental details, how you could use a series of experiments
to determine the order of this reaction with respect to A. In each experiment you
should obtain a measure of the initial rate of reaction.
1a Measure (suitable/known volumes of) A, B and C (ignore quoted values forvolume)1b Use of colorimeter1c into separate container(s) – (allow up to two reagents measured together intoone container) – ignore use of X
Stage 2 Procedure
2a Start clock/timer at the point of mixing2b Take series of colorimeter readings at regular time intervals2c Use of same concentration of B and C / same total volume / (samevolume/amount of X)2d Same temperature2e Repeat with different concentrations of A (can be implied through differentvolumes of A and same total volume)
Stage 3 Use of Results
3a Plot absorbance vs time and measure/calculate gradient at time=03b plot of gradient against volumes/concentrations of A or plot log(1/time) vslog(volume or concentration of A)3c description of interpreting order from shape of 1/time vs volume orconcentration graph / gradient of log plot gives order
What is meant by the term order of reaction with respect to A
Power (or index or shown as x in [ ]x
) of concentration term
(in rate equation)
Suggest an observation used to judge when all the iodine had reacted.
brown colour released
each sample taken from the reaction mixtures is immediately added to an excess of sodium hydrogencarbonate solution before being titrated with sodium thisulfate solution. Suggest the purpose of this addition
stop reaction by removing acid catalyst
graph =
constant gradient
rate doesnt change as conc increases
zero order
rate and temperature
if temperature increases the value of the rate constant increases = faster rate of reaction due to more kinetic energy
but concentration remains constant
iodine clock reaction
H2O2 + 2H+ +2I- –> 2H2O + I2
add sodium thiosulfate and starch (indicator)
sodium thiosulfate immediately reacts with iodine
by varrying the concentration of iodine and keeping everything else constant we can observe the time taken for the blue-black colour to appear
different ways of calculating rate of reaction
-calculate pH at regular intervals using pH meter then calculate H+ ion concentration
-calculate mass loss
-measure volume of gas using syringe
propanone and iodine colour change
brown to colourless
I2 + CH3COCH –> CH3COCH2I +I- + H+
plot calibration curve = known concentration of iodine and absporbance
rate determing step =
reactant doesnt appear in rate equation
e.g rate = k [NO]2 [O2]
NO + NO –> N2O2
N2O2 + O2 –> 2NO2
if we cross off everything in rate equation N2O2 in step 2 remains meaning it is the rate determing step
smaller activiation energy =
higher rate constant
more chance of collisions
increase temp =
increase rate constant = more kinetic energy
graph =
x axis = 1/T
y axis = Ink
