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Chemistry Y13 > Rate Equations, orders of reaction and Kp > Flashcards

Rate Equations, orders of reaction and Kp Flashcards

1
Q

Rate equations

A
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2
Q

What do rate equations tell us?

A

tells us how changing the concentration of any reactant or catalyst affects the rate of reaction

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3
Q

using A, B, C, AND D draw a equation

A

A + B –> C + D

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4
Q

what is the rate equation for any reaction?

A

r = k [A]^m [B]^n
r= rate
k = rate constant
A+B = reactants
m+n = orders of the reaction with respect to A+B
[ ] = concentration

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5
Q

what does the order of a reaction tell us?

A

how the concentration of a reactant affects the rate of reaction

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6
Q

how can orders with respect to a concentration be represented?

A

numerically using the rate equation or graphically on a rate-concentration graph

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7
Q

What is zero order?

A

changing the concentration of a reactant has no effect on the rate of reaction
[A]^0 = 1

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8
Q

describe what a zero order graph shows with respect to each reactant?

A

if A] changes, the rate stays the same, the order of the reaction with respect to A is zero

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9
Q

What is first order?

A

rate and concentration are directly proportional. Doubling the concentration of a reactant doubles the rate
[A]^1 = [A]

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10
Q

describe what a first order graph shows with respect to each reactant?

A

If [A] doubles, the rate doubles, if [A] triples, the rate triples

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11
Q

What is second order?

A

Rate is proportional to concentration squared. Increasing the concentration of A by x will increase the rate by X^2
[A]^2 = [A]^2

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12
Q

describe what a second order graph shows with respect to each reactant?

A

if [A] doubles, the rate will be 2^2 = 4 times faster. if [A] triples, the rate will be 3^2 = 9 times faster

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13
Q

simplify rate equations:

A
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14
Q

r = k [A]^0 [B]^1

A

r = k [B]

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15
Q

r = [A]^2 [B]^1

A

r = k [A]^2 [B]

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16
Q

Calculate the overall order of the reaction:

A
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17
Q

r = k [A]^2 [B]^1

A

2 + 1 = 3

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18
Q

r = k [A]^0 [B]^1

A

0 + 1 = 1

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19
Q

work out the rate equation from a table

A
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20
Q

how to calculate the missing rates on the table?

A
  • first find out the K constant - do this by placing the values from one column into the rate equation (already rearranged)
  • then using the k constant we can find the missing rates from the table
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21
Q

Calculating units?

A

by simplifying - moldm^3 for every reactant and order

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22
Q

Do lots of practice questions on this

A
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23
Q

Rate determining step

A
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24
Q

What is the rate determining step?

A

the slowest step in a reaction mechanism (of a multi-step reaction)

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25
Q

what is an intermediate?

A

is a species that is formed in one step of the reaction mechanism and used in another step

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26
Q

when we measure the rate of reaction, we actually measure the..?

A

rate determining step - as it is the slowest compared to the other steps

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27
Q

what do rate equations not tell us and tell us?

A

rate equations do not tell us about the steps in the mechanism AFTER the rate determining step.
BUT
tells us about the chemicals involved in the rate determining step, and all the steps before it

28
Q

reactants that are zero order must be involved…?

A

in the mechanism after the rate determining step, and so not included in the r.d.s

29
Q

what does the order in the rate equation tell us about the r.d.s?

A

the order of each reactant tells us how many molecules of the reactant are involved in the rate determining step (either directly or by forming an intermediate)

30
Q

what is the suggested mechanism for the overall equation A + 3B —–> AB3

A

step 1: A + B —> AB
step 2: AB + B —-> AB2
Step 3: AB2 + B —–> AB3

31
Q

Given the rate equation is r = k [A] [B]^2 deduce which step is the rate determining step and explain your answer

A

step 2 is the r.d.s, as when step 1 and step 2 are added together one molecule of [A] and two molecules of [B] react

32
Q

What does the rate determining step exactly match to?

A

the rate equation.
the proportion of the reactants is the same as the orders in the rate equation

33
Q

Why can step 1 not be the rate determining step (r.d.s)?

A

if step 1 had beeen the r.d.s then the rate equation would have been r = k [A] [B]

34
Q

why is step two the r.d.s?

A

the rate equation would be r = k [A] [B]^2

35
Q

why can’t the third step be the r.d.s?

A

the rate equation would be r = k [A] [B]^3

36
Q

How can chemist suggest a mechanism?

A
  • the RDS has to have the same number of molecules as the order for the reactant
  • the other steps in the mechanism eventually generates the products in the balanced equation
  • all intermediated which are created are used up in all later steps
37
Q

Arrhenius equation

A
38
Q

why does an increase in temperature, increase the rate of reaction?

A

by increasing the rate constant

39
Q

sketch a graph to show how the rate constant varies with temperature

A

K is on Y axis, temperature on X axis

40
Q

what is the Arrhenius equation?

A

K = Ae^-Ea/RT

41
Q

what does the k, A, e, Ea, R and T symbolise in the equation

A

K = rate constant (reaction specific)
A = Arrhenius constant (reaction specific)
e = exponential (on calculator)
Ea = activation energy (5mol-1)
R = gas constant (8.31 JK-1 mol-1)
T = temperature (K)

42
Q

what does the equation show as the activation energy get’s bigger, and as temperature increases?

A

the equation shows that as the activation energy (Ea) get’s bigger, K get’s smaller, so a large Ea will mean a slower rate.

Also shows that as temperature rises, K increases

43
Q

How can the arrhenius equation be arranged in order to work out on straight line graphs?

A

we convert it to the logarithmic form - and we compare it to the equation for a straight line graph = y= mx + c so

44
Q

what is the logarithmic form of the equation?

A

InK = InA - Ea/RT

45
Q

what does the symbols represent in terms of y = mx + c?

A

InK = y
InA = c
Ea/R = m
1/T = x

46
Q

when we plot the graph, what do we plot it over?

A

InK over 1/T.
InK = Y axis
1/T = x axis

47
Q

On the graph what would the gradient be calculated?

A

change in Y/ change in X

48
Q

How can we rearrange the equation to get the gradient?

A

Gradient = -Ea/R

49
Q

how can we rearrange the equation to get Activation energy?

A

Ea = gradient x - R(8.31) then divide by 1000 to get into KJ mol-1

50
Q

Equilibrium constant Kp

A
51
Q

why is the quantity of each gas in an equilibrium mixture described in terms of the pressure that it exerts (partial pressure)

A

for reactions involving gases, it is difficult to measure the concentration of a particular gas

52
Q

what is the partial pressure?

A

how much pressure a gas exerts in an equilibrium mixture

53
Q

what is the equation for partial pressure?

A

partial pressure = mole fraction x total pressure

54
Q

how can we calculate mole fraction?

A

mole fraction of reactant/product = n of moles of species/ Total number of moles of all species

Xn = n number of moles / total n of moles of all species

55
Q

A + B ⇌ C
At equilibrium, there are 2.5 moles of A, 2.5 moles of B and 1 mol of C. calculate the mole fraction of A?

A

mole fraction of A = 2.5 / (2.5 + 2.5 + 1 )
mole fraction of A = 0.417 moles

56
Q

questions

A
57
Q

A + B ⇌ 2C + D
At equilibrium, there are 0.25 moles of A, 0.63 moles of B, 1.51 mol of C and 0.65 moles of D. calculate the mole fraction of C?

A

mole fraction C = 1.51 / (0.25+ 0.65+0.63+1.51)
Mole fraction C = 0.497 moles

58
Q

2SO2 + O2 ⇌ 2SO3
At equilibrium, mole fractions are: SO2 = 0.46, O2 = 0.23, SO3 = 0.31. the total pressure is 100Kpa. What is the ppSO2?

A

0.46 X 100 = 46pp

59
Q

PCL5 ⇌ PCl3 + Cl2
At equilibrium, there are 0.67 moles of PCl5, 0.33 moles of PCL3 and 0.33 mol of Cl2. The total pressure is 250Kpa. Calculate the ppPCl5?

A
  1. first calculate the mole fraction of PCl5 - 0.67/ (0.67+0.33+0.33)
  2. then we multiply the answer by the pressure 250Kpa
  3. pp = 126Kpa
60
Q

what is the equilibrium constant for gas phase reactions called?

A

Kp

61
Q

what is the general equation given by Kp?

A

Kp = (ppC)^c (ppD)^d / (ppA)^a (ppB)^b

comes from

aA + bB ⇌ cC + dD

62
Q

explain how units for Kp are worked out?

A

Just like Kc

63
Q

Kp calculation

A
64
Q

2H2 + CO ⇌ CH3OH
2.0 moles of H2 and 1 mole of CO were mixed together. At equilibrium the mixture contained 0.175 moles of CH3OH. Calculate a value for Kp and units. given that total partial pressure = 1.75 x 10^4.

A

we can do this by making a table and find out the moles at equilibrium.

then we can work out the mole fractions of each substance

then using the mole fraction ,we can work out the pp of each substance

then we can insert the values into the Kp equation to find an answer

then find units using the equation - measure in Kpa or Pa

  • answers on sheet
65
Q

FINISH

A

Chemistry Y13 (13 decks)

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