Year 13 Chemistry Notes- STANDARD ANSWERS
Rate equations
What is rate?
• The change in concentration per unit time
• Rate can be the rate at which the reactant is lost or the product is formed
What do particles need in to have order to react?
• Correct orientation
• Have energy more than or equal to the activation energy of the reaction
Why does the concentration, surface area and pressure increase rate of reaction?
• More particles have E ≥ Ea
Why does a small increase in temperature massively increase the rate of reaction?
• Many more particles have E ≥ Ea
What is the function of a catalyst?
• It provides an alternative pathway with a lower activation energy so more collisions
are successful as many more particles have E≥Ea
What is meant by the term ‘order of reaction with respects to A”?
• The power to which the concentration o f A in the rate equation affects the rate of
reaction
What is a rate equation?
• A mathematical relationship between rate and the molar concentrations of reactions
• A and b are the order of the reactions (powers) in respect to [A] and [B]
• The overall order of the equation is the sum of the powers (a + b)
What is k?
• The rate constant
• It is different for different temperatures and differs with temperature exponentially
Why can't the rate equation be written from a balanced (stoichiometric) equation?
• A rate equation may include a homogenous (same state as reactants) catalyst which
isn’t in a balanced equation as it is not used up
How is a rate equation determined?
• By experiments
E.g. Determining rate for orders:
A+B+C→D
Rate = K [B] [C]2
This means that the rate is independent of the concentration of A
Initial rate (mol-3s-1) Change in Effects on rate New initial rates
concentration of (mol-3s-1)
reagents
2.5 [A] x 3 None 2.5
0.75 [B] x 4 X4 3
, 12 [C] x 1o X 100 (as C is 1200
squared)
4.8 [B] x 3, [C] / 2 X 3/4 3.6
0.80 [D] / 2 None 0.80
E.g. Finding the units of the rate constant:
Rate equation Rearange to give k Working Units for k
Rate=K[A] K= rate/[A] K= (mol dm-3 )s-1 / (mol dm-3 ) s-1
Rate=K[C][H] K=rate/[C][H] K= (mol dm-3 )s-1 / (mol dm-3 ) Mol-1 dm3 s-1
(mol dm-3 )
Rate=K[D]2[C] K=rate/ [D]2[C] K= (mol dm-3 )s-1 / (mol dm-3) Mol-2 dm6 s-1
(mol dm ) (mol dm )
-3 -3
Remember that (mol dm-3 ) (mol dm-3 ) = mol2 dm-6
Remember that when you have a fraction, in order to bring the denominator to the
numerator, you multiply by –1, so s-1 / (mol dm-3 ) becomes mol-1 dm3 s-1
How would you find rate equations using initial rates data and find the value of k?
E.g. L reacts with M: 2L + 2M → N + 2P
Rate = k[L][M]2
Find the initial rates of [M] on experiment 3 and the overall initial rate for experiment 2
1. Work out the value of k using a complete row (experiment 1)
2. 0.050 = k [0.1][0.1]2
3. K= 50
4. Now we know k, just substitute to find the values of the rate and [M]
Experiment Initial [L] (mol dm-3) Initial [M] (mol dm-3) Initial rate (mol dm-3 s-1
1 0.100 0.100 0.050
2 0.100 0.300 0.450
3 0.100 0.173 0.150
E.g. Find the rate equation and the value of the rate constant, k and its units
Experiment Initial [A] (mol dm-3) Initial [B] (mol dm-3) Initial rate (mol dm-3 s-1
19 1 1 2
20 1 2 8
21 2 2 16
• In experiment 19 and 20, [A]= no change, [B]= x2, initial rate= x4 so [B]= 22, second
order. In experiment 20 and 21, [A]= x2, [B]= no change, initial rate= x2 so [A]= 21,
first order.
• Rate = k[A][B]2
• K= rate/[A][B]2
• Using values from the table, k= 2/[1][1]2
• K= 2 mol-2 dm6 s-1
Experiment Initial [D] (mol dm-3) Initial [E] (mol dm-3) Initial rate (mol dm-3 s-1
22 1 1 0.2
23 2 1 0.2
24 4 4 0.8
• In experiment 22 and 23, [D]= x2, [E]= no change, initial rate= no change so [E]= 21,
first order. In experiment 23 and 24, [D]= x2, [E]= x4, initial rate= x4 so [D] is still
independent of rate.
• Rate = k[E]
• K= rate/[E]
• Using values from the table, k= 0.8/[4]
, • K= 0.2 s-1
What is a mechanism in a reaction?
• The steps/ stages of a reaction
• The mechanism may involve a single step but more commonly, a series of steps
involved
What is the rate determining step (rds)?
• The rate of the slowest step
• Most reactions don’t just form the products straight away. Some reactants may react
to form products which in turn react with other reactants to form the products. These
mini reactions have different rates of reactions and the overall rate for the whole
reaction is the slowest rate of the different reactions, known as the rate determining
step.
What does the rate equation include?
• Only the species involved in the mechanism up to and including the rate determining
step
• The rate equation gives clues about the mechanism which is the rate determining step
• Homogenous catalysts may be in the rate equation
How do we find the rate equation?
• By experiment and then use the rate equation to deduce possible mechanisms
E.g. Overall rate equation: A + 2B → C + D
Rate= k[A][B]
Possible mechanisms consistent with rate equation:
• Step 1: A + B → P (this is the slow rate and so the rate determining step as both [A]
and [B] are in the reaction)
• Step 2: P + B → C + D (faster step)
Or
• Step 1: 2B → B2 (faster step)
• Step 2: B2 + A → AB2 (slower step. Ratee determining step as both 2B and A are in
mechanism)
• AB2 → C + D (faster step)
E.g. Overall rate equation: A + 2B → J + K
Rate= k[H+][B]
Possible mechanisms consistent with rate equation:
• B + H+ → HB+ (slow rate determining step)
• HB+ + A → Q (faster)
• Q + B → QB (faster)
• QB → J + K (faster)
What must you write when identifying the rate determining step?
• BE SPECIFIC: e.g. for the first example, Step 1 is the rate determining step.
• Because it contains 1 mole of A and 1 mole of B as in the rate equation
How can you tell if something is a catalyst?
• It won’t appear in the equation of the reaction but may appear in the rate equation.
Arrhenius equations
What is the Arrhenius equation and what does it stand for?
, Ea = RT (ln A - ln K)
Ln K = ln A – Ea/RT
Ea should be in KJ/mol for final answer but converted to J/mol for calculations
K and A have the same units (s-1)
Example 1- Finding the activation energy for a reaction
Calculate the activation energy for a reaction in KJ mol-1 if A= 5.25 x 105 mol-1 dm3 s-1 and k=
3.36 x 10-4 mol-1 dm3 s-1 at 298 K when R= 8.314 J mol-1
•
Ea = RT (ln A - ln K)
• Ea = 8.314 x 298 (ln 5.25 x 105 – ln 3.36 x 10-4)
• Ea=52423.9 J mol-1 52.4 KJ mol-1
Example 2- Finding rate constant at 2 different temperatures
a.) Calculate the rate constant at 200C constant and 300C in a reaction where Arrhenius
constant is 2.5 x 106 s-1 and the activation energy is 50 KJ mol-1
•
K = Ae– Ea/RT
•
K(293) = 2.5 x 106e-50 000/8.31 x 293
•
K(293)= 3.02 x 10-3 s-1
•
K(303) = 2.5 x 106e-50 000/8.31 x 303
• K(303)= 5.94 x 10-3 s-1
b.) Show that the reaction is roughly twice as fast at 300C than 200C
• 5.94 x 10-3 /3.02 x 10-3 = 1.97
• Therefore the reaction is twice as faster at 30 degrees than 20 degrees
Example 3- Calculating the rate constant with and without a catalyst
A= 5 x 107 mol-1 dm3 s-1, Ea = 100 KJ mol-1 without a catalyst and Ea = 80 KJ mol-1 with a
catalyst at 298 K
• Without catalyst: K = Ae– Ea/RT
• K = 5 x 107e-1o0 000/8.31 x 298
• K = 1.45 x 10-3 mol-1 dm3 s-1
• With catalyst: K = Ae– Ea/RT
• K = 5 x 107e-80 000/8.31 x 298
• K = 4.67 x 10-7 mol-1 dm3 s-1
• 4.67 x 10-.45 x 10-3 = 3220 x faster with a catalyst
Why do we have to multiply the activation energy by 1000?
• If the question gives the activation energy in KJ, then you multiply by 1000 as
activation energy in the calculation should be in J
How would you find the activation energy from a graph of concentration against 1/T?
• Find the gradient of the line.
• The gradient= -Ea/ R and then solve from then. The gradient should always be –ve
and so Ea should always be positive
Rate equations
What is rate?
• The change in concentration per unit time
• Rate can be the rate at which the reactant is lost or the product is formed
What do particles need in to have order to react?
• Correct orientation
• Have energy more than or equal to the activation energy of the reaction
Why does the concentration, surface area and pressure increase rate of reaction?
• More particles have E ≥ Ea
Why does a small increase in temperature massively increase the rate of reaction?
• Many more particles have E ≥ Ea
What is the function of a catalyst?
• It provides an alternative pathway with a lower activation energy so more collisions
are successful as many more particles have E≥Ea
What is meant by the term ‘order of reaction with respects to A”?
• The power to which the concentration o f A in the rate equation affects the rate of
reaction
What is a rate equation?
• A mathematical relationship between rate and the molar concentrations of reactions
• A and b are the order of the reactions (powers) in respect to [A] and [B]
• The overall order of the equation is the sum of the powers (a + b)
What is k?
• The rate constant
• It is different for different temperatures and differs with temperature exponentially
Why can't the rate equation be written from a balanced (stoichiometric) equation?
• A rate equation may include a homogenous (same state as reactants) catalyst which
isn’t in a balanced equation as it is not used up
How is a rate equation determined?
• By experiments
E.g. Determining rate for orders:
A+B+C→D
Rate = K [B] [C]2
This means that the rate is independent of the concentration of A
Initial rate (mol-3s-1) Change in Effects on rate New initial rates
concentration of (mol-3s-1)
reagents
2.5 [A] x 3 None 2.5
0.75 [B] x 4 X4 3
, 12 [C] x 1o X 100 (as C is 1200
squared)
4.8 [B] x 3, [C] / 2 X 3/4 3.6
0.80 [D] / 2 None 0.80
E.g. Finding the units of the rate constant:
Rate equation Rearange to give k Working Units for k
Rate=K[A] K= rate/[A] K= (mol dm-3 )s-1 / (mol dm-3 ) s-1
Rate=K[C][H] K=rate/[C][H] K= (mol dm-3 )s-1 / (mol dm-3 ) Mol-1 dm3 s-1
(mol dm-3 )
Rate=K[D]2[C] K=rate/ [D]2[C] K= (mol dm-3 )s-1 / (mol dm-3) Mol-2 dm6 s-1
(mol dm ) (mol dm )
-3 -3
Remember that (mol dm-3 ) (mol dm-3 ) = mol2 dm-6
Remember that when you have a fraction, in order to bring the denominator to the
numerator, you multiply by –1, so s-1 / (mol dm-3 ) becomes mol-1 dm3 s-1
How would you find rate equations using initial rates data and find the value of k?
E.g. L reacts with M: 2L + 2M → N + 2P
Rate = k[L][M]2
Find the initial rates of [M] on experiment 3 and the overall initial rate for experiment 2
1. Work out the value of k using a complete row (experiment 1)
2. 0.050 = k [0.1][0.1]2
3. K= 50
4. Now we know k, just substitute to find the values of the rate and [M]
Experiment Initial [L] (mol dm-3) Initial [M] (mol dm-3) Initial rate (mol dm-3 s-1
1 0.100 0.100 0.050
2 0.100 0.300 0.450
3 0.100 0.173 0.150
E.g. Find the rate equation and the value of the rate constant, k and its units
Experiment Initial [A] (mol dm-3) Initial [B] (mol dm-3) Initial rate (mol dm-3 s-1
19 1 1 2
20 1 2 8
21 2 2 16
• In experiment 19 and 20, [A]= no change, [B]= x2, initial rate= x4 so [B]= 22, second
order. In experiment 20 and 21, [A]= x2, [B]= no change, initial rate= x2 so [A]= 21,
first order.
• Rate = k[A][B]2
• K= rate/[A][B]2
• Using values from the table, k= 2/[1][1]2
• K= 2 mol-2 dm6 s-1
Experiment Initial [D] (mol dm-3) Initial [E] (mol dm-3) Initial rate (mol dm-3 s-1
22 1 1 0.2
23 2 1 0.2
24 4 4 0.8
• In experiment 22 and 23, [D]= x2, [E]= no change, initial rate= no change so [E]= 21,
first order. In experiment 23 and 24, [D]= x2, [E]= x4, initial rate= x4 so [D] is still
independent of rate.
• Rate = k[E]
• K= rate/[E]
• Using values from the table, k= 0.8/[4]
, • K= 0.2 s-1
What is a mechanism in a reaction?
• The steps/ stages of a reaction
• The mechanism may involve a single step but more commonly, a series of steps
involved
What is the rate determining step (rds)?
• The rate of the slowest step
• Most reactions don’t just form the products straight away. Some reactants may react
to form products which in turn react with other reactants to form the products. These
mini reactions have different rates of reactions and the overall rate for the whole
reaction is the slowest rate of the different reactions, known as the rate determining
step.
What does the rate equation include?
• Only the species involved in the mechanism up to and including the rate determining
step
• The rate equation gives clues about the mechanism which is the rate determining step
• Homogenous catalysts may be in the rate equation
How do we find the rate equation?
• By experiment and then use the rate equation to deduce possible mechanisms
E.g. Overall rate equation: A + 2B → C + D
Rate= k[A][B]
Possible mechanisms consistent with rate equation:
• Step 1: A + B → P (this is the slow rate and so the rate determining step as both [A]
and [B] are in the reaction)
• Step 2: P + B → C + D (faster step)
Or
• Step 1: 2B → B2 (faster step)
• Step 2: B2 + A → AB2 (slower step. Ratee determining step as both 2B and A are in
mechanism)
• AB2 → C + D (faster step)
E.g. Overall rate equation: A + 2B → J + K
Rate= k[H+][B]
Possible mechanisms consistent with rate equation:
• B + H+ → HB+ (slow rate determining step)
• HB+ + A → Q (faster)
• Q + B → QB (faster)
• QB → J + K (faster)
What must you write when identifying the rate determining step?
• BE SPECIFIC: e.g. for the first example, Step 1 is the rate determining step.
• Because it contains 1 mole of A and 1 mole of B as in the rate equation
How can you tell if something is a catalyst?
• It won’t appear in the equation of the reaction but may appear in the rate equation.
Arrhenius equations
What is the Arrhenius equation and what does it stand for?
, Ea = RT (ln A - ln K)
Ln K = ln A – Ea/RT
Ea should be in KJ/mol for final answer but converted to J/mol for calculations
K and A have the same units (s-1)
Example 1- Finding the activation energy for a reaction
Calculate the activation energy for a reaction in KJ mol-1 if A= 5.25 x 105 mol-1 dm3 s-1 and k=
3.36 x 10-4 mol-1 dm3 s-1 at 298 K when R= 8.314 J mol-1
•
Ea = RT (ln A - ln K)
• Ea = 8.314 x 298 (ln 5.25 x 105 – ln 3.36 x 10-4)
• Ea=52423.9 J mol-1 52.4 KJ mol-1
Example 2- Finding rate constant at 2 different temperatures
a.) Calculate the rate constant at 200C constant and 300C in a reaction where Arrhenius
constant is 2.5 x 106 s-1 and the activation energy is 50 KJ mol-1
•
K = Ae– Ea/RT
•
K(293) = 2.5 x 106e-50 000/8.31 x 293
•
K(293)= 3.02 x 10-3 s-1
•
K(303) = 2.5 x 106e-50 000/8.31 x 303
• K(303)= 5.94 x 10-3 s-1
b.) Show that the reaction is roughly twice as fast at 300C than 200C
• 5.94 x 10-3 /3.02 x 10-3 = 1.97
• Therefore the reaction is twice as faster at 30 degrees than 20 degrees
Example 3- Calculating the rate constant with and without a catalyst
A= 5 x 107 mol-1 dm3 s-1, Ea = 100 KJ mol-1 without a catalyst and Ea = 80 KJ mol-1 with a
catalyst at 298 K
• Without catalyst: K = Ae– Ea/RT
• K = 5 x 107e-1o0 000/8.31 x 298
• K = 1.45 x 10-3 mol-1 dm3 s-1
• With catalyst: K = Ae– Ea/RT
• K = 5 x 107e-80 000/8.31 x 298
• K = 4.67 x 10-7 mol-1 dm3 s-1
• 4.67 x 10-.45 x 10-3 = 3220 x faster with a catalyst
Why do we have to multiply the activation energy by 1000?
• If the question gives the activation energy in KJ, then you multiply by 1000 as
activation energy in the calculation should be in J
How would you find the activation energy from a graph of concentration against 1/T?
• Find the gradient of the line.
• The gradient= -Ea/ R and then solve from then. The gradient should always be –ve
and so Ea should always be positive
