### Determine rate law by method of initial rates

Problem #1: Rate data were obtained for following reaction:

A + 2B ---> C + 2D

 Exp. Initial A(mol/L) Initial B(mol/L) Init. Rate of Formationof C (M min-1) 1 0.10 0.10 3.0 x 10-4 2 0.30 0.30 9.0 x 10-4 3 0.10 0.30 3.0 x 10-4 4 0.20 0.40 6.0 x 10-4

What is the rate law expression for this reaction?

Solution:

1) compare exp. 1 and exp. 3. A remains constant and B is tripled. The rate from 1 to 3 remains constant. Conclusion: B is not in the rate law expression.

2) compare exp 1 to exp 2. The concentration of A triples (and we don't care what happens to B). The rate triples. Conclusion: first order in A.

3) we can also show first order in A by comparing exp 1 to exp 4. The concentration of A doubles and the rate doubles. Remember, B is not part of the rate law, so we don't pay any attention to it at all.

rate = k[A]

Problem #2: For the reaction A + B --> products, the following initial rates were found. What is the rate law for this reaction?

Trial 1: [A] = 0.50 M; [B] = 1.50 M; Initial rate = 4.2 x 10-3 M/min

Trial 2: [A] = 1.50 M; [B] = 1.50 M; Initial rate = 1.3 x 10-2 M/min

Trial 3: [A] = 3.00 M; [B] = 3.00 M; Initial rate = 5.2 x 10-2 M/min

Solution:

1) Order with respect to A:

Look at trial 1 and trial 2. B is held constant while A triples. The result is that the rate triples. Conclusion: A is first order.

2) Order with respect to B:

Look at trials 2 and 3. The key to this is that we already know that the order for A is first order.

Both concentrations were doubled from 2 to 3 and the rate goes up by a factor of 4. Since A is first order, we know that a doubling of the rate is due to the concentration of A being doubled.

So, we look at the concentration change for B (a doubling) and the consequent rate change (another doubling - remember the overall increase was a factor of 4 - think of 4 as being a doubled doubling).

Conclusion: the order for B is first order.

The rate law is rate = k [A] [B]

Comment: the above manner of making the rate order determination for B a bit more complex is a common technique. Look for it to be used on your test!

Problem #3: The following data were obtained for this chemical reaction: A + B ---> products

 Exp. Initial A(mmol/L) Initial B(mmol/L) Init. Rate of Formationof products (mM min-1) 1 4.0 6.0 1.60 2 2.0 6.0 0.80 3 4.0 3.0 0.40

(a) Determine the rate law for this reaction.
(b) Find the rate constant.

Solution:

1) Look at experiments 2 and 1. From 2 to 1, we see that A is doubled (while B is held constant). A doubling of the rate with a doubling of the concentration shows that the reaction is first order with respect to A.

2) Now compare experiments 1 and 3. The concentration of A is held constant while the concentration of B is cut in half. When B is cut in half, the overall rate is cut by a factor of 4 (which is the square of 2). This shows the reaction is second order in B.

3) The rate law is this: rate = k [A] [B]2

4) Note that the comparison in (2) can be reversed. Consider that the concentration of B is doubled as you go from exp. 3 to exp. 1. When the concentration is doubled, the rate goes up by a factor of 4 (which is 22).

5) We can use any set of values to determine the rate constant:

rate = k [A] [B]2

1.60 mM min-1 = k (4.0 mM) (6.0 mM)2

k = 0.011 mM-2 min-2

the units on k can be rendered in this manner:

k = 0.067 L2 mmol-2 min-1

Problem #4: The following data were obtained for the chemical reaction: A + B ---> products

 Exp. Initial A(mol/L) Initial B(mol/L) Init. Rate of Formationof products (M s-1) 1 0.040 0.040 9.6 x 10-6 2 0.080 0.040 1.92 x 10-5 3 0.080 0.020 9.6 x 10-6

(a) Determine the rate law for this reaction.
(b) Find the rate constant.
(c) What is the initial rate of reaction when [A]o = 0.12 M and [B]o = 0.015

Solution:

1) Examine exps. 2 and 3. A remains constant while B is doubled in concentration from 3 to 2. The result of this change is that the rate of the reaction doubles. We conclude that the reaction is first order in B.

2) Now we look at exps 1 and 2. B remains constant while the concentration of A doubles. As a result of the doubled concentration, the rate also doubles. Conclusion: the reaction is first order in A.

3) The rate law for this reaction is:

rate = k [A] [B]

4) We can use any set of data to calculate the rate constant:

rate = k [A] [B]

9.6 x 10-6 M s-1 = k (0.040 M) (0.040 M)

k = 0.0060 M-1 s-1

Comment: remember that M-1 s-1 is often written as L mol-1 s-1.

5) We use the rate constant along with the data in part (c) of the question:

rate = (0.0060 M-1 s-1) (0.12 M) (0.015 M)

rate = 1.08 x 10-5 M s-1

Problem #5: Consider the reaction that occurs when a ClO2 solution and a solution containing hydroxide ions (OH¯) are mixed, as shown in the following equation:

2ClO2(aq) + 2OH¯(aq) ---> ClO3¯(aq) + ClO2¯(aq) + H2O(l)

When solutions containing ClO2 and OH¯ in various concentrations were mixed, the following rate data were obtained:

Determination #1: [ClO2]o = 1.25 x 10¯2 M; [OH¯]o = 1.30 x 10¯3 M
Initial rate for formation of ClO3¯ = 2.33 x 10¯4 M s-1
Determination #2: [ClO2]o = 2.50 x 10¯2 M; [OH¯]o = 1.30 x 10¯3 M
Initial rate for formation of ClO3¯ = 9.34 x 10¯4 M s-1
Determination #3: [ClO2]o = 2.50 x 10¯2 M; [OH¯]o = 2.60 x 10¯3 M
Initial rate for formation of ClO3¯ = 1.87 x 10¯3 M s-1

(a) Write the rate equation for the chemical reaction.
(b) Calculate the rate constant, k.
(c) Calculate the reaction rate for the reaction when [ClO2]o = 8.25 x 10¯3 M and [OH¯]o = 5.35 x 10¯2 M.

Solution:

1) Compare #1 and #2. The concentration of ClO2 doubles (hydroxide remains constant) and the rate goes up by a factor of four (think of it as two squared). This means the reaction is second order with respect to ClO2.

2) Compare #2 and #3. The concentration of hydroxide is doubled while the [ClO2] remains constant. The rate doubles, showing that the reaction is first order in hydroxide.

3) The rate law is: rate = k [ClO2]2 [OH¯]

4) Calculation for the rate constant:

1.87 x 10¯3 M s-1 = k (2.50 x 10¯2 M)2 (2.60 x 10¯3 M)

k = 1.15 x 1032 s-1

Often the rate constant unit is rendered thusly: L2 mol-2 s-1.

Note that the overall order of the rate law is third order and that this is reflected in the unit associated with the rate constant.

5) Calculation for part (c):

rate = (1.15 x 1032 s-1 ) (8.25 x 10¯3 M)2 (5.35 x 10¯2 M)

rate = 4.19 x 10¯3 M s-1

Problem #6: 2A + B ---> C + D

The following data about the reaction above were obtained from three experiments:

 Exp. [A] [B] Initial rate offormation of C 1 0.6 0.15 6.3 x 10-3 2 0.2 0.6 2.8 x 10-3 3 0.2 0.15 7.0 x 10-4

Calculate the rate expression in terms of [A] for experiment 1.

Solution:

rate1 / rate3 = k1[A1]x[B1]y / k3[A3]x[B3]y

k's will cancel and [B] will cancel

rate1 / rate3 = [A1]x / [A3]x

0.0063 / 0.0007 = (0.6)x / (0.2)x

9 = 0.6x / 0.2x

9 = 3x

rate = k[A]2

Comment: note that, with a second order, when the concentration increases by a factor of 3, the rate goes up by a factor of 9 (which is 32).

Problem #7: Determine the proper form of the rate law for:

CH3CHO(g) ---> CH4(g) + CO(g)

 Exp. [CH3CHO] [CO] Rate (M s-1) 1 0.30 0.20 0.60 2 0.10 0.30 0.067 3 0.10 0.20 0.067

Solution:

1) Note experiments 2 and 3. The [CH3CHO] remains constant while the [CO] changes. However, no change of the reaction rate is observed. From this, we conclude that CO is not part of the rate law.

2) Examine experiment 3 compared to experiment 1. from 3 to 1, the [CH3CHO] triples and the rate goes up by a factor of 9 (which is 32). Conclusion: the reaction is second order in CH3CHO.

3) The rate expression is this: rate = k [CH3CHO]2

Problem #8: Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided.

Q + X --> products

 Trial [Q] [X] Rate 1 0.12 M 0.10 M 1.5 x 10-3 M/min 2 0.24 M 0.10 M 3.0 x 10-3 M/min 3 0.12 M 0.20 M 1.2 x 10-2 M/min

Solution:

1) Examine trials 1 and 2. [X] is held constant while [Q] is doubled (from 1 to 2). As a result, the rate doubles. We conclude that the reaction is first order in Q.

2) Examine trials 3 and 1. The concentration of Q does not change from 3 to 1 but the concentration for X is doubled. When this happens, we observe an eight-fold increase in the rate of the reaction. We conclude that the reaction is third order in X. This arises from the fact that 23 = 8. In more detail:

rate3 / rate1 = k3[Q3]x[X3]y / k1[Q1]x[X1]y

k's will cancel and [Q] will cancel

rate3 / rate1 = [X3]x / [X1]x

0.012 / 0.0015 = (0.20)x / (0.10)x

8 = 0.20x / 0.10x

8 = 2x

x = 3

Comment: note that, with a third order, when the concentration increases by a factor of 2, the rate goes up by a factor of 8 (which is 23).

3) value of the rate constant:

rate = k [Q] [X]3

0.012 M min-1 = k (0.12 M) (0.20 M)3

k = 12.5 L3 mol-3 min-1

Note that this reaction is overall fourth order.

Problem #9: With the following data, use the method of initial rates to find the reaction orders with respect to NO and O2.:

 Trial [NO]o [O2]o Initial reactionrate, M/s 1 0.020 0.010 0.028 2 0.020 0.020 0.057 3 0.040 0.020 0.227

Solution:

1) Doubling O2 doubles reaction rate (trials 2 and 1).

2) Doubling NO increases reaction rate 8x (trials 2 and 3).

3) Conclusion: first order O2, third order NO.

Problem #10: For the following reaction:

A + B ---> 2C

it is found that doubling the amount of A causes the reaction rate to double while doubling the amount of B causes the reaction rate to quadruple. What is the best rate law equation for this reaction?

(a) rate = k [A]2 [B]
(b) rate = k [A] [B]
(c) rate = k [A] [B]2
(d) rate = k [A]1/2 [B]

The rate doubling when the concentration is doubled is a hallmark of first-order. The rate going up by a factor of 4 (with is two squared) when the concentration is doubled is a hallmark of second-order. Answer choice (c) is the correct answer.

Bonus Problem: A rate law is 1/2 order with respect to a reactant. What is the effect on the rate when the concentration of this reactant is doubled?

Solution:

Let us examine the effect on the rate (symbolized as R) as we double the concentration from step to step:

R = k0.5 = 1
R = k0.5 = 1.4
R = k0.5 = 2
R = k0.5 = 2.8
R = k0.5 = 4