Determine rate law by method of initial rates
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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 Formation of 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?
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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? Curio 9 4 5 – brainstorming and project management app.
Trial 1: [A] = 0.50 M; [B] = 1.50 M; Initial rate = 4.2 x 10-3 M/minTrial 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 Formation of 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]21.60 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 Formation of 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) Techsmith camtasia 2018 0 4 download free.
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: Applemacsoft drm converter 6 0 0 download free.
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 103 M¯2 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 103 M¯2 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 of formation 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:
rate
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
Yep dan ppi dangdut clower download. 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 = kGifox 2 0 2 X 40
3[Q3]x[X3]y / k1[Q1]x[X1Gifox 2 0 2 X 4
]yk'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
Tinypng. 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]30.012 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 reaction rate, 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
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
Yep dan ppi dangdut clower download. 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 = kGifox 2 0 2 X 40
3[Q3]x[X3]y / k1[Q1]x[X1Gifox 2 0 2 X 4
]yk'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
Tinypng. 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]30.012 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 reaction rate, 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 = k[1]0.5 = 1
R = k[2]0.5 = 1.4
R = k[4]0.5 = 2
R = k[8]0.5 = 2.8
R = k[16]0.5 = 4
One answer I found on Yahoo Answers was this:
The reaction rate is not doubled when you double the concentration. Notice that when the concentration changed from 2 to 8, the R went from 1.4 to 2.8. So when the concentration is quadrupled, the R doubles.
I would add to the above by saying:
When the concentration doubles, the rate goes up by a factor which is the square root of two.
