**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?

**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/minTrial 2: [A] = 1.50 M; [B] = 1.50 M; Initial rate = 1.3 x 10

^{-2}M/minTrial 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 2

^{2}).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 L

^{2}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)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 ClO_{2} solution and a solution containing hydroxide ions (OH¯) are mixed, as shown in the following equation:

2ClO_{2}(aq) + 2OH¯(aq) ---> ClO_{3}¯(aq) + ClO_{2}¯(aq) + H_{2}O(l)

When solutions containing ClO_{2} and OH¯ in various concentrations were mixed, the following rate data were obtained:

Determination #1: [ClO_{2}]_{o}= 1.25 x 10¯^{2}M; [OH¯]_{o}= 1.30 x 10¯^{3}M

Initial rate for formation of ClO_{3}¯ = 2.33 x 10¯^{4}M s^{-1}

Determination #2: [ClO_{2}]_{o}= 2.50 x 10¯^{2}M; [OH¯]_{o}= 1.30 x 10¯^{3}M

Initial rate for formation of ClO_{3}¯ = 9.34 x 10¯^{4}M s^{-1}

Determination #3: [ClO_{2}]_{o}= 2.50 x 10¯^{2}M; [OH¯]_{o}= 2.60 x 10¯^{3}M

Initial rate for formation of ClO_{3}¯ = 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 [ClO_{2}]_{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 ClO_{2}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 ClO_{2}.2) Compare #2 and #3. The concentration of hydroxide is doubled while the [ClO

_{2}] remains constant. The rate doubles, showing that the reaction is first order in hydroxide.3) The rate law is: rate = k [ClO

_{2}]^{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 10

^{3}M¯^{2}s^{-1}Often the rate constant unit is rendered thusly: L

^{2}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 10^{3}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 C1 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_{1}/ rate_{3}= k_{1}[A_{1}]^{x}[B_{1}]^{y}/ k_{3}[A_{3}]^{x}[B_{3}]^{y}k's will cancel and [B] will cancel

rate

_{1}/ rate_{3}= [A_{1}]^{x}/ [A_{3}]^{x}0.0063 / 0.0007 = (0.6)

^{x}/ (0.2)^{x}9 = 0.6

^{x}/ 0.2^{x}9 = 3

^{x}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 3^{2}).

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

CH_{3}CHO(g) ---> CH_{4}(g) + CO(g)

Exp. [CH _{3}CHO][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 [CH_{3}CHO] 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 [CH

_{3}CHO] triples and the rate goes up by a factor of 9 (which is 3^{2}). Conclusion: the reaction is second order in CH_{3}CHO.3) The rate expression is this: rate = k [CH

_{3}CHO]^{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/min2 0.24 M 0.10 M 3.0 x 10 ^{-3}M/min3 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 2

^{3}= 8. In more detail:rate_{3}/ rate_{1}= k_{3}[Q_{3}]^{x}[X_{3}]^{y}/ k_{1}[Q_{1}]^{x}[X_{1}]^{y}k's will cancel and [Q] will cancel

rate

_{3}/ rate_{1}= [X_{3}]^{x}/ [X_{1}]^{x}0.012 / 0.0015 = (0.20)

^{x}/ (0.10)^{x}8 = 0.20

^{x}/ 0.10^{x}8 = 2

^{x}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 2

^{3}).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 L

^{3}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 O_{2}.:

Trial [NO] _{o}[O _{2}]_{o}Initial reaction

rate, M/s1 0.020 0.010 0.028 2 0.020 0.020 0.057 3 0.040 0.020 0.227

**Solution:**

1) Doubling O_{2}doubles reaction rate (trials 2 and 1).2) Doubling NO increases reaction rate 8x (trials 2 and 3).

3) Conclusion: first order O

_{2}, 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-line 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.