Entry Task: Nov 15th Friday

Turn in Na2S2O3 Lab on my deskDiscuss Rate and Order ws

Pre-Lab Discussion on Crystal Violet lab

MAYHAN

Clear off DeskHave out:

CalculatorsRates and Order ws

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Answers

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What are the four factors that can affect the rate of a chemical reaction?Temperature- temp RateSurface area- SA RateConcentration- conc RateAdd Catalyst- catalyst Rate

Define chemical kinetics.

The study of how fast (speed) of a reaction will take place. The change in molarity per second.

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Consider the reaction A 2CThe average rate of appearance of C is given by ∆[C]/∆t. How is the average rate of appearance of C relate to the average rate of disappearance of A?

-2∆[A]/∆t

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Consider the reaction: ___N2 (g) + ___ H2 (g) ___ NH3 (g) At the instant N2 is reacting at a rate of 0.25 mol/L •min at what rates are H2 disappearing and NH3 forming?

___N2 (g) + ___ H2 (g) ___ NH3 (g)

Rate = −11

[N2]t

= −13

[H2]t

= 12

[NH3]t

11

0.25= −13

[H2]t

= 31

0.250.75 M/min disappearing

of H2

−

3 2

MAYHAN

Consider the reaction: ___N2 (g) + ___ H2 (g) ___ NH3 (g) At the instant N2 is reacting at a rate of 0.25 mol/L •min at what rates are H2 disappearing and NH3 forming?

___N2 (g) + ___ H2 (g) ___ NH3 (g)

Rate = −11

[N2]t

= −13

[H2]t

= 12

[NH3]t

11

0.25= −12

[NH3]t

= 21

0.250.50 M/min appearance of

NH3

−

3 2

MAYHAN

The reaction between ozone and nitrogen dioxide has been studied at 231 K.___NO2 (g) + ___O3 (g) ___N2O5 (s) + ___O2 (g)

Experiment shows the reaction is first order in O3 and second order in NO2.Write the rate law. How does tripling the concentration of NO2 affect the reaction rate?  

How does halving the concentration of O3 affect the reaction rate? 

2

Rate= k [NO2]2[O3]

[3]2 = 9x rate of reaction

[0.5]1 = ½ rate of reactionMAYHAN

Data are given in the table at 660K for the reaction2 NO(g) + O2 (g) 2 NO2 (g)

a) Write the rate equation for the reaction. b) Calculate the rate constant.

Rate= k [NO]2[O2]

1.0 x 10-4 M/s= k [0.020]2[0.010]1.0 x 10-4 M/s = k [0.020]2[0.010] = 25 M-2s-1

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Data are given in the table at 660K for the reaction2 NO(g) + O2 (g) 2 NO2 (g)

c) Calculate the rate of reaction at the instant [NO] = 0.045 M and [O2] = 0.025 M.

Rate= 25 M-2s-1 [0.045]2[0.025]1.3 x 10-3 M/s

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Data are given in the table at 660K for the reaction2 NO(g) + O2 (g) 2 NO2 (g)

d) At the instant when O2 is reacting at the rate 5.0 x 10-4 mol/L•s, at what rate is NO reacting?

Rate = −12

[NO]t

= −11

[O2]t

= 12

[NO2]t

11

5.0 x 10-4= −12

[NO]t

= 21

5.0 x 10-41.0 x 10-5 M/s =

−

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Given the following equations and experimental data, write the correct A2 + B2 2 AB

a. Rate Law Expression

b. Reaction Law

c. Determine k, the Specific Rate Constant (including units)

Rate = −11

[A]t = −

11

[B]t

= 12

[AB]t

Rate= k [A]2[B]

0.01= k [0.001]2[0.001]

0.01 = k [0.001]2[0.001] = 1 x 107 M-2s-1

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Given the following equations and experimental data, write the correct C + D E

a. Rate Law Expression

b. Reaction Law

c. Determine k, the Specific Rate Constant (including units)

Rate = −11

[C]t = −

11

[D]t

= 11

[E]t

Rate= k [D]1

0.02= k [0.01]1

0.02 = k [0.01] = 2 s-1

MAYHAN

For a 1st order reaction, sketch out what the linear graph would look like and label the axis. - What would the integrated equation be for this graph? What would k be labeled as?_____________

ln[A]

time

-k slope

ln[A]t = -kt + ln[A]0

s-1

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For a 2nd order reaction, sketch out what the linear graph would look like and label the axis. - What would the integrated equation be for this graph? What would k be labeled as?_____________

1/[A]

time

+k slope

1/[A]0 = kt + 1/[A]t

M-1s-1

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Data for the decomposition of N2O5 in a particular solvent at 45°C are as follows: You’ll have to graph or EXCELL this data to determine order. Attach graphs to this paper.

Time (min) Conc ln[Conc] 1/[conc]3.07 2.08 0.732368 0.4807698.77 1.67 0.512824 0.598802

14.45 1.36 0.307485 0.73529431.28 0.72 -0.3285 1.388889

0 5 10 15 20 25 30 350

0.5

1

1.5

2

2.5

f(x) = − 0.0465237300084679 x + 2.12709278414687R² = 0.97679073500298

Conc

Not zero

orderMAYHAN

Data for the decomposition of N2O5 in a particular solvent at 45°C are as follows: You’ll have to graph or EXCELL this data to determine order. Attach graphs to this paper.

Time (min) Conc ln[Conc] 1/[conc]3.07 2.08 0.732368 0.4807698.77 1.67 0.512824 0.598802

14.45 1.36 0.307485 0.73529431.28 0.72 -0.3285 1.388889

Might be 1st order0 5 10 15 20 25 30 35

-0.4

-0.2

0

0.2

0.4

0.6

0.8

f(x) = − 0.0375402819431913 x + 0.846341546096834R² = 0.99994685388098

ln[Conc]

MAYHAN

Data for the decomposition of N2O5 in a particular solvent at 45°C are as follows: You’ll have to graph or EXCELL this data to determine order. Attach graphs to this paper.

Time (min) Conc ln[Conc] 1/[conc]3.07 2.08 0.732368 0.4807698.77 1.67 0.512824 0.598802

14.45 1.36 0.307485 0.73529431.28 0.72 -0.3285 1.388889

Not 2nd order so 1st order it is

0 5 10 15 20 25 30 350

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

f(x) = 0.0330040061921574 x + 0.325928499008066R² = 0.982493905230751

1/[conc]

MAYHAN

Data for the decomposition of N2O5 in a particular solvent at 45°C are as follows: You’ll have to graph or EXCELL this data to determine order. Attach graphs to this paper.

What is the order of this reaction?Time (min) Conc ln[Conc] 1/[conc]3.07 2.08 0.737 0.4818.77 1.67 0.513 0.599

14.45 1.36 0.307 0.73531.28 0.72 -0.329 1.39

1st order

Use the integrate rate equation to confirm the rate constant found in the graph. SHOW YOUR WORK!!!

ln[0.72]t = -k(31.28-3.07= 28.21) + ln[2.08]

ln[A]t = -kt = ln[A]0

k=0.038s-1

-0.329 = -k (28.21) + -0.732

-0.329 -0.732 = -k (28.21)

-1.061= -k (28.21)

-1.061= -k 28.21

MAYHAN

Pre-Lab Discussion

MAYHAN

In this experiment, you will observe the reaction between crystal violet and sodium hydroxide. One objective is to study the relationship between concentration of crystal violet and the time elapsed during the reaction. The equation for the reaction is shown here: A simplified (and less intimidating!) version of the equation is:

CV+ + OH–(aq) CVOH (crystal violet) (hydroxide)

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Absorbance (Concentration) will be used in place of concentration in plotting the following three graphs:

*Concentration vs. time: A linear plot indicates a zero order reaction (k = –slope).

*ln Concentration vs. time: A linear plot indicates a first order reaction (k = –slope).

*1/Concentration vs. time: A linear plot indicates a second order reaction (k = slope).

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The rate law for this reaction is in the form: rate = k[CV+]m[OH–]n, where k is the rate constant for the reaction, m is the order with respect to crystal violet (CV+), and n is the order with respect to the hydroxide ion.

As the reaction proceeds, a violet-colored reactant will be slowly changing to a colorless product. Using the green (565 nm) light source of a Colorimeter, you will monitor the absorbance of the crystal violet solution with time. We will assume that absorbance is proportional to the concentration of crystal violet (Beer’s law).

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