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Kinetics/Rate of Reactions

100

Based on the data below, what are the reaction orders of CH₃Cl and H₂O?

CH₃Cl (g) + H₂O (g) --> CH₃OH (g) + HCl (g)

Exp  [CH₃Cl]₀/M   [H₂O]₀/M  Initial Rate in M/s

1        0.100         0.100            0.182

2        0.200         0.200             1.45

3        0.200         0.400             5.81

CH₃Cl: first order

H₂O: second order

200

DAILY DOUBLE!

Using the following data, which is the correct rate law of the sample reaction?

A + 5B + 6C → 3D + 3E

Exp. [A]    [B]   [C]   Initial Rate (M/s)

1  0.35  0.35  0.35     8.0 x 10^-4

2  0.70  0.35  0.35     3.2 x 10^-3

3  0.70  0.70  0.35     6.4 x 10^-3

4  0.70  0.35  0.70     3.2 x 10^-3

R= k[A]²[B]¹[C]⁰

300

The rate law for the reaction 

A + B --> C + D

is first order in [A] and second order in [B]. If [A] is halved and [B] is doubled, how will the rate of the reaction be affected?

The overall rate will be doubled.

400

What is the rate constant of the reaction, given the following set of experiments? Be sure to give the correct units.

A + B → C + D

Exp     [A]      [B]    Initial Rate (M/s)

1      0.040  0.040     2.2 x 10^-4

2      0.040  0.080     4.4 x 10^-4

3      0.080  0.080     8.8 x 10^-4

rate= k[A]¹[B]¹

k= rate / ([A]¹[B]¹)

k= (2.2 x 10^-4 M s^-1) / (0.040 M x 0.040 M)

k= 0.14 M^-1s^-1

500

The reaction is second order with a rate constant k equal to 5.76×10⁻² L/mol/min under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration remaining after 10.0 min?

2 C₄H₆ → C₈H₁₂

Use the formula 1/[A] = kt + 1/[A]₀

1/[A]_t = (5.76 x 10^-2 L mol^-1 min^-1) (10 min) + 1/(0.200 L mol^-1)

1/[A]_t = 5.576 L mol_-1

[A]_t = 1.79 x 10^-1 M (0.179 M)

Extra easy problems...

200

What is the vapor pressure of an aqueous solution that has a mole fraction of 0.9000? 

The vapor pressure of water is 25.756 mm Hg at 25.00 ºC.

Use  P_a = X_a•P_aº

P_a = 0.9000(25.756 mmHg)

P_a^º= 23.18 mmHg

400

How many grams of pyrazine (C₄H₄N₂) (80.09 g/mole) would have to be dissolved in 1.50 kg of carbon tetrachloride to lower the freezing point by 4.40 °C? The freezing point constant for carbon tetrachloride is 30.0 °C/m.

*HINT: Use the formula: ∆T = i x K_f x m

∆T = i x Kf x m 

i = 1;  ∆T = 4.4ºC; Kf = 30 ºC/m;  kg solvent = 1.5kg

1) Determine the molality:

4.4 ºC = 1 x 30 ºC/m x (m)

m = 0.14667 moles/kg

2) Determine the number of moles: 

0.14667 = #moles ÷ 1.5 kg solvent

moles = 0.22 moles

3) 0.22 moles x 80.09 g/mol = 17.6 g