14.2.3 A Kinetics Problem

Question 1

A Kinetics Problem

Answer 1

• The equations ln[A ]t = –kt + ln[A] 0 and t 1/2 = 0.693/k can be used to solve problems involving first-order reactions

Question 2

note

Answer 2

• The rate law for the hydrolysis of sucrose (C 12 H 22 O 11 ) to form glucose (C 6 H 12 O 6 ) and fructose (C 6 H 12 O 6 ) is rate = k[C 12 H 22 O 11 ].
• Problem: After 2.57 hours at 27°C, 5.00 g/L of sucrose has decreased to 4.50 g/L. What is the rate constant (k)?
• To answer this problem, solve the equation
  ln[A] t = –kt + ln[A] 0 for k and substitute in the values given.
• Problem: What is the half-life (t 1/2 ) of the sucrose at 27°C?
• To answer this problem, use the equation t 1/2 = 0.693/k and the value for k found above.
• This value means that, after 16.9 hours, half of the starting sucrose would be left, or 2.50 g/L. After another 16.9 hours, 1.25 g/L of sucrose would remain.
• Problem: How much sucrose remains after 24.0 hours?
• To answer this problem, solve the equation
  ln[A] t = –kt + ln[A] 0 for [A] t and substitute in the values given and the value for k found above.
• 24.0 hours is between 16.9 hours (one half-life) and 33.8 hours (two half-lives). 1.87 g/L makes sense as an answer, since it is between the concentration after 16.9 hours (2.50 g/L) and the concentration after 33.8 hours (1.25 g/L).

Question 3

How long would it take for the concentration of sucrose in a solution to reach 1.50 g / L if the initial concentration was 4.75 g / L?

C12H22O11(aq) + H2O(l) → 2C6H12O6(aq)

k = 4.10 × 10−2 hr−1 at 300 K

Answer 3

28.1 hours

Question 4

Compund AB decomposes to form A and B in a reaction that is first-order with respect to AB and first order overall. At 28 degrees C, the specific rate constant for the reaction is 3.26*10^-4 s^-1. What is the half-life of AB at 28 degrees C?
AB -> A + B

Answer 4

35.4 minutes

Question 5

Calculate the concentration of a sucrose solution that had an original concentration of 5.00 g / L after 6.00 hours and 20.0 minutes. The rate constant for this reaction at 300 K is 4.10 × 10−2 hr−1.

Answer 5

3.86 g / L

Question 6

Calculate the concentration of a maltose solution that had an original concentration of 6.40 g / L at 25° C after 4 hours and 45 minutes. The rate constant for the decomposition of maltose at 25° C is 4.10 × 10−2 hr−1.

Answer 6

5.27 g / L

Question 7

Compound MN decomposes to form M and N in a reaction that is first-order with respect to MN and first order overall. The half-life for the reaction is 2.3 × 104 s. What is the rate constant of this reaction at 25°C?

MN → M + N

Answer 7

3.0 × 10^−5 s−1

Question 8

For the decomposition of nitrogen dioxide:

2NO2(g) → 2NO(g) + O2(g)

the rate law can be written as: Rate = k[ NO2 ]n

Yet the reverse reaction can also occur:

2NO(g) + O2(g) → 2NO2(g)

Under what type of experimental conditions would the rate = k[ NO2 ]n be considered valid?

Answer 8

When the reaction is studied at a point soon after it begins

Question 9

The hydrolysis of sucrose to form fructose and glucose is considered a first order reaction. Why is this the case, when the term “hydrolysis” clearly indicates that H2O is involved?

Answer 9

[H2O] does not change significantly during the course of the reaction.

Question 10

Compound XY decomposes to form X and Y in a reaction that is first-order with respect to XY and first order overall. The half-life for the reaction is 15.4 s. What is the specific rate constant of this reaction at 25 degrees C
XY -> X + Y

Answer 10

4.50 × 10^−2 s−1

Question 11

How long would it take for the concentration of maltose in a solution to reach 0.250 g / L if the initial concentration was 6.75 g / L? For this reaction at 25°C, k = 4.12 × 10^−2 hr−1.

Answer 11

80 hours

Question 12

A certain first-order reaction has a t1/2 = 78 minutes. How much time is required for the reaction to be 90% complete?

Answer 12

4 hr 20 min