**Problem #1:** Calculate half-life for first-order reaction if 68% of a substance is reacted within 66 s.

**Solution:**

1) 68% reacted means 32% remains:

ln A = -kt + ln A_{o}ln 0.32 = - k (66 s) + ln 1

k = 0.0172642 s

^{-1}

Note that this calculation is done with how much substance remains, not how much is used up. It is very common for the question to give you how much is used up and remain silent that you must use how much remains.

2) for the half-life:

ln 0.5 = - (0.0172642 s^{-1}) (t) + ln 1t = 40. s

or:

t_{1/2}= (ln 2) / kt

_{1/2}= (ln 2) / 0.0172642 s^{-1}t

_{1/2}= 40. s

Note that this question does not ask for the rate constant. You must have the rate constant in order to get the half-life, so that calculation must be done, regardless of the question asking for it or not.

**Problem #2:** A certain first order reaction is 45.0% complete in 65 s. Determine the rate constant and the half-life for this process.

**Solution:**

1) Integrated form of first-order rate law:

ln A = -kt + ln A_{o}

2) 45% complete means 55% remains:

ln 0.55 = - k (65 s) + ln 1k = 0.0091975 s

^{-1}(I kept a few guard digits for the next calculation.)

3) for the half-life:

ln 0.5 = - (0.0091975 s^{-1}) (t) + ln 1t = 75.4 s

You can alse use this:

t_{1/2}= (ln 2) / k

to calculate the half-life.

**Problem #3:** A certain reaction is first order, and 540. seconds after initiation of the reaction, 32.5% of the reactant remains. What is the rate constant for this reaction? At what time after initiation of the reaction of the reaction will 10.0% of the reactant remain?

**Solution:**

1) Integrated form of first-order rate law:

ln A = -kt + ln A_{o}ln 0.325 = - (k) (540. s) + ln 1.00

k = 0.002081352 s

^{-1}To three sig figs, k = 0.00208 s

^{-1}. I will use the one with the guard digits in the next calculation.Note that the problem specified the amount remaining, not the amount decomposed.

2) Integrated form of first-order rate law:

ln A = -kt + ln A_{o}ln 0.100 = - (0.002081352 s

^{-1}) (t) + ln 1.00t = 1106 s (to three sig figs, 1110 s)

**Problem #4:** A reaction of the form

aA ----> Product

gives a plot of ln[A] vs time in seconds which is a straight line with a slope of -7.35 x10^{-3}. Assuming 0.0100 M, calculate the time in seconds required for the reaction to reach 80.5 percent completion.

**Solution #1:**

This answer, was written by m w, a top contributor in the Yahoo Answers chemistry section.

**Solution #2:**

80.5% complete means 19.5% remaining. 19.5% of 0.0100 M equals 0.00195 Mln A = -kt + ln A

_{o}ln 0.00195 = - (7.35 x10

^{-3}) (t) + ln 0.01-6.24 = - (7.35 x10

^{-3}) (t) + (-4.6052)1.6348 = (7.35 x10

^{-3}) (t)t = 222.42 s (to three sig figs, 222 s)

This problem could have been done with A = 0.195 and A

_{o}= 1

**Problem #5:** The reactant concentration in a first-order reaction was 7.30 x 10^{-2} M after 45.0 s and 8.70 x 10^{-3} M after 65.0 s . What is the rate constant for this reaction?

**Solution:**

Set the first concentration to be A_{o} and the second to be A. The time will be 20. seconds.

ln A = -kt + ln A_{o}ln 8.70 x 10

^{-3}= - (k) (20. s) + ln 7.30 x 10^{-2}-4.74443 = - (k) (20. s) + (-2.61730)

2.12713 = (k) (20. s)

k = 0.106 s

^{-1}(to three sig figs)

**Problem #6:** A certain first-order reaction is 75% complete in 69.8 min. What is its rate constant in s^{-1}?

**Solution:**

75% complete means 25% of A remains.ln A = -kt + ln A

_{o}ln 0.25 = - (k) (517.5 s

^{-1}) + ln 1k = 2.68 x 10

^{-3}s^{-1}517.5 comes from 69.8 min times 60 sec / min.

**Problem #7:** The decomposition of aqueous hydrogen peroxide to gaseous oxygen and water is a first-order reaction. If it takes 6.5 hours for the concentration of H_{2}O_{2} to decrease from 0.70 to 0.35, how many hours are required for the concentration to decrease from 0.40 to 0.10 ?

**Solution (the general way):**

1) Find the rate constant:

ln A = -kt + ln A_{o}ln 0.35 = - (k) (6.5 hr) + ln 0.70

-1.0498 = - (k) (6.5 hr) - 0.356675

-0.693125 = - k (6.5 hr)

k = 0.1066346 hr

^{-1}(I kept some guard digits and will round off the time at the end.)

2) Solve for time:

ln A = -kt + ln A_{o}ln 0.10 = - (0.1066346 hr

^{-1}) (t) + ln 0.40-2.3026 = - (0.1066346 hr

^{-1}) (t) - 0.9163-1.3863 = - (0.1066346 hr

^{-1}) (t)t = 13 hr

**Solution (specific to this problem):**

0.70 to 0.35 is one half-life and 0.40 to 0.10 is two half-lives. Since one half-life equals 6.5 hrs, two half-lives would take 13.0 hrs.

**Problem #8:** The decomposition of hydrogen peroxide is a first-order reaction. The half-life of the reaction is 17.0 minutes. (Which is different from the problem above. These are made-up values.)

(a) What is the rate constant of the reaction?

(b) If you had a bottle of H_{2}O_{2}, how long would it take for 86% to decompose?

(c) If you started the reaction with [H_{2}O_{2}] = 0.1 M, what would be the hydrogen peroxide concentration after 15.0 minutes?

**Solution:**

Part (a)

k = (ln 2) / t_{1/2}k = (ln 2) / 17.0 min = 0.04077 min

^{-1}(0.0408 to three sig figs)

Part (b)

86% decomposed means 14% remains.ln A = -kt + ln A

_{o}ln 0.14 = - (0.04077 min

^{-1}) (t) + ln 1t = 48.2 min

Part (c)

0.1 M is A_{o}ln A = - (0.04077 min

^{-1}) (15.0 min) + ln 0.1ln A = -2.914135

A = 0.0542 M

Comment: one half-life is 17.0 min, so the [H_{2}O_{2}] would be 0.05 M at the end of 17.0 min. Note that part c involves a time frame slightly less than one half-life, so the ending concentration is slightly more than 0.05 M.

**Problem #9:** In a first-order decomposition reaction, 50.0% of a compound decomposes in 17.5 min. What is the rate constant of the reaction

**Solution:**

The 50.0% decomposed tells us that the 17.5 min is the half-life.k = (ln 2) / 17.5 min = 0.0396 min

^{-1}This:

ln A = -kt + ln A_{o}can also be used. Its use is left to the reader.

**Problem #10:** The half-life for a first-order reaction is 32.0 s. What was the original concentration if, after 2.00 minutes, the reactant concentration is 0.062 M?

**Solution:**

1) Find k:

k = (ln 2) / 32.0 s = 0.021661 s^{-1}

2) Find A_{o}:

ln A = -kt + ln A_{o}ln 0.062 = - (0.021661 s

^{-1}) (120. s) + ln A_{o}-2.78062 = -2.59932 + ln A

_{o}ln A

_{o}= -0.1813A

_{o}= 0.834 M