shay asked in Science & MathematicsChemistry · 1 month ago

A first order reaction has a rate law

of K= 8.23 x 10-2 1/sec .

If the reaction starts with a concentration of A [A]=3.24

M and is stopped at [A] = 0.79 M, how much time has

passed?

Relevance
• 1 month ago

First order kinetics is described by the integrated rate law:

[A] = [A]o e^(-kt) ...... where t is the elapsed time and k is the rate constant.

The equation can also be written a:

ln[A] = -kt + ln[A]o

ln[A] - ln[A]o = -kt

t = ln([A] / [A]o) / -k

t = ln(0.79 / 3.24) / -8.23x10^-2 s⁻¹

t = 17.1 s

• Dr W
Lv 7
1 month ago

I cover the derivation of rate equations in my answer here

note this table... (and MEMORIZE IT!!!)

.. .. .order.. .. .. . non-integrated.. .. .. . ..integrated

.. .. . .. 0.. . .. . ...rate = k x [A]°... .. .. . . ..[At] = -kt + [Ao]

.. . .. .. 1.. . .. . ...rate = k x [A]¹... .. .. . . ln[At] = -kt + ln[Ao]

.. . .. . .2... . .. .. .rate = k x [A]²... .. .. . . 1/[At] = +kt + 1/[Ao]

note the difference between non-integrated and integrated

.. non-integrated... relates RATE with CONCENTRATION

... .. . .integrated... relates TIME with CONCENTRATION

*********

This problem of yours states

.. first order

.. concentration

.. time

so we use this equation

.. ln[At] = -kt + ln[Ao]

*************

rearranging that 1st order integrated rate equation

.. kt = ln[Ao] - ln[At]

.. kt = ln([Ao]/[At])...  recall that ln(a) - ln(b) = ln(a/b)

.. t = ln([Ao]/[At]) / k

solving

.. t = ln(3.24M / 0.79M) / (8.23x10^-2/s) = 17.1 sec

• 1 month ago

Since the reaction is first order, it has a half life (t 1/2) = 0.693/k

So, t 1/2 = 0.693/8.23x10^-2/sec

t 1/2 = 8.42 sec

Next, we can use a simple expression for the first order decay:

fraction remaining = 0.5^n where n = the number of half lives elapsed

In this problem we are given (or can find) the fraction remaining....

0.79 M / 3.24 M = 0.2438 = fraction remaining

Next, using this value, we can find how many half lives have elapsed....

fraction remaining = 0.5^n

0.2438 = 0.5^n

log 0.2438 = n log 0.5

n = 2.036 half lives

Finally, we simply convert 2.036 half lives to seconds, minutes...whatever.....

2.036 half lives x 8.42 sec/half life = 17.1 seconds has passed