The half-life of the following first order reaction is 231 seconds. What percent of the original concentration of nitrite ion remains after 10.0 minutes?
NH4+ + NO2- --> N2 + 2H2O
how many half lives?
10 * 60 = 600
600/231 = 2.6
(1/2)^2.6 = .165
.165*100 = 16.5 %
To determine the percent of the original concentration of nitrite ion remaining after 10.0 minutes, we need to use the concept of half-life.
The half-life of a first-order reaction can be calculated using the following formula:
t1/2 = (0.693) / k
Where:
t1/2 = half-life of the reaction
k = rate constant of the reaction
Given that the half-life (t1/2) is 231 seconds, we can rearrange the formula to solve for k:
k = (0.693) / t1/2
Plugging in the values, we can calculate the rate constant (k):
k = 0.693 / 231
k ≈ 0.003
Now, we can use the integrated rate law equation to determine the remaining concentration of nitrite ion after 10.0 minutes (600 seconds):
ln([A]t/[A]0) = -kt
Where:
[A]t = concentration at time t
[A]0 = initial concentration
k = rate constant
t = time
Rearranging the equation to solve for [A]t, we get:
[A]t = [A]0 * e^(-kt)
Plugging in the values:
[A]t = [A]0 * e^(-0.003 * 600)
To calculate the percent remaining, we divide [A]t by [A]0 and multiply by 100:
Percent Remaining = ([A]t / [A]0) * 100
Now you can follow these steps and plug in the appropriate values to find the answer.