The decomposition of N2O5(g) --> NO2(g) + NO3(g) proceeds as a first order reaction with a half-life of 30.0 seconds at a certain temperature. If the initial concentration [N2O5]0 = 0.400 M, what is the concentration after 120 seconds?
See your other post. Same procedure.
To solve this problem, we can use the equation for a first order reaction:
ln([A]t/[A]0) = -kt
Where [A]t is the concentration of reactant A at time t, [A]0 is the initial concentration of reactant A, k is the rate constant, and t is the time.
Given that the half-life of the reaction is 30.0 seconds, we can determine the rate constant (k) using the following equation:
t 1/2 = ln(2)/k
Substituting the given value for the half-life:
30.0 seconds = ln(2)/k
Solving for k:
k = ln(2)/30.0 seconds
Now that we have the rate constant, we can determine the concentration of N2O5 at 120 seconds.
Substituting the given values into the equation for a first order reaction and solving for [N2O5]120:
ln([N2O5]120/0.400 M) = -k(120 seconds)
Simplifying, we have:
ln([N2O5]120/0.400 M) = -ln(2)/30.0 seconds * 120 seconds
Now, let's solve for [N2O5]120 by isolating the term [N2O5]120:
[N2O5]120/0.400 M = e^(-ln(2) * 120 seconds / 30.0 seconds)
[N2O5]120/0.400 M = e^(-ln(2) * 4)
[N2O5]120/0.400 M = e^(-ln(16))
[N2O5]120/0.400 M = 1/16
[N2O5]120 = (1/16) * 0.400 M
[N2O5]120 = 0.025 M
Therefore, the concentration of N2O5 after 120 seconds is 0.025 M.