The thermal decomposition of dinitrogen pentoxide in the gas phase to give nitrogen dioxide and oxygen is a first order reaction. If the half life at 55 C is 410 s, what is the value of the rate constant k? If the initial concentration of dinitrogen pentoxide in an expt at 55 C was 0.200 M what will be the concentration of dinitrogen pentoxide, nitrogen dioxide and oxygen after 4 half-lives?
donna, gina, jen, or whoever --
To quote one of our very good math and science tutors: “You will find here at Jiskha that long series of questions, posted with no evidence of effort or thought by the person posting, will not be answered. We will gladly respond to your future questions in which your thoughts are included.”
It's also helpful if you post under only one name -- less confusion.
To find the value of the rate constant, we can use the half-life formula for a first-order reaction:
t1/2 = (0.693 / k)
Given that the half-life (t1/2) is 410 s, we can rearrange the formula to solve for the rate constant (k):
k = 0.693 / t1/2
k = 0.693 / 410
k ≈ 0.00169 s^(-1)
Now, let's calculate the concentration of dinitrogen pentoxide, nitrogen dioxide, and oxygen after 4 half-lives.
Initially, the concentration of dinitrogen pentoxide is 0.200 M.
After 1 half-life:
Concentration of dinitrogen pentoxide remaining = 0.200 * 0.5 = 0.100 M
Concentration of nitrogen dioxide formed = 0.200 - 0.100 = 0.100 M
No oxygen is formed yet.
After 2 half-lives:
Concentration of dinitrogen pentoxide remaining = 0.100 * 0.5 = 0.050 M
Concentration of nitrogen dioxide formed = 0.200 - 0.100 - 0.050 = 0.050 M
No oxygen is formed yet.
After 3 half-lives:
Concentration of dinitrogen pentoxide remaining = 0.050 * 0.5 = 0.025 M
Concentration of nitrogen dioxide formed = 0.200 - 0.100 - 0.050 - 0.025 = 0.025 M
No oxygen is formed yet.
After 4 half-lives:
Concentration of dinitrogen pentoxide remaining = 0.025 * 0.5 = 0.0125 M
Concentration of nitrogen dioxide formed = 0.200 - 0.100 - 0.050 - 0.025 - 0.0125 = 0.0125 M
No oxygen is formed yet.
Therefore, after 4 half-lives, the concentration of dinitrogen pentoxide is approximately 0.0125 M, the concentration of nitrogen dioxide is approximately 0.0125 M, and the concentration of oxygen is still zero.
To find the value of the rate constant "k" for the thermal decomposition of dinitrogen pentoxide, we can use the half-life equation for a first-order reaction:
t1/2 = (0.693 / k)
Given that the half-life at 55°C is 410 seconds, we can rearrange the equation to solve for k:
k = 0.693 / t1/2
k = 0.693 / 410
k ≈ 0.0016902 (rounded to 7 significant figures)
Now, let's move on to finding the concentration of dinitrogen pentoxide, nitrogen dioxide, and oxygen after 4 half-lives.
Since it's a first-order reaction, we can use the following equation to calculate the concentration of a reactant after a certain number of half-lives:
C = C0 * (1/2)^(n)
Where:
C is the concentration after n half-lives
C0 is the initial concentration
n is the number of half-lives
Given:
Initial concentration of dinitrogen pentoxide (C0) = 0.200 M
Number of half-lives (n) = 4
Concentration of dinitrogen pentoxide after 4 half-lives:
C = 0.200 M * (1/2)^(4)
C ≈ 0.200 M * (1/16)
C ≈ 0.0125 M
Concentration of nitrogen dioxide after 4 half-lives:
C = 0.200 M * (1 - (1/2)^(4))
C ≈ 0.200 M * (1 - 1/16)
C ≈ 0.200 M * (15/16)
C ≈ 0.1875 M
Concentration of oxygen after 4 half-lives:
C = 0.200 M * (1 - (1/2)^(4))
C ≈ 0.200 M * (1 - 1/16)
C ≈ 0.200 M * (15/16)
C ≈ 0.1875 M
Therefore, after 4 half-lives, the concentration of dinitrogen pentoxide would be approximately 0.0125 M, while the concentrations of nitrogen dioxide and oxygen would both be approximately 0.1875 M.