For the reaction A $\to$ Products,successive half-lives are observed to be 10.0 min and 40.0 min. -The reaction is first order in N₂O₅.For this reaction at 45^oC,the rate constant k = 1.0 $\times$ 10^-⁵ s^-¹,where the rate law is defined as For a particular experiment ([N₂O₅]₀ = 1.0 $\times$ 10^-³M),calculate [N₂O₅] after 3.5 $\times$ 10⁵ seconds.

Question

Quizplus · Accepted Answer

We can use the first-order integrated rate law to solve for [NS1U1B12S1U1B0OS1U1B15S1U1B0] at a given time:

ln([NS1U1B12S1U1B0OS1U1B15S1U1B0]S1U1B10/[NS1U1B12S1U1B0OS1U1B15S1U1B0]t) = -kt

where [NS1U1B12S1U1B0OS1U1B15S1U1B0]S1U1B10 is the initial concentration and [NS1U1B12S1U1B0OS1U1B15S1U1B0]t is the concentration at a given time t.

Plugging in the values given in the problem:

ln([NS1U1B12S1U1B0OS1U1B15S1U1B0]S1U1B10/[NS1U1B12S1U1B0OS1U1B15S1U1B0]3.5S1S1P0M) = -(1.0 F1F1F1S1F1F1F10 10S1U1P1-S1S1P0S1U1P15S1S1P0)(3.5 F1F1F1S1F1F1F10 10S1U1P15S1S1P0)

ln([NS1U1B12S1U1B0OS1U1B15S1U1B0]S1U1B10/[NS1U1B12S1U1B0OS1U1B15S1U1B0]3.5S1S1P0M) = -3.5

[NS1U1B12S1U1B0OS1U1B15S1U1B0]3.5S1S1P0M = [NS1U1B12S1U1B0OS1U1B15S1U1B0]S1U1B10 e^(-3.5)

[NS1U1B12S1U1B0OS1U1B15S1U1B0]3.5S1S1P0M = (1.0 F1F1F1S1F1F1F10 10S1U1P13 S1S1P0M) e^(-3.5)

[NS1U1B12S1U1B0OS1U1B15S1U1B0]3.5S1S1P0M = 3.0 F1F1F1S1F1F1F10 10S1U1P15S1S1P0M (rounded to one significant figure)

Therefore, the answer is (C) 3.0 F1F1F1S1F1F1F10 10S1U1P1-S1S1P0S1U1P15S1S1P0 M.

For the Reaction A →\to→ Products,successive Half-Lives Are Observed to Be 10

For the Reaction A $\to$ Products,successive Half-Lives Are Observed to Be 10