N2O5 gives 2NO2 plus 0.5 O2.

Rate for first order reaction.
K = 5 * 10/10000
R = 0.30 mol/liter
after 40 min. What is the [C] of N2O5 and NO2 ?

To determine the concentration of N2O5 and NO2 after 40 minutes, we need to use the integrated rate law for a first-order reaction:

ln([A]t/[A]0) = -kt

Where:
[A]t is the concentration of the reactant or product at time t
[A]0 is the initial concentration of the reactant or product
k is the rate constant of the reaction
t is the time

Let's calculate the concentration of N2O5 first. We are given that the rate constant (k) is 5 * 10/10000, and the initial concentration ([N2O5]0) is unknown. The concentration of N2O5 at 40 minutes ([N2O5]t) is also unknown.

However, we are given that the rate of the reaction (R) is 0.30 mol/liter. The rate formula for this reaction is:

R = -d[N2O5]/dt = k[N2O5]

We can rearrange this equation to solve for [N2O5]0:

[N2O5]0 = R/k

Substituting the given values:

[N2O5]0 = 0.30 mol/liter / (5 * 10/10000)

[N2O5]0 = 60 mol/liter

Now we can use the integrated rate law to calculate [N2O5]t:

ln([N2O5]t / [N2O5]0) = -kt

ln([N2O5]t / 60) = -(5 * 10/10000)t

Since the reaction is first order, the natural logarithm of the concentration ratio is equal to the negative rate constant multiplied by the time. Let's plug in the values.

ln([N2O5]t / 60) = -(5 * 10/10000) * 40

ln([N2O5]t / 60) = -0.02

Now, solving for [N2O5]t:

[N2O5]t / 60 = e^(-0.02)

[N2O5]t = 60 * e^(-0.02)

Using a scientific calculator, we find [N2O5]t ≈ 59.27 mol/liter.

To determine the concentration of NO2, we need to consider the stoichiometry of the reaction. We're given that N2O5 reacts to produce 2NO2 and 0.5O2. This means that the moles of NO2 formed would be twice the moles of N2O5 consumed.

Since the initial concentration of N2O5 ([N2O5]0) is 60 mol/liter and the concentration at 40 min is [N2O5]t ≈ 59.27 mol/liter, the change in concentration is:

Δ[N2O5] = [N2O5]t - [N2O5]0

Δ[N2O5] = 59.27 mol/liter - 60 mol/liter

Δ[N2O5] ≈ -0.73 mol/liter

Since 2 mol of NO2 is produced for every 1 mol of N2O5 consumed, the change in NO2 concentration ([NO2]t) would be:

Δ[NO2] = 2 * Δ[N2O5]

Δ[NO2] ≈ 2 * (-0.73 mol/liter)

Δ[NO2] ≈ -1.46 mol/liter

To find the concentration of NO2 at 40 minutes, we need to subtract the change in NO2 concentration from the initial concentration of NO2 ([NO2]0). However, we don't have the initial concentration of NO2. Without that information, we cannot determine the exact concentration of NO2 at 40 minutes.