# Dinitrogen tetroxide decomposes to form nitrogen dioxide in a second order reaction:

N2O4(g)-> 2NO2(g)
At 400K, the rate constant for this reaction has been measured to be 2.9x10^8 L/mol*s.suppose 0.222 mol of N2O4(g) is placed in a sealed 41.7L container at 400K and allowed to react. What is the total pressure inside the vessel after 32.9ns has elapsed? (R=0.0821(L*atm)/(K*mol).

## Use the second order rate equation of

(1/A) - (1/Ao) = kt and solve for A at the allotted time. That will give you NO2 also. Convert to mols NO2 and mols N2O4 and use PV = nRT to solve for P. Post your work if you get stuck.

## To solve this problem, we'll follow these steps:

Step 1: Calculate the initial concentration of N2O4.
Step 2: Determine the concentration of N2O4 after 32.9 ns.
Step 3: Use the stoichiometry of the reaction to calculate the concentration of NO2.
Step 4: Calculate the total number of moles of gas after the reaction.
Step 5: Use the ideal gas law to calculate the total pressure inside the vessel.

Step 1: Calculate the initial concentration of N2O4.
We are given that 0.222 mol of N2O4 is placed in a 41.7 L container. Therefore, the initial concentration of N2O4 can be calculated as follows:

Initial concentration = moles/volume
Initial concentration = 0.222 mol / 41.7 L
Initial concentration = 0.00533 M

Step 2: Determine the concentration of N2O4 after 32.9 ns.
To find the concentration of N2O4 after a certain time has elapsed, we can use the second-order rate equation:

Rate = k * [N2O4]^2

We need to rearrange this equation to solve for [N2O4]:

Rate / ([N2O4]^2) = k

Plugging in the given values, we have:

2.9x10^8 L/mol*s = k * [N2O4]^2

Now we solve for [N2O4]:

[N2O4]^2 = (2.9x10^8 L/mol*s) / (0.00533 mol/L)
[N2O4]^2 = 5.44x10^10 s^-1

Taking the square root of both sides, we find:

[N2O4] = √(5.44x10^10 s^-1)
[N2O4] = 7.38x10^5 s^-1

Step 3: Use the stoichiometry of the reaction to calculate the concentration of NO2.
From the balanced equation, we know that for every 1 mole of N2O4 that reacts, 2 moles of NO2 are formed. Therefore, the concentration of NO2 after 32.9 ns is equal to twice the concentration of N2O4:

[NO2] = 2 * [N2O4]
[NO2] = 2 * 7.38x10^5 s^-1
[NO2] = 1.48x10^6 s^-1

Step 4: Calculate the total number of moles of gas after the reaction.
Since the volume of the container remains constant, we know that the total number of moles of gas inside the container remains the same throughout the reaction. Therefore, the total number of moles is equal to the sum of the moles of N2O4 and NO2:

Total moles = moles of N2O4 + moles of NO2
Total moles = 0.222 mol + (2 * 0.222 mol)
Total moles = 0.666 mol

Step 5: Use the ideal gas law to calculate the total pressure inside the vessel.

PV = nRT

Rearranging the equation to solve for pressure, we have:

P = (nRT) / V

Plugging in the given values, we get:

P = (0.666 mol * 0.0821 (L*atm)/(K*mol) * 400 K) / 41.7 L

P = 2.45 atm

Therefore, the total pressure inside the vessel after 32.9 ns has elapsed is 2.45 atm.

## To solve this problem, we need to use the rate constant and the given information to determine the amount of N2O4 that has reacted. Then we can use the ideal gas law to calculate the total pressure inside the vessel.

Step 1: Calculate the amount of N2O4 that has reacted.
Given:
Rate constant (k) = 2.9x10^8 L/mol*s
Initial amount of N2O4 (n0) = 0.222 mol
Time (t) = 32.9 ns = 32.9 x 10^(-9) s

The integrated rate equation for a second-order reaction is given by:
1/[N2O4] - 1/[N2O4]0 = kt

Rearranging the equation, we can solve for [N2O4]:
[N2O4] = 1 / (1/[N2O4]0 + kt)

Substituting the given values, we have:
[N2O4] = 1 / (1/0.222 + (2.9x10^8 L/mol*s) * (32.9x10^(-9) s))
[N2O4] = 0.222 mol

So, all of the initial N2O4 has reacted.

Step 2: Calculate the final pressure inside the container.
We can use the ideal gas law to calculate the final pressure inside the container:
PV = nRT

Given:
Volume (V) = 41.7 L
Number of moles (n) = 0.222 mol
Gas constant (R) = 0.0821 L*atm/(K*mol)
Temperature (T) = 400 K

P = nRT / V
P = (0.222 mol) * (0.0821 L*atm/(K*mol)) * (400 K) / (41.7 L)
P ≈ 1.20 atm

Therefore, the total pressure inside the vessel after 32.9 ns has elapsed is approximately 1.20 atm.