Question

NOTE: This is a difficult question involving gas laws in addition to reaction rates. 8. N2O4 decomposes to form NO2 in a second-order reaction: N2O4(g) ––> 2NO2(g) At 400.0 K, the rate constant for this reaction is 2.9 x 108 L/(mol · s). Suppose 0.222 mol of N2O4(g) is placed in a sealed 41.7-L container at 400.0 K and allowed to react. What is the total pressure inside the vessel after 32.9 ns has elapsed? A) 0.183 atm B) 0.175 atm C) 0.166 atm D) 0.524 atm E) 0.350 atm

Answer 1

Solution :-

Reaction is second order

Lets first calculate the amount of the N2O4 remain after the time 32.9 ns

32.9 ns * 1 s / 1*10^9 ns = 3.29*10^-8 s

Second order integrated formula

1/[A]t = kt + 1/[A]o

1/[A]t = 2.9*10^8 L mol – 1 s-1 * 3.29*10^-8 s + 1/[0.222]

1/[A]t = 9.541 + 4.505

1/[A]t= 14.046

1/14.046 = [A]t

0.0712 mol = [A]t

So the moles of the N2O4 reacted = 0.222 mol – 0.0712 mol = 0.1508 mol

Moles of NO2 formed = 0.1508 * 2 = 0.3016 mol

Now total moles in the container = moles of NO2 + moles of N2O4

                                                      = 0.3016 mol + 0.0712 mol

                                                      = 0.3728 mol

Now lets calculate the total pressure after the reaction completes after 32.9 ns

PV= nRT

P = nRT /V

   = 0.3728 mol * 0.08206 L atm per mol K * 400 K / 41.7 L

   = 0.2934 atm

We can round it to 0.30 atm

So the total pressure is 0.30 atm

The closest answer is option E) 0.350 atm