Answer to Question #226643 in General Chemistry for Eldon

Question

Dinitrogen tetroxide decomposes to form nitrogen dioxide in a second-order reaction: N2O4(g) → 2NO2(g) At 300.0 K, the rate constant for this reaction has been measured to be 1.3 × 108 L/(mol.s). Suppose 0.454 mol of N2O4(g) is placed in a sealed 11.0 L container at 300.0 K and allowed to react, what is the total pressure inside the vessel after 77.0 ns has elapsed? (R = 0.0821 (L atm)/(K mol)) 

Answer 1

PV= nRT

P=?

V=11.0L

n= 2 × 0.454=0.908moles

R=0.0821

T=300K

$P=\frac{nRT}{V}$

$P=\frac{0.908×0.0821×300}{11}=2.033atm$