Problem 14.117 Dinitrogen pentoxide (N2O5) decomposes in chloroform as a solvent
ID: 1020428 • Letter: P
Question
Problem 14.117
Dinitrogen pentoxide (N2O5) decomposes in chloroform as a solvent to yield NO2 and O2. The decomposition is first order with a rate constant at 45 C of 1.0×105s1.
Part A
Calculate the partial pressure of O2 produced from 1.38 L of 0.637 M N2O5 solution at 45 C over a period of 23.4 h if the gas is collected in a 11.7-L container. (Assume that the products do not dissolve in chloroform.)
Problem 14.117
Dinitrogen pentoxide (N2O5) decomposes in chloroform as a solvent to yield NO2 and O2. The decomposition is first order with a rate constant at 45 C of 1.0×105s1.
Part A
Calculate the partial pressure of O2 produced from 1.38 L of 0.637 M N2O5 solution at 45 C over a period of 23.4 h if the gas is collected in a 11.7-L container. (Assume that the products do not dissolve in chloroform.)
P= ???? atm
Explanation / Answer
2N2O5(g) ---------- > 4NO2(g) + O2(g)
given rate constant, k = 1.0x10-5 s-1
Time t = 23.4 hr = 23.4 hr x (60 min / 1 hr) x (60 s / 1 min) = 84240 s
Initial concentration of N2O5, [N2O5]o = 0.637 M
Let the concentration of N2O5 after 23.4 hr be [N2O5]t
Applying integrated rate equation for 1st order reaction:
kt = ln([N2O5]o / [N2O5]t)
=> 1.0x10-5 s-1 x 84240 s = ln(0.637 M / [N2O5]t)
=> ln(0.637 M / [N2O5]t) = 0.8424
=> 0.637 M / [N2O5]t = exp (0.8424) =
=> [N2O5]t = 0.637 M / exp (0.8424) = 0.27434 M
i.e the concentration of N2O5 after 23.4 hr is 0.27434 M
=> decrease in the concentration of N2O5 = 0.637 M - 0.27434 M = 0.36266 M
Hence moles of N2O5 reacted = Decrease in concentration xV = 0.36266 mol/L x 1.38 L = 0.5005 mol N2O5
2 mole of N2O5 produce 1 mol of O2(g).
=> 0.5005 mol N2O5 that will produce the moles of O2(g)
= 0.5005 mol N2O5 x (1 mol O2 / 2 mol N2O5)
= 0.25025 mol O2(g)
Hence moles of O2(g) = 0.25025 mol
temperature, T = 45 DegC = 45+273 = 318 K
Volume of container, V = 11.7 L
Applying ideal gas equation,
PV = nRT
=> P = nRT / V
=> P = (0.25025 mol x 0.0821 L.atm.mol-1K-1 x 318 K) / 11.7 L
=> P = 0.5584 atm or 0.558 atm (answer)