Question

The decomposition of N2O5 is first order reaction. N2O5(g) decomposes to yield NO2 (g) and O2(g). At 48 deg C the rate constant for the reaction is 1.2x10^-5 s^-1. Calculate the partial pressure of NO2(g) produced from 1.0L of 0.700M N2O5 solution at 48 degC over a period of 22 hours if the gas is collected in a 10.0L container

Answer 1

Formula for first order reaction: k = (2.303/t)*log(a/(a-x)

k is rate constant, t is time, a is initial concentration, and x is change.

t = 22 h =22*60*60 = 7.92*10^4 /s

a = 1.0 L*0.700 M = 0.700 mol

1.2*10^-5 /s = (2.303/7.92*10^4 /s)*log(0.700/(0.700-x)

log(0.700/(0.700-x)) = 0.52

0.700/(0.700-x) = 3.28

3.28x = 1.6

x = 0.49 mol

Balanced equation: 2N2O5 ------>4NO2 + O2

mol ration of N2O5 to NO2 = 2:4 = 1:2

So, 2x = 0.98 moles of NO2 produced.

Ideal gas equation: PV = nRT

P is pressure, V is volume, n is number of moles, R is constant, and T is temperature

T = 48+273.15 = 321.15 K

P = (nRT)/V

   = (0.98 mol*0.0821 atm L/mol K*321.15)/10.0 L

   = 2.58 atm

So, the partial pressure of NO2 is 2.58 atm