Question

The decomposition of n2o5 is a first order reaction. N2o5 decomposes to yield no2 and o2.

At 48deg C the rate constant for the reaction is 1.2x10^-5s^-1. Calculate the partial pressure of no2 produced from 1.0L of 0.700M n2o5 solution at 48deg C over a period of 22 hours if the gas is collected in a 10.0L container. Show work please.

Answer 1

Given that; At 48deg C the rate constant for the reaction is 1.2x10^-5s^-1 and order of reaction is 1^st.

2 N₂O₅ → 4 NO₂ + O₂

d[N₂O₅]/dt = - k∙[N₂O₅]

or [N₂O₅] = [N₂O₅]₀ ∙e^(-k∙t)

Here[N₂O₅]₀ = 0.700 M

time t= 22 hours = 79200 s, now calculate the concentration after 79200 s:

[N₂O₅] = 0.700 mol/L ∙ e^(-1.2×10⁻⁵s⁻¹ ∙ 79200 s)

= 0.271mol/L

Now calculate the∆n(N₂O₅) :

= ∆n(N₂O₅) = ∆[N₂O₅] ∙ V
= (0.700mol/L - 0.271 mol/L) ∙ V
= 0.429 mol

According to reaction equation ;

2 N₂O₅ → 4 NO₂ + O₂
4mmole of NO2 is formed per two moles of N₂O₅ decomposed, so the amount of NO2 formed in the 22 hours is:
n(NO₂) = 4 ∙∆n(N₂O₅)

= 4*0.429 mol

= 1.716 mol
Here the gas is collected in a 10.0L container.
Partial pressure in collection container can be found from ideal gas law:
p(NO₂)∙V = n(NO₂)∙R∙T
=>
p(NO₂) = n(NO₂)∙R∙T / V
= 1.716 mol ∙0.0821 L atm /molK ∙ (273.15 + 48)K / 10.0 L

=1.716 mol ∙0.0821 L atm /molK ∙ (321.15)K / 10.0 L

= 4.52 atm