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Nitrogen dioxide decomposes at 300°C via a second-order process to produce nitrogen monoxide and oxygen according to the following chemical equation.
2 NO2(g) → 2 NO(g) + O2(g).
A sample of NO2(g) is initially placed in a 2.50-L reaction vessel at 300°C. If the half-life and the rate constant at 300°C are 11 seconds and 0.54 M-1 s-1, respectively, how many moles of NO2 were in the original sample?

I just need an idea on how to approach this problem. ...?

Respuesta :

MissPhiladelphia
This problem can be simplified by assuming that the reaction took place in a rigid vessel, that is, the volume is held constant. It is also safe to assume that the reaction is isothermal. The first step here is to derive an expression for the dependency of the reactant concentration with time. Since it is stated that the reaction is second order we start with

-dC/dt = k C^2

where

C is the amount of reactant (NO2)
t is time 
k is the reaction rate constant

The negative sign indicates that the concnetration is decreasing with time. Solving the equation, we get

1/Co - 1/C = kt

where Co is the initial amount of NO2

Now we are given the half life which is the time in which the amount of NO2 is halved, that is, C = 0.5Co. Therefore we can solve the initial amount of NO2 (Co) by substituting t = 11 s, k = 0.54 M-1s-1 and C = 0.5Co

Hope this helps
okpalawalter8

We have that the moles of NO2  in the original sample is mathematically given as

moles of NO2=8.4e-2mol

Moles of NO2 in the original sample

Question Parameters:

Nitrogen dioxide decomposes at 300°C

A sample of NO2(g) is initially placed in a 2.50-L reaction vessel at 300°C.

The rate constant at 300°C are 11 seconds and 0.54 M-1 s-1.

Generally the equation for the Half life  is mathematically given as

[tex]t_[1/2}=1/k[A][/tex]

Where

initial conc of NO2=[tex]1/k*t_[1/2}[/tex]

initial conc of NO2=1/55*0.54

NO2=0.0336

moles of NO2=0.0336*2.50

moles of NO2=8.41*10^{-2}

moles of NO2=8.4e-2

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