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 33 seconds and 0.54 M-1 s-1, respectively, how many moles of NO2 were in the original sample?

Question

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 33 seconds and 0.54 M-1 s-1, respectively, how many moles of NO2 were in the original sample?

anonymous · Answer

I am really interested in HOW you get the answer too. Please help

cuanchi · Answer

for a second order reaction the half-life is
$$t _{\frac{ 1 }{ 2 }}= \frac{ 1 }{ k[A]_{0} }$$
you know t1/2 and the k , then you can calculate the [A]0 the initial concentration. If you know the concentration you can calculate the amount of sample for it relationship with the volume.