# Thread: Rate of Reaction

1. ## Rate of Reaction

Experiments show that if the chemical reaction
N2O5 --> 2NO2 + 1/2 O2
takes place at 45 C, the rate of reaction of dinitrogen pentoxide is proportional to its concentration as follows:
-d[N2O5]/dt=0.0002[N2O5]
How long will the reaction take to reduce the concentration of [N2O5] to 90% of its original value?

2. Originally Posted by asnxbbyx113
Experiments show that if the chemical reaction
N2O5 --> 2NO2 + 1/2 O2
takes place at 45 C, the rate of reaction of dinitrogen pentoxide is proportional to its concentration as follows:
-d[N2O5]/dt=0.0002[N2O5]
How long will the reaction take to reduce the concentration of [N2O5] to 90% of its original value?
ok. whenever i see chemistry, my brain freezes up, so don't take this answer too seriously.

$- \frac {d [N_2 O_5]}{dt} = 0.0002[N_2 O_5]$

$\Rightarrow \frac {d [N_2 O_5]}{dt} = - 0.0002[N_2 O_5]$

$\Rightarrow \frac {1}{[N_2 O_5]} \frac {d[N_2 O_5]}{dt} = -0.0002$

$\Rightarrow \ln [N_2 O_5] = -0.0002t + C$

$\Rightarrow [N_2 O_5] = Ae^{-0.0002t}$

$A$ is the initial concentration of $N_2 O_5$. We want the time when we have 90% of this concentration.

So, we want:

$0.9A = Ae^{-0.0002t}$

$\Rightarrow 0.9 = e^{-0.0002t}$

Now we solve for $t$ and that should be your answer

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# rate of reaction of dinitrogen pentoxide

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