Nitrogen dioxide decomposes at 300°C via a second-order process to produce nitrogen monoxide and oxygen according to the followi
ng 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. ...?
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
Since the concentration of A is doubling the rate of reaction will also double, since in this scenario concentration is proportional to rate of reaction. So if the concentration of A were to triple the rate of reaction would also triple.
