If the initial concentration of N2O5 is 5.65×10-2 M, the concentration of N2O5 will be _____________M after 475s have passed.

Question

The gas phase decomposition of dinitrogen pentoxide at 335 K

N₂O₅(g)2 NO₂(g) + ½ O₂(g)

is first order in N₂O₅ with a rate constant of 4.70×10^-3s^-1.

If the initial concentration of N₂O₅ is 5.65×10^-2 M, the concentration of N₂O₅ will be _____________M after 475s have passed.

Answer 1

To approach this problem, look at what is known and what is unknown and see if you can find an equation that relates these values to each other. 

Known: rate constant k, initial [N2O5], time t

Unknown: final [N2O5]

When a kinetics problem involves changing concentrations and relates them to time, you will use the integrated rate laws. We are told the decomposition is first order so we will use that particular integrated rate law which is

ln [A] = -kt + ln [A]o

where [A] is [N2O5] at time t and [A]o is initial [N2O5]..

Plugging in our values and ensuring time units match (e.g., k in s-1, time in s), we will solve for [A]:

ln [A] = -(0.00470 s^-1)(475 s) + ln (0.0565 M)

ln [A] = -5.106

Taking the inverse natural log of both sides,

[A] = e^-5.106 = 0.00606 M N2O5