Question

Under certain conditions the rate of this reaction is zero order in dinitrogen monoxide with a rate constant of ·0.0080Ms−1: 2N2O(g)→2N2(g)+O2(g) Suppose a 5.0L flask is charged under these conditions with 150.mmol of dinitrogen monoxide. After how much time is there only 75.0mmol left? You may assume no other reaction is important.

Answer 1

`t=1.875s`

Step-by-step explanation:

Hello,

In this case, for the given reaction, the concentration of dinitrogen oxide is quantified as:

`(dC_(N_2O))/(dt)=r_(N_2O)`

Considering that in molar units (M), the initial and final concentration of dinitrogen oxide are:

`C_(N_2O,initial)=(150.0mmol)/(5.0L)*(1mol)/(1000mmol) =0.03M\\\\C_(N_2O,final)=(75.0mmol)/(5.0L)*(1mol)/(1000mmol) =0.015M`

Now, since the rate is zeroth order in dinitrogen oxide, it depends on the rate constant only:

`(dC_(N_2O))/(dt)=-k`

Thus, by integrating from the initial concentration to the final concentration:

`\int\limits^(0.015M)_(0.03M) {dC_(N_2O)} \, dx =-k\int\limits^t_0 {} \, dt`

The time finally results:

`t=(0.015M-0.03M)/(-0.0080M/s) \\\\t=1.875s`

Best regards.