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

Answer:

$t=1.875s$

Explanation:

Hello,

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

$\frac{dC_{N_2O}}{dt}=r_{N_2O}$

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

$C_{N_2O,initial}=\frac{150.0mmol}{5.0L}*\frac{1mol}{1000mmol} =0.03M\\\\C_{N_2O,final}=\frac{75.0mmol}{5.0L}*\frac{1mol}{1000mmol} =0.015M$

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

$\frac{dC_{N_2O}}{dt}=-k$

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:

$t=\frac{0.015M-0.03M}{-0.0080M/s} \\\\t=1.875s$

Best regards.