Answer:
[tex]t=1.875s[/tex]
Explanation:
Hello,
In this case, for the given reaction, the concentration of dinitrogen oxide is quantified as:
[tex]\frac{dC_{N_2O}}{dt}=r_{N_2O}[/tex]
Considering that in molar units (M), the initial and final concentration of dinitrogen oxide are:
[tex]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[/tex]
Now, since the rate is zeroth order in dinitrogen oxide, it depends on the rate constant only:
[tex]\frac{dC_{N_2O}}{dt}=-k[/tex]
Thus, by integrating from the initial concentration to the final concentration:
[tex]\int\limits^{0.015M}_{0.03M} {dC_{N_2O}} \, dx =-k\int\limits^t_0 {} \, dt[/tex]
The time finally results:
[tex]t=\frac{0.015M-0.03M}{-0.0080M/s} \\\\t=1.875s[/tex]
Best regards.
