Question

The decomposition of N2O5in CCl4 at 318K has been studied by monitoring the concentration of N2O5 in the solution. Initially the concentration of N2O5is 2.33mol/L and after 184 minutes, it is reduced to 2.08mol/L. The reaction takes place according to the equation 2N2O5(g)----> 4NO2(g) +O2(g)
a) Calculate the average rate of this reaction in terms of hours, minutes and seconds.
b) What is the rate of production of NO2 during this period.?

Answer 1

$$\begin{gathered} 2N_2{O}_5\to 4N{O}_2+O_2 \\ AverageRate=\frac{1}{2}\frac{N_2{O}_5}{t} \\ =\frac{1}{2}\frac{(2.33-2.08)}{184}=6.79\times {10}^{-4}mol{L}^{-1}mi{n}^{-1} \\ =6.79\times {10}^{-4}mol{L}^{-1}mi{n}^{-1}\times 60min/1hr \\ =4.07\times {10}^{-2}mol{L}^{-1}h{r}^{-1} \\ =6.79\times {10}^{-4}mol{L}^{-1}mi{n}^{-1}\times 1min/60s \\ =1.13\times {10}^{-5}mol{L}^{-1}{s}^{-1} \\ Rate=\frac{1}{4}\frac{N{O}_2}{t} \\ \frac{N{O}_2}{t}=6.79\times {10}^{-4}mol{L}^{-1}mi{n}^{-1}\times 4=2.72\times {10}^{-3}mol{L}^{-1}mi{n}^{-1} \end{gathered}$$