Question

The rate of decomposition of N2O5 in CCl4 at 317 K has been studied by monitoring the concentration of N2O5 in the solution. 2 N2O5(g) → 4 NO2(g) + O2(g) Initially the concentration of N2O5 is 2.36 M. At 177 minutes, the concentration of N2O5 is reduced to 2.16 M. Calculate the average rate of this reaction in M/min.

Answer 1

Answer:

Average rate of reaction is 0.000565 M/min

Explanation:

Applying law of mass action for the given reaction:

Average rate = $-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO_{2}]}{\Delta t}=\frac{\Delta [O_{2}]}{\Delta t}$

Where, $-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}$ represents average rate of disappearance of $N_{2}O_{5}$, $\frac{1}{4}\frac{[NO_{2}]}{\Delta t}$ represents average rate of appearance of $NO_{2}$ and $\frac{[O_{2}]}{\Delta t}$ represents average rate of appearance of $O_{2}$

Here,$-\frac{[N_{2}O_{5}]}{\Delta t}$ = $-\frac{(2.16-2.36)}{(177-0)}M/min=0.00113M/min$

So average rate of reaction = $[tex]-\frac{1}{2}\frac{[N_{2}O_{5}]}{\Delta t}$[/tex] = $\frac{1}{2}\times (0.00113M/min)=0.000565M/min$