Question

At a certain temperature this reaction follows first-order kinetics with a rate constant of 0.0197s. 2N2O5 (g) right arrow 2N2O4 (g) + O2 (g)Suppose a vessel contains N2O5 at a concentration of 0.280 M. Calculate how long it takes for the concentration of N2Os to decrease by 0.0476 M. You may assume no other reaction is important.

Answer 1

`t=90.0s`

Step-by-step explanation:

Hello,

In this case, for first-order kinetics we have an integrated rate law of this reaction as:

`ln(([N_2O_5])/([N_2O_5]_0) )=-kt`

Thus, we compute the time for an initial concentration of 0.280 M which ends up in 0.0476 M as shown below:

`t=(ln(([N_2O_5])/([N_2O_5]_0) ))/(-k)=(ln((0.0476M)/(0.280M)))/(-0.0197s)\\ \\t=90.0s`

Best regards.