Question

The gas phase decomposition of hydrogen peroxide at 400 °C

H2O2(g)H2O(g) + ½ O2(g)

is second order in H2O2 with a rate constant of 0.650 M-1 s-1.

If the initial concentration of H2O2 is 8.00×10-2 M, the concentration of H2O2 will be 1.70×10-2 M after
seconds have passed.

Answer 1

t = 71.3 s

Step-by-step explanation:

Hello there!

In this case, since the second-order integrated law is given by the following equation:

`(1)/([H_2O_2]) =(1)/([H_2O_2]_0)+kt`

Thus, given the initial and final concentration of hydrogen peroxide and the rate constant, we obtain the following time:

`(1)/([0.0170M])-(1)/(0.0800M)=0.650M^(-1)s^(-1)t\\\\t=(46.32M^(-1))/(0.650M^(-1)s^(-1)) \\\\t=71.3s`

Best regards!