Answer:
4.8% of the H2O2 has decomposed
Step-by-step explanation:
The first-order decomposition of H2O2 is described by the following rate equation:
rate = k[H2O2]
Where k is the rate constant, which is given as 0.00020 s-1 at 35.0oC.
To determine the percent of H2O2 that has decomposed after 266 s, we need to use the integrated form of the rate equation:
ln[H2O2]0 = -kt + ln[H2O2]t
Where [H2O2]0 is the initial concentration of H2O2, [H2O2]t is the concentration of H2O2 at time t, and t is the time in seconds.
To find the percent of H2O2 that has decomposed, we can use the following formula:
% H2O2 decomposed = (1 - [H2O2]t / [H2O2]0) x 100%
By plugging in the given values, we can calculate the percent of H2O2 that has decomposed:
% H2O2 decomposed = (1 - e^(-kt)) * 100%
% H2O2 decomposed = (1 - e^(-0.00020 s^-1 * 266 s)) * 100%
% H2O2 decomposed = (1 - e^(-0.0532)) * 100%
% H2O2 decomposed = (1 - 0.048) * 100%
% H2O2 decomposed = 4.8%
So after 266 seconds, 4.8% of the H2O2 has decomposed.