Question

• Determine % H2O2 decomposed in first 500 s in a first order decomposition reaction of H2O2? k=7.30x10-4 s-1

Answer 1

Given rate constant of the first-order reaction is:

K=7.30x10^-4 s-1

Time t=500s

Determine % H2O2 decomposed in first 500 s in a first-order decomposition reaction of H2O2?

Step-by-step explanation:

The expression for the rate constant of the first-order reaction is:

`k=(2.303)/(t) log (ao)/(a-x)`

where,

k=rate constant

t=time period

ao=initial amount of the reactant

a-x=amount of a remained after time t.

Substitute the given values in the above formula to get ao/a-x value.

`7.30x10^-4 s^-1 = (2.303)/(500s) log (ao)/(a-x) \\log (ao)/(a-x) =7.30x10^-4 s^-1 * (500s)/(2.303) \\log (ao)/(a-x) =0.158\\(ao)/(a-x)=10^(0.158) \\(ao)/(a-x)=1.438`

% of H2O2 decomposed is:

`(ao)/(a-x) = 1.438 *100\\ =143.8`