Answer:
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?
Explanation:
The expression for the rate constant of the first-order reaction is:
[tex]k=\frac{2.303}{t} log \frac{ao}{a-x}[/tex]
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.
[tex]7.30x10^-4 s^-1 = \frac{2.303}{500s} log \frac{ao}{a-x} \\log \frac{ao}{a-x} =7.30x10^-4 s^-1 * \frac{500s}{2.303} \\log \frac{ao}{a-x} =0.158\\\frac{ao}{a-x}=10^{0.158} \\\frac{ao}{a-x}=1.438[/tex]
% of H2O2 decomposed is:
[tex]\frac{ao}{a-x} = 1.438 *100\\ =143.8[/tex]
