Respuesta :
This is an incomplete question, here is a complete question.
Hydrogen peroxide decomposes to water and oxygen according to the following reaction:
[tex]H_2O_2(aq)\rightarrow H_2O+\frac{1}{2}O_2(g)[/tex]
It’s rate of decomposition is measured by titrating samples of the solution with potassium permanganate at certain intervals.
Initial rate determinations at 40°C for the decomposition give the following data:
[H₂O₂] (M) Initial Rate (mol/L min)
0.10 1.93 × 10⁻⁴
0.20 3.86 × 10⁻⁴
0.30 5.79 × 10⁻⁴
Calculate the half-life for the reaction at 40°C?
Answer : The half life for the reaction is, 3590.7 minutes
Explanation :
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:
[tex]H_2O_2(aq)\rightarrow H_2O+\frac{1}{2}O_2(g)[/tex]
Rate law expression for the reaction:
[tex]\text{Rate}=k[H_2O_2]^a[/tex]
where,
a = order with respect to [tex]H_2O_2[/tex]
Expression for rate law for first observation:
[tex]1.93\times 10^{-4}=k(0.10)^a[/tex] ....(1)
Expression for rate law for second observation:
[tex]3.86\times 10^{-4}=k(0.20)^a[/tex] ....(2)
Expression for rate law for third observation:
[tex]5.79\times 10^{-4}=k(0.30)^a[/tex] ....(3)
Dividing 2 by 1, we get:
[tex]\frac{3.86\times 10^{-4}}{1.93\times 10^{-4}}=\frac{k(0.20)^a}{k(0.10)^a}\\\\2=2^a\\a=1[/tex]
Thus, the rate law becomes:
[tex]\text{Rate}=k[H_2O_2]^1[/tex]
[tex]\text{Rate}=k[H_2O_2][/tex]
Now, calculating the value of 'k' by using any expression.
Putting values in equation 1, we get:
[tex]1.93\times 10^{-4}=k(0.10)^1[/tex]
[tex]k=1.93\times 10^{-3}min^{-1}[/tex]
Now we have to calculate the half-life for the reaction.
The expression used is:
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]1.93\times 10^{-3}min^{-1}=\frac{0.693}{t_{1/2}}[/tex]
[tex]t_{1/2}=3590.7min[/tex]
Thus, the half life for the reaction is, 3590.7 minutes
