Respuesta :
Answer : The half-life is, [tex]1.17\times 10^4s[/tex]
Explanation :
First we have to calculate the rate constant for zero order reaction.
The expression used is:
[tex]\ln [A]=-kt+\ln [A_o][/tex]
where,
[tex][A_o][/tex] = let initial concentration = 100
[tex][A][/tex] = final concentration = 25
t = time = 324 s
k = rate constant = ?
Now put all the given values in the above expression, we get:
[tex]\ln (25)=-k\times 324+\ln (100)[/tex]
[tex]k=4.28\times 10^{-3}Ms^{-1}[/tex]
Now we have to calculate the half-life.
The expression used is:
[tex]t_{1/2}=\frac{[A_o]}{2k}[/tex]
[tex]t_{1/2}=\frac{100}{2\times 4.28\times 10^{-3}}[/tex]
[tex]t_{1/2}=11682.24s=1.17\times 10^4s[/tex]
Thus, the half-life is, [tex]1.17\times 10^4s[/tex]
The half-life for a zero order reaction is [tex]1.17*10^4s[/tex].
Zero order reaction:
Zero-order reaction is a chemical reaction wherein the rate does not vary with the increase or decrease in the concentration of the reactant.
Let's calculate the rate constant for zero order reaction.
The expression used is:
[tex]ln[A]=-kt+ln[A_0][/tex]
where,
[tex][A_0][/tex] = initial concentration = 100
[tex][A][/tex] = final concentration = 25
t = time = 324 s
k = rate constant = ?
On substituting the values in the above formula:
[tex]ln[A]=-kt+ln[A_0]\\\\ln(25)=-k*324+ln(100)\\\\k=4.28*10^{-3}Ms^{-1}[/tex]
In order to calculate the half-life.
[tex]t_{1/2}=\frac{[A_0]}{2k} \\\\t_{1/2}=\frac{100}{2*4.28*10^{-3}}\\\\t_{1/2}=11682.24s=1.17*10^4s[/tex]
Therefore, the half-life is, [tex]1.17*10^4s[/tex]
Find more information about Zero order reaction here:
brainly.com/question/13314785
