Answer:
[tex]0.005734 s^{-1}[/tex] and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]a_o=a\times e^{-kt}[/tex]
where,
k = rate constant
t = age of sample
[tex]a_o[/tex] = let initial amount of the reactant
a = amount left after decay process
We have :
[tex]a_o=x[/tex]
[tex]a=58\%\times x=0.58x[/tex]
t = 95 s
[tex]0.58x=x\times e^{-k\times 95 s}[/tex]
[tex]\k= 0.005734 s^{-1}[/tex]
Half life is given by for first order kinetics::
[tex]t_{1/2}=\frac{0.693}{k}[/tex]
[tex]=\frac{0.693}{0.005734 s^{-1}}=120.86 s[/tex]
[tex]0.005734 s^{-1}[/tex] and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..
