Respuesta :
Answer : The initial rate of the reaction is, [tex]1.739\times 10^{-3}s^{-1}[/tex]
Explanation :
First we have to calculate the rate constant of the reaction.
Expression for rate law for first order kinetics is given by :
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = ?
t = time taken for the process = 44 s
[tex][A_o][/tex] = initial amount or concentration of the reactant = 0.1108 M
[tex][A][/tex] = amount or concentration left time 44 s = 0.0554 M
Now put all the given values in above equation, we get:
[tex]k=\frac{2.303}{44}\log\frac{0.1108}{0.0554}[/tex]
[tex]k=0.0157[/tex]
Now we have to calculate the initial rate of the reaction.
Initial rate = K [A]
At t = 0, [tex][A]=[A_o][/tex]
Initial rate = 0.0157 × 0.1108 = [tex]1.739\times 10^{-3}s^{-1}[/tex]
Therefore, the initial rate of the reaction is, [tex]1.739\times 10^{-3}s^{-1}[/tex]
