Respuesta :
Answer:
Explanation:
Consider the general chemical reaction
[tex]\mathrm{A} \ \overset{k}{\longrightarrow} \ \mathrm{product}[/tex] .
If [A] is the concentration of A (reactant) at any time t and n is the reaction order for the whole equation, the rate is then related to the concentration of reactant A with the following differential form of equation
[tex]Rate \ = \ -\displaystyle\frac{d[\mathrm{A}]}{dt} \ = \ k[\mathrm{A}]^{n}[/tex] .
where k is the rate constant.
*Note that the differential term [tex]\displaystyle\frac{d[\mathrm{A}]}{dt}[/tex] has a negative sign to denote that the concentration of A is decreasing over time t.
Since the chemical reaction between tert-butyl bromide and water is given to be a first-order reaction, hence n = 1, and the resulting differential equation becomes
[tex]Rate \ = \ -\displaystyle\frac{d[\mathrm{A}]}{dt} \ = \ k[\mathrm{A}]^{1} \ = \ k[\mathrm{A}][/tex]
To solve this first-order linear homogenous differential equation, the method of separation of variables can be used.
[tex]\-\hspace{1cm} \displaystyle\frac{d[\mathrm{A}]}{dt} \ = -\ k[\mathrm{A}] \\ \\ \-\hspace{0.5cm} \displaystyle\frac{1}{[\mathrm{A}]} \, d[\mathrm{A}] \ = -\ k \, dt \\ \\ \int\ {\displaystyle\frac{1}{[\mathrm{A}]} \, d[\mathrm{A}] \ = \ -\int {k} \, {dt}[/tex]
[tex]\ln{[\mathrm{A}]} \ = \ -kt \ + \ C \\ \\ \-\hspace{0.45cm} $[A]$ \ = \ e^{-kt \ + \ C} \\ \\ \-\hspace{0.45cm} $[A]$ \ = \ e^{-kt}e^{C} \ \ \ \ \ \ \ \ \ (e^{a \ + \ b} \ = \ e^{a}e^{b} \ \ \ \mathrm{by \ the \ law \ of \ indices})[/tex]
Since the term [tex]e^{C}[/tex] is a constant, let [tex]\alpha \ = \ e^{C}[/tex], hence [tex][\mathrm{A}] \ = \ \alpha e^{-kt}[/tex] or [tex]C \ = \ \alpha e^{-kt}[/tex] according to the question. Given that the initial concentration (t = 0) of tert-butyl bromide is 0.2 mol/L and k = 0.0537[tex]\mathrm{h^{-1}}[/tex], so
[tex]0.2 \ = \ \alpha e^{-0.0537 \ \times \ 0} \\ \\ 0.2 \ = \ \alpha \ \ \ \ \ (e^{0} \ = \ 1)[/tex]
Therefore, the rate equation is [tex]C \ = \ 0.2e^{-0.0537t}[/tex].
The concentration of tert-butyl alcohol after 1 hour is
[tex]C \ = \ 0.2e^{-0.0537 \ \times \ 1} \\ \\ C \ = \ 0.2e^{-0.0537} \\ \\ C \ = \ 0.190 \ \mathrm{mol/L} \ \ \ (3 \ \ \mathrm{s.f.})[/tex]
