Question

A chemical engineer must be able to predict the changes in chemical concentration in a reaction. dC/dt = -kCn where C is chemical concentration and k is rate constant. Order of reaction is the value of n. The first order reaction that combines tert-butyl bromide and water to produce tert-butyl alcohol and hydrogen bromide is shown below; (CH3)3CBr + H2O  (CH3)3COH +HBr From the experimental data, k was estimated to be k = 0.0537 (h -1 ). Determine concentration after 1 hour, if C(0) = 0.2 mol/L

Answer 1

Step-by-step explanation:

Consider the general chemical reaction

`\mathrm{A} \ \overset{k}{\longrightarrow} \ \mathrm{product}` .

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

`Rate \ = \ -\displaystyle\frac{d[\mathrm{A}]}{dt} \ = \ k[\mathrm{A}]^(n)` .

where k is the rate constant.

*Note that the differential term
`\displaystyle\frac{d[\mathrm{A}]}{dt}` 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

`Rate \ = \ -\displaystyle\frac{d[\mathrm{A}]}{dt} \ = \ k[\mathrm{A}]^(1) \ = \ k[\mathrm{A}]`

To solve this first-order linear homogenous differential equation, the method of separation of variables can be used.

`\-\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}`

`\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})`

Since the term
`e^(C)` is a constant, let
`\alpha \ = \ e^(C)`, hence
`[\mathrm{A}] \ = \ \alpha e^(-kt)` or
`C \ = \ \alpha e^(-kt)` according to the question. Given that the initial concentration (t = 0) of tert-butyl bromide is 0.2 mol/L and k = 0.0537
`\mathrm{h^(-1)}`, so

`0.2 \ = \ \alpha e^(-0.0537 \ * \ 0) \\ \\ 0.2 \ = \ \alpha \ \ \ \ \ (e^(0) \ = \ 1)`

Therefore, the rate equation is
`C \ = \ 0.2e^(-0.0537t)`.

The concentration of tert-butyl alcohol after 1 hour is

`C \ = \ 0.2e^(-0.0537 \ * \ 1) \\ \\ C \ = \ 0.2e^(-0.0537) \\ \\ C \ = \ 0.190 \ \mathrm{mol/L} \ \ \ (3 \ \ \mathrm{s.f.})`