| rate = –d[A]/dt = d[B]/dt = k1[A] - k-1[B]
|
[A]₀ =
|
| [1 mM . . . 5 mM]
|
|
k1 =
|
| [0 s-1 . . . 2 s-1]
|
|
k-1 =
|
| [0 s-1 . . . 2 s-1]
|
In this example we consider a chemical process that leads to equilibrium between reactant, $A$ and product, $B$, according to the following scheme $$A\xrightleftharpoons[k_{-1}]{k_1} B$$ where two reactions $$\begin{array}{c}A\xrightarrow[]{k_1}B& &(1)\\B\xrightarrow[]{k_{-1}}A& &(2)\end{array}$$take place concurrently. The rate of reaction (1), the decomposition of $A$ and the formation of $B$, is proportional to the concentration of $A$ $$rate_1\propto[A]$$ or $$rate_1=k_1[A]$$ whereas the rate of reaction (2), the disappearance of $B$ also the production of $A$, is proportional to the concentration of $B$ $$rate_2\propto[B]$$ or $$rate_2=k_{-1}[B]$$with $k_1$ and $k_{-1}$ being the rate constants for reaction (1) and (2), respectively. Initially $rate_1\gg rate_2$ because $[B]\approx0$, but as [$A$] decreases the first process gradually becomes slower whilst the rate of reaction (2) steadily increases. The net (overall) rate of formation of $A$ and $B$ in the two simultaneous reactions are
net rate of
formation of $A$ |
= |
rate of
production of $A$ |
– |
rate of
decomposition of $A$ |
| $\dfrac{d[A]}{dt}$ |
= |
$k_{-1}[B]$ |
– |
$k_1[A]$ |
| | | |
net rate of
formation of $B$ |
= |
rate of
appearance of $B$ |
– |
rate of
dissapearance of $B$ |
| $\dfrac{d[B]}{dt}$ |
= |
$k_1[A]$ |
– |
$k_{-1}[B]$ |
Following the definition of the rate of a chemical reaction we may write, using the net rate of formation of $A$,
$$-\dfrac{d[A]}{dt}=k_1[A]-k_{-1}[B]$$ which is identical to writing $$\dfrac{d[B]}{dt}=k_1[A]-k_{-1}[B].$$ The latter two equations highlight the
conservation of mass in chemical transformations.
In order to be able to draw concentration–time curves we first need to find the expression linking time to the concentration of $A$. We realize that it is enough to establish the $[A]\sim t$ relation because we will always be able to calculate the concentration of $B$ for any given moment as $[B]=[A]_0-[A]$. Substituting this into the rate equation for the net production of $A$, we obtain $$-\dfrac{d[A]}{dt}=k_1[A]-k_{-1}([A]_0-[A])=(k_1+k_{-1})[A]-k_{-1}[A]_0.$$ To find the expression linking time to $[A]$ we use integration but first we need to separate variables. Rearrangement yields
$$-\dfrac{d[A]}{(k_1+k_{-1})[A]-k_{-1}[A]_0}=dt.$$
Both sides can now be integrated between the limits below:
|
at the start |
some time later |
at equlibrium |
| time |
$0$ |
$t$ |
$\infty$ (theoretically; in practice, when $[A]\approx[A]_{\infty}$ and $[B]\approx[B]_{\infty}$) |
| conc. of $A$ |
$[A]_0$ |
$[A]$ |
$[A]_{\infty}$ or $[A]_e$ |
| conc. of $B$ |
$0$ |
$[B]$ |
$[B]_{\infty}$ or $[B]_e$ |
as
$$\int^{[A]}_{[A]_0}-\dfrac{d[A]}{(k_1+k_{-1})[A]-k_{-1}[A]_0}=\int^t_0dt$$
The right hand side becomes
$$\int^t_0 dt=t.$$
The left hand side rearranges into
$$-\int^{[A]}_{[A]_0}\dfrac{1}{u[A]-v}\,d[A]$$
where we used $u=k_1+k_{-1}$ and $v=k_{-1}[A]_0$ to make the equation look simpler. Substituting $q=u[A]-v$ leads to $dq=\frac{dq}{d[A]}d[A]=u\,d[A]$ which in turn gives $$-\int\dfrac{1}{u[A]-v}\,d[A]=-\int\dfrac{1}{u\,q}\,dq=-\dfrac{1}{u}\int\dfrac{1}{q}\,dq.$$ Recall that $\int \frac{1}{x}dx=ln|x|+const$; therefore, (switching now to definite integration)
$$-\int^{[A]}_{[A]_0}\dfrac{d[A]}{(k_1+k_{-1})[A]-k_{-1}[A]_0}=-\dfrac{1}{u}\Big[ln\, q\Big]^{q}_{q_0}=-\dfrac{1}{u}ln\left(\dfrac{u[A]-v}{u[A]_0-v}\right)=-\dfrac{1}{k_1+k_{-1}}ln\left(\dfrac{(k_1+k_{-1})[A]-k_{-1}[A]_0}{k_1[A]_0}\right).$$
Thus, the integrated rate equation is
$$-\dfrac{1}{k_1+k_{-1}}ln\left(\dfrac{(k_1+k_{-1})[A]-k_{-1}[A]_0}{k_1[A]_0}\right)=t.$$
Taking exponentials of both sides then rearranging for [$A$] yields
$$[A]=\dfrac{k_1[A]_0\,e^{-(k_1+k_{-1})t}+k_{-1}[A]_0}{k_1+k_{-1}},$$
which enables us to calculate the concentration of $A$ (and $B$) at any given time (see Figure above) and that is what we sought to establish. Note that as time goes to infinity the value of the above expression approaches the equilibrium concentration $$[A]_{\infty}=\dfrac{k_{-1}[A]_0}{k_1+k_{-1}}.$$Expressions $$[B]=\dfrac{k_1[A]_0(1-e^{-(k_1+k_{-1})t})}{k_1+k_{-1}}$$ and $$[B]_{\infty}=\dfrac{k_1[A]_0}{k_1+k_{-1}}$$linking the concentration of $B$ to time and giving the equliblium concentration of $B$ can be derived similarly.
The ratio of equilibrium concentrations is a useful measure of the reversibility of a reaction. Despite all reactions being reversible, many appear irreverible in practice if $[B]_{\infty}\gg [A]_{\infty}$.$$\dfrac{[B]_{\infty}}{[A]_{\infty}}=\dfrac{\dfrac{k_1[A]_0}{k_1+k_{-1}}}{\dfrac{k_{-1}[A]_0}{k_1+k_{-1}}}=\dfrac{k_1}{k_{-1}}=K$$ where $K$ denotes the equilibrium constant. This expression can be also derived from the observation that concentrations do not change in equilibrium, which translates into the reaction rate and the rate of change of concentrations all being zero.$$-\dfrac{d[A]}{dt}=\dfrac{d[B]}{dt}=k_1[A]_{\infty}-k_{-1}[B]_{\infty}=0~~\Rightarrow~~\dfrac{[B]_{\infty}}{[A]_{\infty}}=\dfrac{k_1}{k_{-1}}=K$$
Exercise 1: By changing the rate constants, design chemical equilibria where the product concentration is (i) half of, (ii) quarter of, (iii) three times, (iv) five times, (v) equal to the concentration of the reactant.
Exercise 2: Keeping the initial reactant concentration unchanged, are equlibrium concentrations dependent on rate constants? (Try setting $k_1=0.4,~k_{-1}=0.1$ then $k_1=2,~k_{-1}=0.5$; with $[A]_0=5$ mM)