Deriving kinetic equations for reversible reactions

### Problem



> 
> Given the reaction $\ce{A <=>[$k\_\mathrm f$][$k\_\mathrm b$] B}$ with rate constants $k\_\mathrm f = \pu{4E-2 s-1}$ and $k\_\mathrm b = \pu{10^{-2} s-1}$. Initially, $\pu{2 mol}$ of $\ce{A}$ and $\pu{0 mol}$ of $\ce{B}$ are taken in $\pu{1 L}$ of solution. Find:
> 
> 
> 1. The equilibrium concentrations of $\ce{A}$ and $\ce{B}$.
> 2. Time taken for $\ce{B}$ to reach $50\%$ of its equilibrium concentration.
> 3. Time taken for completion of $10\%$ of the reaction.
> 
> 
> 


### Answer



> 
>  1. $[\ce{B}] = \pu{1.6 M},$ $[\ce{A}] = \pu{0.4 M}$  
> 
>  2. $\pu{14 s}$  
> 
>  3. $\pu{2.8 s}$
> 
> 
> 


### Question


Finding the equilibrium concentrations was a trivial task. However, I cannot find the concentrations of $\ce{A}$ and $\ce{B}$ as a function of time. I wrote the relevant differential equations:


$$
\begin{align}
\frac{\mathrm d[\ce{A}]}{\mathrm dt} &= -4×10^{-2}[\ce{A}] + 10^{-2}[\ce{B}]\\
\frac{\mathrm d[\ce{B}]}{\mathrm dt} &= 4×10^{-2}[\ce{A}] - 10^{-2}[\ce{B}]\\
\end{align}
$$


I am unable to proceed from here, as I have so far only learnt application of chemical kinetics for one-way reactions.




Actually you need concept of coupled differential equations but I suppose you can do a 'trick' to avoid it:


$$ \frac{dA}{dt} =- a A + b B$$


$$ - \frac{dB}{dt} = - aA + bB$$


Are the form of the two equations, you may subtract the two equation to find $ \frac{dA}{dt} = - \frac{dB}{dt}$, and by integration: $ A(t)+B(t)=C$, setting $t=0$ , we find $A(0)+B(0)=2=C$. Hence, we have the following law governing the equation:


$$A(t) + B(t) = 2$$


The above equation is interpreted 'conservation of moles'.


Using this new equation , plug it into any of the first order diffy eqtn and make it single variable diffy equation which you have standard methods to solve.


 10

I'll just post what I did after @Buraian's answer for a general case -


Let the reaction be $\ce{A <=>[$k\_\mathrm f$][$k\_\mathrm b$] B}$


Let the concentration of $\ce{A}$, $\ce{[A]} = a$, and $\ce{[B]} = b$. If initially $c$ moles of $\ce{A}$ and $0$ moles of $\ce{B}$ are taken, then the relation holds: $a + b = c$.


Also, by using equilibrium conditions
$$\frac{a\_\mathrm{eq}}{b\_\mathrm{eq}} = \frac{k\_\mathrm{b}}{k\_\mathrm{f}}$$


which gives
$$a\_\mathrm{eq} = \frac{k\_\mathrm{b}c}{k\_\mathrm{f} + k\_\mathrm{b}} , b\_\mathrm{eq} = \frac{k\_\mathrm{f}c}{k\_\mathrm{f} + k\_\mathrm{b}} $$


Now the differential equation gives


$$
\begin{align}
\frac{\mathrm d[\ce{A}]}{\mathrm dt} &= -k\_\mathrm{f}[\ce{A}] + k\_\mathrm{b}[\ce{B}]\\
\frac{\mathrm da}{\mathrm dt} &= -k\_\mathrm{f}a + k\_\mathrm{b}b\\
\frac{\mathrm da}{\mathrm dt} &= -k\_\mathrm{f}a + k\_\mathrm{b}(c-a)\\
\frac{\mathrm da}{\mathrm dt} &+ (k\_\mathrm{f} + k\_\mathrm{b})a = k\_\mathrm{b}c\\
\end{align}
$$


The RHS is a constant, making it a first order linear differential equation which can be solved easily using standard methods.


The initial condition, $a = c$ at $t = 0$ can be used for the particular solution.


The solution comes out to be
$$ a = \left(\frac{k\_\mathrm{b}c}{k\_\mathrm{f} + k\_\mathrm{b}}\right) + \left(\frac{k\_\mathrm{f}c}{k\_\mathrm{f} + k\_\mathrm{b}}\right)\mathrm{e}^{-(k\_\mathrm{f}+k\_\mathrm{b})t}$$


which can also be written as
$$ a = a\_\mathrm{eq} + b\_\mathrm{eq}\mathrm{e}^{-(k\_\mathrm{f}+k\_\mathrm{b})t}$$


 4

This is a classic example of the kinetics of reversible reactions.


From the mechanism $\ce{A <=>[$k\_\mathrm f$][$k\_\mathrm b$] B}$ you can write the kinetic equations


\begin{align}
\frac{\mathrm d[\ce{A}]}{\mathrm dt} &= - k\_a [\ce{A}] + k\_b [\ce{B}]\\
\frac{\mathrm d[\ce{B}]}{\mathrm dt} & = k\_a [\ce{A}] - k\_b [\ce{B}]\\
\end{align}


To obtain the dependence of the concentration with time, you must solve this set of differential equations, but its solution is
trivial since both concentrations are not independent


\begin{equation}
 [\ce{A}]\_0 + [\ce{B}]\_0 = constant = [\ce{A}]\_\mathrm{t} + [\ce{B}]\_\mathrm{t}
\end{equation}


thus


\begin{equation}
 [\ce{B}]\_\mathrm{t} = constant - [\ce{A}]\_\mathrm{t}
\end{equation}


In the case you describe $[\ce{B}]\_0 = 0$ so


\begin{equation}
 [\ce{B}]\_\mathrm{t} = [\ce{A}]\_0 - [\ce{A}]\_\mathrm{t}
\end{equation}


If the initial concentration of the species A is $[\ce{A}]\_0 = a$, and $[\ce{B}]\_\mathrm{t} = x$, $\mathrm{[A]\_t = a -x} $


\begin{equation}
 \mathrm{ \frac{d[\ce{A}]}{dt} = -\frac{d x}{dt} = - k\_a (a-x) + k\_b x}
\end{equation}


The solution of this differential equation is simple and the result is


\begin{equation}
 \mathrm{ x\_{eq} - x = x\_{eq} \, e^{-(k\_a + k\_b) t}}
\end{equation}


where $\mathrm{x\_{eq} = |B|\_{eq}}$.


You write this expression as a function of $\mathrm{|A|\_t}$ and $\mathrm{|B|\_t}$ as


\begin{align}
 \mathrm{|A|\_t \,=\,} & \mathrm{|A|\_{eq} + |B|\_{eq}e^{-(k\_a + k\_b) t}}\\
 \mathrm{|B|\_t \,=\,} & \mathrm{|B|\_{eq} (1 -e^{-(k\_a + k\_b) t})}\\
\end{align}


Thus this reaction behaves as a first order reaction with $\mathrm{k\_1 = k\_a + k\_b}$ and with $\mathrm{|A|\_\infty = |A|\_{eq}}$ and $\mathrm{|B|\_\infty = |B|\_{eq}}$


[![enter image description here](https://i.stack.imgur.com/GqAyL.png)](https://i.stack.imgur.com/GqAyL.png)


[![enter image description here](https://i.stack.imgur.com/cpLqd.png)](https://i.stack.imgur.com/cpLqd.png)


Furthermore, once the reaction has reached the equilibrium


\begin{equation}
 \mathrm{v\_a} \,=\, \mathrm{v\_b}
\end{equation}


\begin{equation}
 \mathrm{k\_a\,|A|\_{eq}} \,=\, \mathrm{k\_b\,|B|\_{eq} }
\end{equation}


so


\begin{equation}
 \mathrm{K\_{eq}} \,=\, \frac{\mathrm{|B|\_{eq}}}{\mathrm{|A|\_{eq}}} \,=\, \frac{\mathrm{k\_a}}{\mathrm{k\_b}}
\end{equation}


Using the equilibrium equations and the integrated rate equations you can answer the questions.


**Bibliography.**
This particular mechanism is a *classic* in chemical kinetics.


You can find more information on general physical chemistry or introductory chemical kinetics textbooks. For example:


T. Engel et al., *Physical Chemistry, 3rd Edition*, Pearson (2013)


M. J. Pilling and P. W. Seakins, *Reaction Kinetics, 2nd Edition*, Oxford University Press (1996)
K. J. Laidler, *Chemical Kinetic*, 3rd edition, Pearson (1997)


 3

Using [SymPy](https://www.sympy.org):



```
>>> from sympy import *
>>> t = Symbol('t', real=True)
>>> M = Matrix([[ -0.04,  0.01],
                [  0.04, -0.01]])
>>> exp(t * M) * Matrix([2,0])
Matrix([[0.4 + 1.6*exp(-0.05*t)],
        [1.6 - 1.6*exp(-0.05*t)]])

```

In order to find when the concentration of species $\ce{B}$ reaches $50\%$ of its equilibrium concentration, we solve the following transcendental equation


$$1.6 - 1.6 \exp(-0.05 t) = \frac{1.6}{2}$$


and obtain


$$t\_{50\%} := 20 \ln(2) \approx 13.86 \approx 14 \,\pu{seconds}$$


 2