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.


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


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.

  • $\begingroup$ Look up coupled differential equations , maths.surrey.ac.uk/explore/vithyaspages/coupled.html $\endgroup$
    – Babu
    Commented Aug 18, 2021 at 11:08
  • $\begingroup$ @andselisk - I realize my fault in using sec for seconds, but I couldn't spot what had been changed each time in kf and kb (they look similar before and after). Could you please clarify, so that I'll format them properly in my future posts? Thanks :) $\endgroup$
    – TRC
    Commented Aug 19, 2021 at 7:53
  • 1
    $\begingroup$ @TRC In these notations "f" and "b" are descriptive indices, referring to "forward" and "backward" labels (not variables), which should be upright. See Which symbols are written in roman (upright) font and which are italicized? I cannot find a reference at the moment, but preferred symbols in kinetics are actually $k_\mathrm{fwd}$ and $k_\mathrm{rev}$ ("forward" and "reverse", correspondingly) to avoid ambiguity. $\endgroup$
    – andselisk
    Commented Aug 19, 2021 at 7:56

4 Answers 4


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.

  • 1
    $\begingroup$ Pls note: I took the given equation for dB/dt and multiplied by negative one , a= 4* 10^{-2} and b=10^{-2} $\endgroup$
    – Babu
    Commented Aug 18, 2021 at 11:12
  • $\begingroup$ Thank you so much - Really dumb of me not to think of A+B = 2! $\endgroup$
    – TRC
    Commented Aug 18, 2021 at 13:33
  • 1
    $\begingroup$ I wouldn't say it is dumb , you can't be sure of anything unless you can prove it /experimentally verify ;) $\endgroup$
    – Babu
    Commented Aug 18, 2021 at 13:34

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 $$ a = a_\mathrm{eq} + b_\mathrm{eq}\mathrm{e}^{-(k_\mathrm{f}+k_\mathrm{b})t}$$


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}


\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 $\mathrm{|B|_\infty = |B|_{eq}}$

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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}


\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)


Using SymPy:

>>> 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

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


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