I am having trouble with the following question:

Consider the following reversible reaction in which the reaction is first order in both directions: $$\ce{[A] <=> [B]}$$ $k_\mathrm a$ is the rate constant for the forward reaction and $k_\mathrm b$ is the rate constant for the reverse reaction.

If only reagent $\ce{A}$ is present at $t = 0$, such that $[\ce{A}](t = 0)$ = $[\ce{A}]_0$ (and likewise for $[\ce{B}]_0$), then at all subsequent times $[\ce{A}]+[\ce{B}] = [\ce{A}]_0$. Write down a differential equation for $\frac{\mathrm d[\ce{A}]}{\mathrm dt}$ where $[\ce{A}] \equiv [\ce{A}](t)$ is the concentration of the reagent $\ce{A}$, and hence show that:

$$[\ce{A}]=[\ce{A}]_0\frac{k_\mathrm b+k_\mathrm a \exp[-(k_\mathrm a+k_\mathrm b)t]}{k_\mathrm a+k_\mathrm b}$$

Here's my attempt:

$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\mathrm a[\ce{A}]+k_\mathrm b[\ce{A}]\tag{1}$$

Then using the identity in the question:

$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\mathrm a[\ce{A}]+k_\mathrm b([\ce{A}]_0-[\ce{A}])\tag{2}$$


$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}+(k_\mathrm a+k_\mathrm b)[\ce{A}]=k_\mathrm b[\ce{A}]_0\tag{3}$$

I believe that this is a first order linear differential equation with the integrating factor: $\mathrm e^{\int (k_\mathrm a+k_\mathrm b)\,\mathrm dt}=\mathrm e^{(k_\mathrm a+k_\mathrm b)t}$ Thus:

$$\frac{\mathrm d}{\mathrm dt}\left([\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\right)=k_\mathrm b[\ce{A}]_0\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\tag{4}$$

$$[\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}=\int k_\mathrm b[\ce{A}]_0\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\,\mathrm dt\tag{5}$$

This gives me:

$$[\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}=k_\mathrm b[\ce{A}]_0\frac{\mathrm e^{(k_\mathrm a+k_\mathrm b)t}}{k_\mathrm a+k_\mathrm b}+c\tag{6}$$

But this doesn't give me the final result if I put in the boundary conditions. Where have I gone wrong?


Third equation looks right to me. That's a non-homogeneous (some would say heterogeneous) first-order differential equation.

I think the heterogeneity is what is giving you problems. Those equations have a general solution which is a sum of a "complementary" solution and a "particular" solution. Don't try to use any boundary conditions until after you have combined particular and complementary solutions.

  1. $$\frac{d[\ce{A}]}{dt}+(k_\mathrm a+k_\mathrm b)[\ce{A}]=k_\mathrm b[\ce{A}]_0$$

The "complementary solution" is the solution to the corresponding homogeneous equation $\frac{d[\ce{A}]}{dt}+(k_\mathrm a+k_\mathrm b)[\ce{A}]=0$, which is $[\ce{A}]_\mathrm c=K e^{-(k_\mathrm a+k_\mathrm b)t}$, where $K$ is an unknown constant of integration.

The "particular solution" is a polynomial in $t$ of the same degree as the hetereogeneous term, which in this case is a polynomial of degree 0, $[\ce{A}]_\mathrm p=C$, where $C$ is an undetermined constant. We solve for this constant by substitution into the original equation:

  1. $$0 + (k_\mathrm a+k_\mathrm b)C=k_\mathrm b[\ce{A}]_0$$

This provides the value of $C$ as $C=\frac{k_\mathrm b[\ce{A}]_\mathrm 0}{k_\mathrm b+k_\mathrm a}$.

The general solution is thus the sum of the particular and complementary solutions:

  1. $$[\ce{A}]=[\ce{A}]_\mathrm c + [\ce{A}]_\mathrm p=K e^{-(k_\mathrm a+k_\mathrm b)t} + \frac{k_\mathrm b[\ce{A}]_0}{k_\mathrm b+k_\mathrm a}$$

Now we can apply boundary conditions. At time 0, $[\ce{A}]=[\ce{A}]_0$, which gives an algebraic equation we can solve for $K$

  1. $$[\ce{A}]_0=K + \frac{k_\mathrm b[\ce{A}]_0}{k_\mathrm b+k_\mathrm a}$$

If I did my algebra right, then the solution is $K=[\ce{A}]_\mathrm 0 \left (1-\frac{k_b}{k_\mathrm b+k_\mathrm a} \right ) $. Substituting this value back into the general solution from my equation 5 gives:

  1. $$[\ce{A}]=[\ce{A}]_0 \left (1-\frac{k_\mathrm b}{k_\mathrm b+k_\mathrm a} \right )e^{-(k_a+k_b)t} + \frac{k_\mathrm b}{k_\mathrm b+k_\mathrm a}[\ce{A}]_0$$

This solution satisfies the boundary condition and also the original equation. It can be simplified a bit by algebraic manipulation to:

  1. $$[\ce{A}]=[\ce{A}]_0\frac{k_\mathrm a e^{-(k_\mathrm a+k_\mathrm b)t}+k_\mathrm b}{k_\mathrm a+k_\mathrm b}$$

From this expression it is easy to check that (i) the condition at $t=0$ is satisfied, that (ii) the expected exponential dependence in time is obtained, and (iii) that as $t \rightarrow \infty $, the reaction should go to equilibrium. The last point about equilibrium can be checked by applying the two chemical definitions for the equilibrium constant (the ratio of product $B$ to reactant $A$ vs. the ratio of rate constants $k_a/k_b$), i.e. $\frac{\ce{B}_{\infty}}{\ce{A}_{\infty}}=\frac{\ce{A}_0-\ce{A}_{\infty}}{\ce{A}_{\infty}}=\frac{k_\mathrm a}{k_\mathrm b}$ and using equation 8 to check consistency between our obtained expression for $\ce{A}_{\infty}$ and the chemically intuitive definition we just made.

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The simplest and most general way to solve this type of scheme is to calculate the effect of an amount x that reacts with initial amounts $A_0$ and $B_0$, thus $$\ce{A~~~ <=>[k_a][k_b] ~~~B} \\ A_0-x ~~~~ B_0 +x $$ which produces the rate equation $$ \frac{dx}{dt} = k_a(A_0-x)-k_b(B_0-x)$$

the variables can be split to give a straightforward integration instead of having to solve a complicated pair of differential equations. $$\int _0^x \frac{dx}{k_aA_0-k_bB_0-(k_a+k_b)x} = \int _0^t dt$$

which, after some rearrangement, gives $$\ln\left(1-\frac{(k_a+k_b)}{k_aA_0-k_bB_0}\right) = -(k_a+k_b)t$$ and this can be rearranged to $$x= \left( 1-e^{-(k_a+k_b)t} \right)\frac{(k_aA_0-k_bB_0)}{k_a+k_b}$$

with the initial condition that at $t=0$, $B_0=\ce{[B]=0 }$ this becomes $$x= \left( 1-e^{-(k_a+k_b)t} \right)\frac{k_aA_0}{k_a+k_b}$$ The concentration of A is $A_0-x$ which is

$$ \ce{[A]}_t =A_0 \frac{k_b+k_ae^{-(k_a+k_b)t}}{k_a+k_b}$$

This expression has the correct form at all times. $\ce{[B]}_t$ can be calculated similarly from $B_0+x$.

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We know, that at time $t = 0$, $[A] = A_0$, utting these values in the equation you derived, we will get $$c=\frac{K_a A_0}{K_a + K_b}. \tag{Constant of Integration}$$

And voilà your answer is in front of you, you just need to transpose the exponential term in the left hand side to the right hand side.

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