9
$\begingroup$

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:

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

Rearranging...

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

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?

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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 $B_0+x$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.