# Chemical Kintetics - Sequential reactions

1. Jul 6, 2013

### Saitama

Chemical Kinetics - Sequential reactions

1. The problem statement, all variables and given/known data
For a reaction $A \rightarrow B \rightarrow C$, $t_{1/2}$ for A and B are 4 and 2 minutes respectively. How much time would be required for the B to reach maximum concentration.

2. Relevant equations

3. The attempt at a solution
I can't form the differential rate equations until I know the order of the reaction or am I missing something?

Last edited: Jul 6, 2013
2. Jul 6, 2013

### Curious3141

It's implied that the half-lives are constant. First-order reactions are the only ones with half-lives that remain constant as the reaction progresses (or, equivalently, have half-lives that are independent of the reactant concentration).

Once you know it's first order, it should be easy to set up the d.e.s. If you need more help you can search under my username for "sequential decay", where I helped someone else solve a sequential radioactive decay problem years ago. Same principle.

3. Jul 6, 2013

### Saitama

Thanks, missed it. :)

Setting up the D.Es
$$-\frac{dA}{dt}=k_1A$$
$$\frac{dB}{dt}=k_1A-k_2B$$
$$\frac{dC}{dt}=k_2B$$

At maximum concentration of B, $dB/dt=0$, that gives $k_1/k_2=B/A$. I don't see how to proceed from here.

4. Jul 6, 2013

### Curious3141

That's why this simple approach is a nonstarter. Did you take my suggestion and look up my earlier post? Search for "sequential decay" under my username.

5. Jul 6, 2013

### Staff: Mentor

$$A=A_0e^{-k_1t}$$
$$\frac{dB}{dt}+k_2B=k_1A_0e^{-k_1t}$$
$$\frac{d(e^{k_2t}B)}{dt}=k_1A_0e^{(k_2-k_1)t}$$
$$e^{k_2t}B=\frac{k_1A_0}{(k_2-k_1)}(e^{(k_2-k_1)t}-1)$$
$$B=\frac{k_1A_0}{(k_2-k_1)}(e^{-k_1t}-e^{-k_2t})$$
B is maximum when $$e^{(k_2-k_1)t}=\frac{k_2}{k_1}$$
or when
$$t=\frac{\ln{k_2}-\ln{k_1}}{k_2-k_1}$$

6. Jul 7, 2013

### Saitama

Thanks Curious. I found out the concentration of A as a function of time and then formed a linear differential equation to find the concentration of B.

Thank you Chet! That helped me to check my working. :)