[camariela]

Hello, I'm having trouble deriving expressions for two different questions.

1) there are two consecutive 1st order rxns with rate constants k1 and k2 having k2 = 2k1. I have to derive an expression for the time when the intermediates rises to a maximum concentration, and find the value of that maximum concentration in terms of the initial concentration of the starting material.

*im thinking about doing a steady state approximation for this one, but i'm not quite sure how to solve and derive this expression.

Also,
2) there is an equilibrium between P and Q, so P <--> Q and this is first order in both directions. I named the rate constant of the forward reaction k1 and the backward reaction k2. i have to derive an expression for the concentration of P as a function of time, if the initial molar concentrations of P and Q are Po and Qo.

*I'm also thinking about doing a steady state approximation for both the forward rxn and backward and then equalling it to 0? I'm not sure though...

I'd really appreciate any help with this.
Best regards to all,
camariela

[madscientist]

Have a go at the question, show your working and I may be able to see where your getting stuck.

cheers,

madscientist :albert:

[Donaldson Tan]

1) there are two consecutive 1st order rxns with rate constants k1 and k2 having k2 = 2k1. I have to derive an expression for the time when the intermediates rises to a maximum concentration, and find the value of that maximum concentration in terms of the initial concentration of the starting material.

*im thinking about doing a steady state approximation for this one, but i'm not quite sure how to solve and derive this expression.

Consider the reaction A -> B -> C

You are almost on the right track, but this is no steady state behavior. It is transient behavior. it relates to the steady state approximation, because when the concentration of the intermediate B is maximum, d[]/dt = 0

Here's the method:

r1 = k1.[A]

r2 = 2.k1.[]

d[A]/dt = -r1 = -k1.[A]  <-- derive the concentration profile of A with respect to time

d[]/dt = r1 - r2 = k1.[A] - 2.k1.[]  <-- substitute the underlined expression with the concentration profile of A into this equation. You will end up with homogenous ordinary differential equation (ODE), which you are required to solve, by using appropriate boundary conditions, ie. conc of B at time 0 = 0 M. Solve the homogeneous ODE, and you will end up with the concentration profile of intermediate B with respect to time.

When [] is max, d[]/dt = 0     <-- Differentiate the concentration profile of B to find d[]/dt. Then set d[]/dt = 0 and solve for the corresponding time required

[Donaldson Tan]

2) there is an equilibrium between P and Q, so P <--> Q and this is first order in both directions. I named the rate constant of the forward reaction k1 and the backward reaction k2. i have to derive an expression for the concentration of P as a function of time, if the initial molar concentrations of P and Q are Po and Qo.

This is similar to to the 1st question.

r1 = k1.[P]
r2 = k2.[Q]

d[P]/dt = r2 - r1
d[Q]/dt = r1 - r2
this looks more complicated than the first question.

fortunately the material balance simplifies the problem.
P <-> Q
[P] + [Q] = [P]₀ + [Q]₀ = N
where N is the total number of moles in the system.

The material balance allows us to express [P] in terms of [Q] and vice-versa. --> [Q] = N - [P]

d[P]/dt = k2.[Q]  - k1.[P] = k2.(N-[P]) - k1.[P] = k2.N - (k1 + k2)[P]

Solve the homogenous ODE below and you arrive at your answer.
d[P]/dt =  k2.N - (k1 + k2)[P]

[plu]

Solve the homogenous ODE below and you arrive at your answer.

Hello,

Sorry to butt in on this topic but how does one go about doing this?

[Donaldson Tan]

Solve the homogenous ODE below and you arrive at your answer.
d[P]/dt =  k2.N - (k1 + k2)[P]

Is anyone not familiar with calculus?

This is a first order homogeneous ODE, so we can use the Integrating Factor (IF) technique.

d[P]/dt =  k2.N - (k1 + k2)[P]
d[P]/dt + (k1 + k2)[P] =  k2.N
IF = exp(int (k1+k2) dx) = exp( (k1+k2)t )
int [[ exp( (k1+k2)t ).d[P]/dt + exp( (k1+k2)t ).(k1 + k2)[P] ]] dt=  int [[ exp( (k1+k2)t ).k2.N ]] dt
exp( (k1+k2)t ).[P] = k2.N.exp( (k1+k2)t )/(k1+k2) + C
where C is the constant of integration

when t = 0, [P] = P₀
exp(0).P₀ = k2.N.exp(0)/(k1+k2) + C
P₀ = k2.N/(k1+k2) + C
C = P₀ - k2.N/(k1+k2)

Solving the above equation,
P = [[ (k2.N.exp((k1+k2)t)+P₀.(k1+k2) - k2.N)]] /[[ (k1+k2).exp((k1+k2).t) ]]

note: exp(x) = e^x