[A3H4zw]

Hello together,

currently, I can not continue with my task. Maybe you can help me or give me a little hint.
The following information is given:

The chemical equation is the following:
B + C  A
The rate law is the following:
dc(A)/dt=-k₁*c(A)+k₂*c(B)*c(C)

The task is now to indicate the equation for the concentration of A.

Unfortunately, I can not get on there.

[Babcock_Hall]

It is a forum rule (see red link) that you must show your attempt before we can help you.  Have you tried anything yet?

[A3H4zw]

Oh Sorry,

principally I would follow the next steps:
1. *dt
=>dc(A)=(-k1*c(A)+k2*c(B)*c(C))*dt
2. c(A) to the left side
=>This is my current problem because I do not know how to separate this equation.

But maybe someone of you has a different idea for solving this task.

[chenbeier]

Use substitution method for differentials.


Phi = B0-B = C0-C = A-A0

d(phi)  = d(A)

d(phi)/dt = -k1 (A0 + phi)  + k2 (B0- phi)*(C0-phi)

Can you continue now?

[A3H4zw]

Use substitution method for differentials.


Phi = B0-B = C0-C = A-A0

d(phi)  = d(A)

d(phi)/dt = -k1 (A0 + phi)  + k2 (B0- phi)*(C0-phi)

Can you continue now?

Sorry I have never done this before. On the internet, I have only found school mathematician explanations (unfortunately mostly unusable in my case).

I am pretty sure that I wrong but are these first steps, right?

dφ/dt=-k₁(A₀+φ)+k₂(B₀-φ)(C₀-φ)
u=(A₀+φ)
φ=u-A₀
dφ/dt=-k₁(u)+k₂(B₀-φ)(C₀-φ)
dφ/dt=-k₁(u)+k₂(B₀-u-A₀)(C₀-u-A₀)

[chenbeier]

No you have to convert to dφ/φ=..dt

It means to solve the binomial equation    φ² + aφ +b and integration

[A3H4zw]

No you have to convert to dφ/φ=..dt

It means to solve the binomial equation    φ² + aφ +b and integration

dφ/dt=-k₁(A₀+φ)+k₂(B₀-φ)(C₀-φ)
dφ=(-k₁(A₀+φ)+k₂(B₀-φ)(C₀-φ))dt
dφ=((k₂φ²)+((-k₁-k₂B₀-k₂C₀)φ)-(k₁A₀+k₂B₀C₀))dt
dφ=((φ²)+(((-k₁-k₂B₀-k₂C₀)φ)/k₂)-((k₁A₀-B₀C₀k₂)/(k₂)))dt

But how I have to solve this equation now? Because for the pq-Formula dφ need to be 0. Or am I being wrong here?

[chenbeier]

dφ=((φ2)+(((-k1-k2B0-k2C0)φ)/k2)-((k1A0-B0C0k2)/(k2)))dt

dφ/(((φ2)+(((-k1-k2B0-k2C0)φ)/k2)-((k1A0-B0C0k2)/(k2))) = dt

Integral solve it like here:


https://www.quora.com/How-do-I-integrate-dy-y-2-y-1

[A3H4zw]

First of all I would like to thank you for assistance.

Unfortunately, I can‘t find a way to factorize the denominator of my equation. Sure I can see some parallels between a²+2ab+b²=(a+b)² and
dφ/(((φ²)+(((-k1-k2B0-k2C0)φ)/k2)-((k1A0-B0C0k2)/(k2))) = dt but nothing concrete. Also special calculators for the factorization of terms couldn't help.

[mjc123]

Your denominator is a quadratic of the form φ² + aφ + b. Factorise it as (φ - X)(φ - Y) where X and Y are the solutions of the quadratic equation φ² + aφ + b = 0.

[A3H4zw]

Your denominator is a quadratic of the form φ² + aφ + b. Factorise it as (φ - X)(φ - Y) where X and Y are the solutions of the quadratic equation φ² + aφ + b = 0.

dφ/(((φ2)+(((-k1-k2B0-k2C0)φ)/k2)-((k1A0-B0C0k2)/(k2))) = dt
Now I can reduce the denominator:
φ²+(((-k₁/k₂)-B₀-C₀)φ)-((k₁A₀)/k₂)-B₀C₀)


Now I set the term equal to 0:
0=φ²+(((-k₁/k₂)-B₀-C₀)φ)-((k₁A₀)/k₂)-B₀C₀)


And solve to φ:
φ_1/2=((-(-k₁/k₂)-B₀-C₀)/(2))±(((((-k₁/k₂)-B₀-C₀)²)/4)+(((k₁A₀)/k₂)-B₀C₀))^0,5


Factories like described:
dt=(dφ)/(φ-((-(-k₁/k₂)-B₀-C₀)/(2))+(((((-k₁/k₂)-B₀-C₀)²)/4)+(((k₁A₀)/k₂)-B₀C₀))^0,5)(φ-((-(-k₁/k₂)-B₀-C₀)/(2))-(((((-k₁/k₂)-B₀-C₀)²)/4)+(((k₁A₀)/k₂)-B₀C₀))^0,5)

But I don't know what to do next - or rather how...