December 06, 2019, 12:42:27 PM
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### Topic: Kinetics (B+C ->A)  (Read 607 times)

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#### A3H4zw

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##### Kinetics (B+C ->A)
« on: August 09, 2019, 10:25:54 AM »
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

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##### Re: Kinetics (B+C ->A)
« Reply #1 on: August 09, 2019, 03:02:02 PM »
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

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##### Re: Kinetics (B+C ->A)
« Reply #2 on: August 10, 2019, 03:23:37 AM »
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

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##### Re: Kinetics (B+C ->A)
« Reply #3 on: August 10, 2019, 09:22:03 AM »
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

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##### Re: Kinetics (B+C ->A)
« Reply #4 on: August 11, 2019, 05:13:39 AM »
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

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##### Re: Kinetics (B+C ->A)
« Reply #5 on: August 11, 2019, 06:51:50 AM »
No you have to convert to dφ/φ=..dt

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

#### A3H4zw

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##### Re: Kinetics (B+C ->A)
« Reply #6 on: August 11, 2019, 01:22:50 PM »
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

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##### Re: Kinetics (B+C ->A)
« Reply #7 on: August 12, 2019, 06:17:22 AM »

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

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##### Re: Kinetics (B+C ->A)
« Reply #8 on: August 13, 2019, 05:43:40 PM »
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

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##### Re: Kinetics (B+C ->A)
« Reply #9 on: August 14, 2019, 04:40:12 AM »
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

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##### Re: Kinetics (B+C ->A)
« Reply #10 on: August 15, 2019, 04:52:34 AM »
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:
φ1/2=((-(-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...