Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: A3H4zw on August 09, 2019, 10:25:54 AM
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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 :rarrow: 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.
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It is a forum rule (see red link) that you must show your attempt before we can help you. Have you tried anything yet?
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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.
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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?
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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)
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No you have to convert to dφ/φ=..dt
It means to solve the binomial equation φ2 + aφ +b and integration
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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?
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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
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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.
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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.
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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...