The problem statement says to derive the concentration as a function of time for the general three species first order reactions. I have attached an image of the problem, so see the image for the reaction.(adsbygoogle = window.adsbygoogle || []).push({});

I have written the reaction rates for all the reactions but I am having trouble after that.

The reaction rates that I have are,

[itex]\frac{dCA}{dt}[/itex]=-k[itex]_{2}[/itex]C[itex]_{A}[/itex]+k[itex]_{2}[/itex]C[itex]_{Q}[/itex]-k[itex]_{3}[/itex]C[itex]_{A}[/itex]+k[itex]_{4}[/itex]C[itex]_{S}[/itex]

[itex]\frac{dCQ}{dt}[/itex]=k[itex]_{1}[/itex]C[itex]_{A}[/itex]-k[itex]_{2}[/itex]C[itex]_{Q}[/itex]-k[itex]_{5}[/itex]C[itex]_{Q}[/itex]+k[itex]_{6}[/itex]C[itex]_{S}[/itex]

[itex]\frac{dCS}{dt}[/itex]=k[itex]_{3}[/itex]C[itex]_{A}[/itex]-k[itex]_{4}[/itex]C[itex]_{S}[/itex]+k[itex]_{5}[/itex]C[itex]_{Q}[/itex]-k[itex]_{6}[/itex]C[itex]_{S}[/itex]

Where k1 and k2 are the forward and reverse rate coefficient for A[itex]\rightleftharpoons[/itex]Q reaction

k4 and k3 are the forward and reverse rate coefficient for A[itex]\rightleftharpoons[/itex]S reaction

and

k5 and k6 are the forward and reverse rate coefficient for Q[itex]\rightleftharpoons[/itex]S reaction

from stoichiometry, I also know that C[itex]_{A}[/itex]+C[itex]_{Q}[/itex]+C[itex]_{S}[/itex]=C[itex]_{A0}[/itex]+C[itex]_{Q0}[/itex]+C[itex]_{S0}[/itex]=C[itex]_{T0}[/itex] where C[itex]_{T0}[/itex] is the total mols

Initial Conditions:

C[itex]_{A}[/itex](t=0)=C[itex]_{A0}[/itex]

C[itex]_{Q}[/itex](t=0)=C[itex]_{Q0}[/itex]

C[itex]_{S}[/itex](t=0)=C[itex]_{S0}[/itex]

So I have three differential equations which I am having a hard time solving, If anyone could help it would be greatly appreciated.

Thanks In advance.

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# Three species reversible reaction rate

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