Webpage title: What are the Rate Equations for a Parallel Reversible Reaction?

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The discussion focuses on deriving rate equations for a parallel reversible reaction involving species A, B, and C. The rate of change of concentration for A is influenced by both the forward and reverse reactions, leading to the equation d[A]/dt = -k1[A] + k2[B] - k3[A] + k4[C]. For species B, the equation is d[B]/dt = k1[A] - k2[B], while for species C, it is d[C]/dt = k3[A] - k4[C]. The concentration of A is affected by all reactions involving it, highlighting the complexity of parallel reversible reactions. Understanding these equations is crucial for analyzing reaction dynamics in chemical kinetics.
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I know the rate equations for a parallel reaction are
d[A]/dt =-k1[A]-k2[A]
d/dt =k1[A]
d[C]/dt =k2[A]

and I know that the rate equations for a reversible reaction are
d[A1]/dt =-k1[A1]+k2[A2]
d[A2]/dt =k1[A1]-k2[A2]-k3[A2]+k4[A3]
d[A3]/dt =k3[A2]-k4[A3]

But what would d[A]/dt, d/dt, and d[C]/dt be for a parallel reversible reaction be such as
A<--->B where --->k1 and <---k2
A<--->C where --->k3 and <---k4

where t is time of course and it's a first order reaction
 
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The concentration of A is going to depend on the rates of all of the reactions that involve A. In your reversible case, A is used up by both the k1 and k3 reactions, and produced by both the k2 and k4 reactions.

d[A]/dt = -k1[A] + k2 - k3[A] + k4[C]

Similarly, the concentration of B will depend on the k1 and k2 steps:

d/dt = k1[A] - k2

and the concentration of C will depend on the k3 and k4 steps:

d[C]/dt = k3[A] - k4[C]
 
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