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TL;DR; Addition in a chemical reaction does not mean mathematical addition, and translating this operation to a rate expression makes no physical sense.

Elaboration: Your idea $\ce{C = A + B}$ changes the order of reaction, and has no physical basis.

The product of two concentrations, $\ce{c_{a}}$ and $\ce{c_{b}}$ in the equation $${\partial c_\ce{AB} \over \partial t} = -k_\mathrm{off} c_\ce{AB} + k_\mathrm{on} c_\ce{A} c_\ce{B}$$ has the physical interpretation of $\ce{A}$ and $\ce{B}$ must collide for reaction to occur, and is hence a second order reaction.

Your proposed solution would change that to first order in a fictional component, $\ce{C}$, and leads you astray.

To help clarify, consider the following reaction: $$\ce{Br^{\cdot} + Br^{\cdot} <=> Br2}$$

In this case A would be $\ce{Br^{\cdot}}$ and B would be $\ce{Br^{\cdot}}$, and thus your C would be $\ce{[Br^{\cdot}] + [Br^{\cdot}]}$. The rate expressions would therefore be first order in $\ce{Br^{\cdot}}$ as opposed to second order.

Other comments:

• You also forgot to include the other reactions for the bimolecular case: $${\partial c_\ce{A} \over \partial t} = {\partial c_\ce{B} \over \partial t} = k_\mathrm{off} c_\ce{AB} - k_\mathrm{on} c_\ce{A} c_\ce{B}$$Omitting these may be leading to some confusion.

• Little k is a rate constant. Equilibria are described with big K. It is unclear what $\ce{k_{d}}$ means in your question.

TL;DR; Addition in a chemical reaction does not mean mathematical addition, and translating this operation to a rate expression makes no physical sense.

Elaboration: Your idea $\ce{C = A + B}$ changes the order of reaction, and has no physical basis.

The product of two concentrations, $\ce{c_{a}}$ and $\ce{c_{b}}$ in the equation: $${\partial c_\ce{AB} \over \partial t} = -k_\mathrm{off} c_\ce{AB} + k_\mathrm{on} c_\ce{A} c_\ce{B}$$ has the physical interpretation that $\ce{A}$ and $\ce{B}$ must collide for reaction to occur, and is hence a second order reaction.

Your proposed solution would change that to first order in a fictional component, $\ce{C}$, and leads you astray.

To help clarify, consider the following reaction: $$\ce{Br^{\cdot} + Br^{\cdot} <=> Br2}$$

In this case A would be $\ce{Br^{\cdot}}$ and B would be $\ce{Br^{\cdot}}$, and thus your C would be $\ce{[Br^{\cdot}] + [Br^{\cdot}] = 2 [Br^{\cdot}]}$. The rate expressions would therefore be first order in $\ce{Br^{\cdot}}$ as opposed to second order.

Other comments:

• You also forgot to include the other reactions for the bimolecular case: $${\partial c_\ce{A} \over \partial t} = {\partial c_\ce{B} \over \partial t} = k_\mathrm{off} c_\ce{AB} - k_\mathrm{on} c_\ce{A} c_\ce{B}$$Omitting these may be leading to some confusion.

• Little k is a rate constant. Equilibria are described with big K. It is unclear what $\ce{k_{d}}$ means in your question.

Your idea $\ce{C = A + B}$ changes the order of reaction, and has no physical basis.

The product of two concentrations, $\ce{c_{a}}$ and $\ce{c_{b}}$ in the equation $${\partial c_\ce{AB} \over \partial t} = -k_\mathrm{off} c_\ce{AB} + k_\mathrm{on} c_\ce{A} c_\ce{B}$$ has the physical interpretation of $\ce{A}$ and $\ce{B}$ must collide for reaction to occur, and is hence a second order reaction.

Your proposed solution would change that to first order in a fictional component, $\ce{C}$, and leads you astray.

You also forgot to include the other reactions for the bimolecular case: $${\partial c_\ce{A} \over \partial t} = {\partial c_\ce{B} \over \partial t} = k_\mathrm{off} c_\ce{AB} - k_\mathrm{on} c_\ce{A} c_\ce{B}$$

As an aside, little k is a reaction constant. Equilibria are described with big K. It is unclear what $\ce{k_{d}}$ means in your question.

TL;DR; Addition in a chemical reaction does not mean mathematical addition, and translating this operation to a rate expression makes no physical sense.

Elaboration: Your idea $\ce{C = A + B}$ changes the order of reaction, and has no physical basis.

The product of two concentrations, $\ce{c_{a}}$ and $\ce{c_{b}}$ in the equation $${\partial c_\ce{AB} \over \partial t} = -k_\mathrm{off} c_\ce{AB} + k_\mathrm{on} c_\ce{A} c_\ce{B}$$ has the physical interpretation of $\ce{A}$ and $\ce{B}$ must collide for reaction to occur, and is hence a second order reaction.

Your proposed solution would change that to first order in a fictional component, $\ce{C}$, and leads you astray.

To help clarify, consider the following reaction: $$\ce{Br^{\cdot} + Br^{\cdot} <=> Br2}$$

In this case A would be $\ce{Br^{\cdot}}$ and B would be $\ce{Br^{\cdot}}$, and thus your C would be $\ce{[Br^{\cdot}] + [Br^{\cdot}]}$. The rate expressions would therefore be first order in $\ce{Br^{\cdot}}$ as opposed to second order.

Other comments:

• You also forgot to include the other reactions for the bimolecular case: $${\partial c_\ce{A} \over \partial t} = {\partial c_\ce{B} \over \partial t} = k_\mathrm{off} c_\ce{AB} - k_\mathrm{on} c_\ce{A} c_\ce{B}$$Omitting these may be leading to some confusion.

• Little k is a rate constant. Equilibria are described with big K. It is unclear what $\ce{k_{d}}$ means in your question.

Your idea $\ce{C = A + B}$ changes the order of reaction, and has no physical basis.

The product of two concentrations, $\ce{c_{a}}$ and $\ce{c_{b}}$ in the equation $${\partial c_\ce{AB} \over \partial t} = -k_\mathrm{off} c_\ce{AB} + k_\mathrm{on} c_\ce{A} c_\ce{B}$$ has the physical interpretation of $\ce{A}$ and $\ce{B}$ must collide for reaction to occur, and is hence a second order reaction.

Your proposed solution would change that to first order in a fictional component, $\ce{C}$, and leads you astray.

Your idea $\ce{C = A + B}$ changes the order of reaction, and has no physical basis.

The product of two concentrations, $\ce{c_{a}}$ and $\ce{c_{b}}$ in the equation $${\partial c_\ce{AB} \over \partial t} = -k_\mathrm{off} c_\ce{AB} + k_\mathrm{on} c_\ce{A} c_\ce{B}$$ has the physical interpretation of $\ce{A}$ and $\ce{B}$ must collide for reaction to occur, and is hence a second order reaction.

Your proposed solution would change that to first order in a fictional component, $\ce{C}$, and leads you astray.

You also forgot to include the other reactions for the bimolecular case: $${\partial c_\ce{A} \over \partial t} = {\partial c_\ce{B} \over \partial t} = k_\mathrm{off} c_\ce{AB} - k_\mathrm{on} c_\ce{A} c_\ce{B}$$

As an aside, little k is a reaction constant. Equilibria are described with big K. It is unclear what $\ce{k_{d}}$ means in your question.