Correct Rate Equation for a Reversible Reaction I'm helping my son with Matlab as he is having trouble learning differential equations as they apply to chemical kinetics.   His task is to solve for the equilibrium concentrations for a reaction
$$\ce{A + B &lt;=&gt;[$k_{\mathrm{f}}$][$k_{\mathrm{b}}$] C}.$$
We are given the forward and backward rates and the initial concentrations of the reactants and products.   I helped him to set up the required equations and ran them through the ode integrator in R and Matlab but the results don't look right.   I'm pretty sure the integration process is working so I think maybe the actual formulae are wrong:
\begin{align}
\frac{\mathrm{d[A]}}{\mathrm{d}t} &amp;= {k}_\mathrm{b}[C]-{k}_\mathrm{f}[A][B]\\
\frac{\mathrm{d[B]}}{\mathrm{d}t} &amp;= {k}_\mathrm{b}[C]-{k}_\mathrm{f}[A][B]\\
\frac{\mathrm{d[C]}}{\mathrm{d}t} &amp;= {k}_\mathrm{f}[A][B] -{k}_\mathrm{b}[C]
\end{align}
When I run this through the integrator with $k_\mathrm{f}=3$, $k_\mathrm{b}=1.5$,
$[A]=\pu{1M}$, $[B]=\pu{1M}$, and $[C]=0$,
the concentrations all reach equilibrium at the same level of $\pu{0.5M}$.  Is that correct?  I would have expected them to be different with the forward rate being bigger.
Really sorry for the basic question but as I said, I'm a programmer not a chemist.  I got the formulae from this How to set up an equation to solve a rate law computationally? but assumed that I didn't need to square any of the concentrations since the reaction is $\ce{1+1&lt;=&gt;1}$.
If all you want to find is the final concentration of all the species at equilibrium than you don't need to solve these differential equations.
To solve this types of questions easily in chemistry we define a quantity $K_{eq}$. Where $K_{eq}=\frac{K_f}{K_b}$ and $K_{eq}=\frac{\text{product of concentrations of products at equilibrium to the power of their stoichiometric coefficient}}{\text{product of concentrations of reactants at equilibrium to the power of their stoichiometric coefficient}}$.
So for your case we can set the equations as such:
$$
\begin{align}
\begin{array}{cccccc}
&amp; \ce{A} &amp; \ce{+} &amp; \ce{B} &amp; \ce{&lt;=&gt;} &amp; \ce{C}\\
\text{Initial} &amp; \pu{1 M} &amp; &amp; \pu{1 M} &amp; &amp; \pu{0 M}\\
\text{Final} &amp; (1-x)\pu{M} &amp; &amp; (1-x)\pu{M} &amp; &amp; x\pu{M}
\end{array}
\end{align}
$$
$$
K_{eq}=\frac{K_f}{K_b}=\frac{[\ce{C}]}{[\ce{A}][\ce{B}]}\\
2=\frac{x}{(1-x)^2}\\
2x^2-5x+2=0\\
(2x^2-4x)+(-x+2)=0\\
(2x-1)(x-2)=0\\
x=\{\pu{0.5M}, \pu{2M}\}\\
$$
But as $[\ce{B}]\leq \pu{1M}$, $x$ has to be equal to $\pu{0.5 M}$.
Which gives, $[\ce{A}]=[\ce{B}]=[\ce{C}]=\pu{0.5 M}$.
This also shows that your solving of differential equations was also correct!