User:Matthionine

Autocatalysis Mathematical Equations
$$A+B \rightleftharpoons 2B$$

$$\frac{\mathrm{d}[A]}{\mathrm{d}t}=-k_+[A][B]+k_-[B]^2$$

$$\frac{\mathrm{d}[B]}{\mathrm{d}t}=k_+[A][B]-2k_-[B]^2+k_-[B]^2=k_+[A][B]-k_-[B]^2$$

$$[A]=[A]_0+[B]_0-[B]$$

$$\frac{\mathrm{d}[B]}{\mathrm{d}t}=k_+([A]_0+[B]_0-[B])[B]-k_-[B]^2$$

$$\frac{\mathrm{d}[B]}{\mathrm{d}t}=(k_+([A]_0+[B]_0)-(k_++k_-)[B])[B]$$

$$\frac{\mathrm{d}[B]}{\mathrm{d}t}=(k_++k_-)(\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B])[B]$$

$$\int\frac{1}{(\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B])[B]}\frac{\mathrm{d}[B]}{\mathrm{d}t}\mathrm{d}t=\int(k_++k_-)\mathrm{d}t$$

$$\frac{1}{(\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B])[B]}\equiv \frac{1}{\frac{k_+}{k_++k_-}([A]_0+[B]_0)(\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B])}+\frac{1}{\frac{k_+}{k_++k_-}([A]_0+[B]_0)[B]}$$

$$\int(\frac{1}{\frac{k_+}{k_++k_-}([A]_0+[B]_0)(\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B])}+\frac{1}{\frac{k_+}{k_++k_-}([A]_0+[B]_0)[B]})\frac{\mathrm{d}[B]}{\mathrm{d}t}\mathrm{d}t=\int(k_++k_-)\mathrm{d}t$$

$$\frac{1}{\frac{k_+}{k_++k_-}([A]_0+[B]_0)}\int(\frac{1}{\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B]}+\frac{1}{[B]})\frac{\mathrm{d}[B]}{\mathrm{d}t}\mathrm{d}t=\int(k_+ + k_-)\mathrm{d}t$$

$$\frac{1}{\frac{k_+}{k_++k_-}([A]_0+[B]_0)}\ln{(\frac{[B]}{\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B]})}=(k_++k_-)t+c$$

Substituting initial conditions of $$[B]=[B]_0$$ and $$t=0$$,

$$c=\frac{1}{\frac{k_+}{k_++k_-}([A]_0+[B]_0)}\ln{(\frac{[B]_0}{\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B]_0})}$$

$$\frac{1}{\frac{k_+}{k_++k_-}([A]_0+[B]_0)}\ln{(\frac{(\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B]_0)[B]}{(\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B])[B]_0})}=(k_++k_-)t$$

$$\frac{(\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B]_0)[B]}{(\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B])[B]_0}=e^{k_+([A]_0+[B]_0)t}$$

$$(\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B]_0)[B]=(\frac{k_+}{k_++k_-}([A]_0+[B]_0)-[B])[B]_0e^{k_+([A]_0+[B]_0)t}$$

$$(\frac{[A]_0}{[B]_0}+1-\frac{k_++k_-}{k_+})[B]=([A]_0+[B]_0-\frac{k_++k_-}{k_+}[B])e^{k_+([A]_0+[B]_0)t}$$

$$(\frac{[A]_0}{[B]_0}-\frac{k_-}{k_+})e^{-k_+([A]_0+[B]_0)t}[B]=(([A]_0+[B]_0)-\frac{k_++k_-}{k_+}[B])$$

$$((\frac{[A]_0}{[B]_0}-\frac{k_-}{k_+})e^{-k_+([A]_0+[B]_0)t}+\frac{k_++k_-}{k_+})[B]=[A]_0+[B]_0$$

$$[B]=\frac{[A]_0+[B]_0}{(\frac{[A]_0}{[B]_0}-\frac{k_-}{k_+})e^{-k_+([A]_0+[B]_0)t}+1+\frac{k_-}{k_+}}$$

$$[A]=\frac{([A]_0+[B]_0)((\frac{[A]_0}{[B]_0}-\frac{k_-}{k_+})e^{-k_+([A]_0+[B]_0)t}+1+\frac{k_-}{k_+}-1)}{(\frac{[A]_0}{[B]_0}-\frac{k_-}{k_+})e^{-k_+([A]_0+[B]_0)t}+1+\frac{k_-}{k_+}}$$

$$[A]=\frac{([A]_0+[B]_0)((\frac{[A]_0}{[B]_0}-\frac{k_-}{k_+})e^{-k_+([A]_0+[B]_0)t}+\frac{k_-}{k_+})}{(\frac{[A]_0}{[B]_0}-\frac{k_-}{k_+})e^{-k_+([A]_0+[B]_0)t}+1+\frac{k_-}{k_+}}$$