Chemical kinetics


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> Consider a reversible reaction $\ce{A}$ converts to $\ce{B}$ and $\ce{B}$ converts to $\ce{A}$ with forward and backward rate constants $k\_\mathrm{fwd} = k\_\mathrm{rev }= \pu{1 s-1}.$ Suppose we start with a 1 molar solution of $\ce{A}$. How long will the concentration of $\ce{A}$ take to reach 0.75 molar?
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Here's what I have done:


$$\ce{A <=>[$k\_\mathrm{fwd}$][$k\_\mathrm{rev}$] B}$$


$$k\_\mathrm{fwd} = k\_\mathrm{rev} = \pu{1 s-1}$$


$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k\_\mathrm{fwd}[\ce{A}]^n - k\_\mathrm{rev}[\ce{B}]^m$$


For the first order reaction $n = m =1$:


$$
\begin{align}
\frac{\mathrm d[\ce{A}]}{\mathrm dt} &= k\_\mathrm{rev}[\ce{B}] - k\_\mathrm{fwd}[\ce{A}] \\
 &= k\_\mathrm{rev}([\ce{A}]\_0 - [\ce{A}]) - k\_\mathrm{fwd}[\ce{A}] &\qquad \{[\ce{B}] = [\ce{A}]\_0 - [\ce{A}]\} \\
 &= k\_\mathrm{rev}[\ce{A}]\_0 - k\_\mathrm{rev}[\ce{A}] - k\_\mathrm{fwd}[\ce{A}] \\
 &= k\_\mathrm{rev}[\ce{A}]\_0 - [\ce{A}](k\_\mathrm{rev} + k\_\mathrm{fwd}) \\
 &= [\ce{A}]\_0 -2[\ce{A}] &\qquad \{k\_\mathrm{rev} = k\_\mathrm{fwd} = 1\}
\end{align}
$$


$$\frac{\mathrm d[\ce{A}]}{\mathrm dt} = [\ce{A}]\_0 -2[\ce{A}]$$


$$\int\_{[\ce{A}]\_0}^{[\ce{A}]\_t}\frac{\mathrm d[\ce{A}]}{[\ce{A}]\_0 - 2[\ce{A}]} = \int\_0^t\mathrm dt$$


$$\frac 1 2 \ln{\left(\frac{0.5 - 0.75}{0.5 - 1}\right)} = t$$


$$t = \pu{-0.35 s}$$


My answer is coming in negative. I am a biology student and not good at calculus. Is something wrong with the math or the whole process is wrong?




You have made an error in the final integration:


$$\begin{align}
t &= \int\_{[\ce{A}]\_0}^{[\ce{A}]\_t}\frac{\mathrm d[\ce{A}]}{[\ce{A}]\_0 - 2[\ce{A}]} \\
&= \frac{-1}{2} {\ln|{A\_{0}-2[A]}|\_{A\_0}}^{A\_t} \\
&= \frac{-1}{2} \ln\left|\frac{1-2\cdot 0.75}{1-2\cdot 1}\right| \\
&= \frac{-1}{2}\ln(0.5) \\
&= \ln(2)/2 \\
&= \pu{0.35 s}
\end{align}$$


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