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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?

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?

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1 Answer 1

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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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  • $\begingroup$ How do you get -1/2 $\endgroup$ Oct 2, 2019 at 3:56
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    $\begingroup$ Make a substitution say $x=[A]_0 - 2[A], dx = -2d[A]$ and change the limits of integration accordingly. In general, $\int \frac{dx}{a+bx} = \frac{1}{b} \ln|a+bx| + C$ $\endgroup$
    – user600016
    Oct 2, 2019 at 5:02

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