# Chemical kinetics

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?

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