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?
Consider a reversible reaction A converts to B and B converts to A with forward and backward rate constants Kf=Kb= 1sec raised to the power minus 1. Suppose we start with a 1 molar solution of A. How long will the concentration of 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 mathsmath or the whole process is wrong .?
