The effect of reactant concentration on when equilibrium is established
I you have an equilibrium with significant amounts of $\ce{A}$ and $\ce{B}$ you need to consider the backwards reaction. If it is first order in $\ce{B}$, the rate law is

$$\frac{\mathrm d[\mathrm{A}]}{\mathrm dt}= -k_1 \cdot [\mathrm{A}] + k_2 \cdot [\mathrm{B}] \tag{1}$$

If the initial concentration of $\ce{A}$ is $\mathrm{[A]}_0$ and the initial concentration of $\ce{B}$ is $0$ the solution of the differential equation is

$$\frac{\mathrm{[A]}}{\mathrm{[A]}_0} = \frac{1}{k_1 + k_2} \cdot \left(k_2 + k_1 \cdot \mathrm e^{-\left(k_1 + k_2\right) t}\right) \tag{2}$$

The equilibrium concentration of $\ce{A}$ is obtained as the limit for $t \to \infty$:

$$\frac{\mathrm{[A]_{eq}}}{\mathrm{[A]}_0} = \frac{k_2}{k_1 + k_2} \tag{3}$$

From $(2)$ and $(3)$ it follows that

$$\frac{\mathrm{[A]} - \mathrm{[A]_{eq}}}{\mathrm{[A]}_0} = k_1 \cdot \mathrm e^{-(k_1 + k_2) t} \tag{4}$$

or, logarithmically,

$$t = -\frac{1}{k_1 + k_2} \left(\ln\left(\mathrm{[A]} - \mathrm{[A]_{eq}}\right) - \ln\left(\mathrm{[A]}_0\right) - \ln\left(k_1\right)\right) \tag{6}$$

The criterion for equilibrium $\left|\mathrm{[A]} - \mathrm{[A]_{eq}}\right| \ll 1$ lets us neglect the term $\ln\left(\mathrm{[A]}_0\right)$ in equation $(6)$.

This shows that the time it takes to reach the equilibrium is independent of the initial concentration of $\ce{A}$.