Answer: The following statement is incorrect: The slope of the straight line in the graph of $\;ln[A_t]\;vs\;t_0$ is $\tan\theta=\large\frac{-kt}{2.303}$

Let's look at the given statements in detail:

The slope of the graph between $\log k$ and $\large\frac{1}{t}$ is $ \;\tan\theta=-\large\frac{E_a}{2.303R}$

$\quad$ By Arrhenius equation, $k=Ae^{\large\frac{-E_a}{RT}}$

$\quad$ Taking log, $ln\; k=ln A-\large\frac{E_a}{RT}$ which when represented in base-10:

$\quad\; \log_{10} k = \large\frac{-E_a}{2.303R}\frac{1}{T}$$+\log_{10} A$

$\quad$ which when compared to $y = mx + c$ gives us the slope: $\large\frac{-E_a}{2.303R}$

For a second order reaction, the graph between $\;\large\frac{A_0-A_t}{A_t}\;\normalsize vs\;t\;$ is a straight line with slope $k_{A_0}$

$\quad$ From the integrated rate law for a second order reaction,$\large\frac{1}{[A]}$$ = \large\frac{1}{[A]_0}$$ + kt$

$\quad\; \Rightarrow kt =,\large\frac{1}{[A]}$$ - \large\frac{1}{[A]_0}$$ \rightarrow kt =\large\frac{[A_0] - [A]}{[A_0][A]}$

$\quad\; \Rightarrow kA_0t = \large\frac{[A_0]-[A_t]}{A_t}$

The slope of the straight line in the graph of $\;ln[A_t]\;vs\;t_0$ is $\tan\theta=\large\frac{-kt}{2.303}$

$\quad$ The integrated first-order equation is the equation of a straight line.

$\quad$ In this case the y-value is $ln [A]$, m equals negative $k$, the x-value is $t$, and the y-intercept is $ln [A_0]$. A plot of $ln [A] $ versus $t$ will yield a line with slope equal to negative $k$.

$\quad$ So, this statement is wrong.

For zero order reaction as reaction proceeds the subsequent half lives of reactant decreases

$\quad$ Half-life is the time taken for the concentration of a reactant to drop to half its original value $[A] = \large\frac{[A_0]}{2}$

$\quad$ Using the integrated form of the rate law, we can develop a relationship between zero-order reactions and the half-life, $\rightarrow \; [A] = [A_0] - kt$

$\quad$ Solving for $t_{1/2}$, we get $t_{1/2} = \large\frac{[A_0]}{2k}$