Browse Questions

# Which of following is incorrect?

$\begin{array}{1 1}(a)\;The\;slope\;of \;graph\;between\;\log k\;vs\;1/t\; is \;\tan\theta=-\large\frac{E_a}{2.303R}\\(b)\;for\; second\; order\; reaction\; graph\; between\;\large\frac{a_0-a_t}{a_t}\;\normalsize vs\;t\;is\; straight\;line\;with\;slope\;k_{a_0}\\(c)\;The\;slope\;of\;straight\;line\;in\;graph\;of\;ln[A_t]\;vs\;t_0\;is\tan\theta=\large\frac{-kt}{2.303}\\(d)\;\text{for zero order reaction as reaction proceeds the subsequent half lives of reactant decreases}\end{array}$

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