Question

On introducing a catalyst at $500 K$ the rate of a first order reaction increases by $1.718$ times. The activation energy in presence of catalyst is $4.15 kJ mol^{-1}$ The slope of the plot of In $k\left(in \,sec^{-1}\right)$ versus $\frac{1}{T}\left(T\, in\, Kelvin\right)$ in absence of catalyst is $\left(R=8.3 J mol^{-1}K^{-1}\right)$

• A. $+1$
• B. $-1$
• C. $+ 1000$
• D. $-1000$

Answer 1

Answer: (D)

$k=Ae^{-E_{a/RT}}$
$k'=Ae^{-E_{a'/RT}}$
(where $k$ = rate constant for non catalysed reaction and $k$ = rate constant for catalysed reaction. $E_a$ = activation energy for non-catalysed reaction and $E_a =$ activation energy for catalysed reaction)
$\frac{k'}{k}=e^{E_{a}-E_{a/RT}}$
Also given $k' = k+ 1.718k = 2.718k$
$\therefore 2.718=e\frac{E_{a}-E_{a}}{RT}log_{e}2.718=\frac{E_{a}-E_{a}'}{8.314\times10^{-3}\times500}$
$E_{a}-E_{a}=4.11$
$E_{a}=4.15$
$\therefore E_{a}=8.3kJ/mol^{-1}$
$ k=Ae^{-E_{a}/RT}$
$\therefore log_e k=loge A-\frac{E_{a}}{RT}$
This is an equation for straight line with slope
$=-\frac{E_{a}}{R}=-\frac{8.3}{8.3\times10^{-3}}=-1000$