Question

On introducing a catalyst at $500K$ the rate of a first order reaction increases by $1.718$ times. The activation energy in presence of catalyst is $4.15kJmo{l}^{-1}$ The slope of the plot of In $k\left(inse{c}^{-1}\right)$ versus $\frac{1}{T}\left(TinKelvin\right)$ in absence of catalyst is $\left(R=8.3Jmo{l}^{-1}{K}^{-1}\right)$

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

Answer 1

Answer: (D)

$k=A{e}^{-E_{a/RT}}$
$k^{\prime }=A{e}^{-E_{a^{\prime }/RT}}$
(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)
$\frac{k^{\prime }}{k}=e^{E_a-E_{a/RT}}$
Also given $k^{\prime }=k+1.718k=2.718k$
$\therefore 2.718=e\frac{E_a-E_a}{RT}lo{g}_{e}2.718=\frac{E_a-E_a^{\prime }}{8.314\times 10^{-3}\times 500}$
$E_a-E_a=4.11$
$E_a=4.15$
$\therefore E_a=8.3kJ/mo{l}^{-1}$
$k=A{e}^{-E_a/RT}$
$\therefore lo{g}_{e}k=logeA-\frac{E_a}{RT}$
This is an equation for straight line with slope
$=-\frac{E_a}{R}=-\frac{8.3}{8.3\times 10^{-3}}=-1000$