JEE MAIN - Chemistry (2019 - 11th January Morning Slot - No. 9)
If a reaction follows the Arrhenius equation, the plot ln k vs $${1 \over {\left( {RT} \right)}}$$ gives straight line with a gradient ($$-$$ y) unit.
The energy required to active the reactant is :
The energy required to active the reactant is :
y unit
y/R unit
yR unit
$$-$$y unit
Explanation
According to Arrhenius equation,
k = A$${e^{ - {{{E_a}} \over {RT}}}}$$
or ln k = ln A - $${{{{E_a}} \over {RT}}}$$
Comparing the above equation with straight line equation,
y = mx + c,
we get, slope or gradient (m) = –Ea
and Intercept (c) = ln A
Also given that slope or gradient (m) = -y
$$ \therefore $$ -y = –Ea
$$ \Rightarrow $$ Ea = y
So the activation energy of the reactant, Ea = y unit
k = A$${e^{ - {{{E_a}} \over {RT}}}}$$
or ln k = ln A - $${{{{E_a}} \over {RT}}}$$
Comparing the above equation with straight line equation,
y = mx + c,
we get, slope or gradient (m) = –Ea
and Intercept (c) = ln A
Also given that slope or gradient (m) = -y
$$ \therefore $$ -y = –Ea
$$ \Rightarrow $$ Ea = y
So the activation energy of the reactant, Ea = y unit
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