JEE MAIN - Chemistry (2020 - 8th January Evening Slot - No. 2)
Consider the following plots of rate constant
versus $${1 \over T}$$ for four different reactions. Which
of the following orders is correct for the
activation energies of these reactions?
_8th_January_Evening_Slot_en_2_1.png)
Ec > Ea > Ed > Eb
Ea > Ec > Ed > Eb
Eb > Ea > Ed > Ec
Eb > Ed > Ec > Ea
Explanation
We know
$$k = A{e^{ - {{{E_a}} \over {RT}}}}$$
$$ \Rightarrow $$ $${\log _e}k = {\log _e}A - {{{E_a}} \over {2.303RT}}$$
According to Arhenius equation plot of $${\log _e}k$$ versus $${1 \over T}$$ is linear.
Slope = $$ - {{{E_a}} \over {2.303R}}$$
From plot we conclude :
Slope : c $$>$$ a $$>$$ d $$>$$ b
So correct order of activation energies
Ec $$>$$ Ea $$>$$ Ed $$>$$ Eb
$$k = A{e^{ - {{{E_a}} \over {RT}}}}$$
$$ \Rightarrow $$ $${\log _e}k = {\log _e}A - {{{E_a}} \over {2.303RT}}$$
According to Arhenius equation plot of $${\log _e}k$$ versus $${1 \over T}$$ is linear.
Slope = $$ - {{{E_a}} \over {2.303R}}$$
From plot we conclude :
Slope : c $$>$$ a $$>$$ d $$>$$ b
So correct order of activation energies
Ec $$>$$ Ea $$>$$ Ed $$>$$ Eb
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