JEE MAIN - Chemistry (2020 - 5th September Evening Slot - No. 14)
The rate constant (k) of a reaction is measured at differenct temperatures (T), and the data are
plotted in the given figure. The activation energy of the reaction in kJ mol–1 is :
(R is gas constant)_5th_September_Evening_Slot_en_14_1.png)
(R is gas constant)
R
2R
$${1 \over R}$$
$${1 \over {2R}}$$
Explanation
K = A$${e^{ - {{{E_a}} \over {RT}}}}$$
lnK = lnA $${ - {{{E_a}} \over {RT}}}$$
$$ \Rightarrow $$ lnK = lnA $${ - {{{E_a}{{10}^3}} \over {{{10}^3}RT}}}$$
$$ \therefore $$ Slope = $${ - {{{E_a}} \over {{{10}^3}R}}}$$ = $$ - {{10} \over 5}$$
$$ \Rightarrow $$ $${{{{E_a}} \over R}}$$ = 2000 J/mol
$$ \Rightarrow $$ Ea = 2R KJ/mol
lnK = lnA $${ - {{{E_a}} \over {RT}}}$$
$$ \Rightarrow $$ lnK = lnA $${ - {{{E_a}{{10}^3}} \over {{{10}^3}RT}}}$$
$$ \therefore $$ Slope = $${ - {{{E_a}} \over {{{10}^3}R}}}$$ = $$ - {{10} \over 5}$$
$$ \Rightarrow $$ $${{{{E_a}} \over R}}$$ = 2000 J/mol
$$ \Rightarrow $$ Ea = 2R KJ/mol
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