1. Tardigrade
  2. Question
  3. Chemistry
  4. The rate of a reaction decreased by 3.555 times when the temperature was changed from 40° C to 30° C. The activation energy (in kJ mol -1 ) of the reaction is--------. Take; R=8.314 J mol-1 K-1 In 3.555 = 1.268

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Q. The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ} C$ to $30^{\circ} C$. The activation energy (in $kJ mol ^{-1}$ ) of the reaction is--------.
Take; $R=8.314 \,J \,mol^{-1}\, K^{-1}$ In $3.555 = 1.268$

JEE MainJEE Main 2020Chemical Kinetics

Solution:

$\ell n \left(\frac{ K _{ T _{2}}}{ K _{ T _{ i }}}\right)=\frac{ E _{ a }}{ R }\left[\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right]$

$T _{1}=303 K ; T _{2}=313 K$

$\frac{ K _{ T _{2}}}{ K _{ T _{ i }}}=3.555$

$\ell (3.555)=\frac{ E _{ a }}{8.314}\left[\frac{1}{303}-\frac{1}{313}\right]$

$E _{ a }=99980.715$

$E _{ a }=99.98 \frac{ kJ }{ mole }$