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  • The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from 27°C to 42°C. Its energy of activation in J/mol is _______. (Take ln 5 =1.6094 ; R=8.314  J&

The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from 27°C to 42°C. Its energy of activation in J/mol is _______. (Take ln 5 =1.6094 ; R=8.314  J mol-1 K-1)
 

Answers (1)

best_answer

Given,

T1 = 300K

T2 = 315K

As per question KT2 = 5KT2 as molecules activated are increased five times so k will increases five time.

Now

\ln \left(\frac{\mathrm{K}_{\mathrm{T}_{2}}}{\mathrm{~K}_{\mathrm{T}_{1}}}\right)=\frac{\mathrm{Ea}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right)

\ln 5 = \frac{\mathrm{Ea}}{\mathrm{R}}\left(\frac{1}{300}-\frac{1}{315}\right)

\ln 5=\frac{\mathrm{Ea}}{\mathrm{R}}\left(\frac{15}{300 \times 315}\right)

So, 

\mathrm{Ea}=\frac{1.6094 \times 8.314 \times 300 \times 315}{15}

\mathrm{Ea}=84297.47\textup{ Joules/mole}

Ans = 84297

Posted by

Kuldeep Maurya

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