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  • A first-order reaction is 50 % completed in 20 min at 27oC and in 5 min at 47oC. The energy of activation (in kJ/mol)  of reaction is:

A first-order reaction is 50 % completed in 20 min at 27oC and in 5 min at 47oC. The energy of activation (in kJ/mol)  of reaction is:

Option: 1

55.14


Option: 2

43.85


Option: 3

11.97


Option: 4

6.65


Answers (1)

best_answer

Given,

A first-order reaction is 50 % completed in 20 min at 27oC and in 5 min at 47oC.

Case 1 ; T1= 27oC = 300 K , t1/2 = 20 min

\mathrm{k_1=\frac{ln 2}{t_{1/2}} = \frac{ln2}{20}}

Case 1 ; T2= 47oC = 320 K , t1/2 = 5 min


\mathrm{k_2=\frac{ln 2}{t_{1/2}} = \frac{ln2}{5}}

Now,

Arrhenius Equation -

\mathrm{k = A e^{-E_{a}/RT}}

\mathrm{ln k=lnA-\frac{E_{a}}{RT}}

k = Rate constant

So,

 \mathrm{\frac{k_{2}}{k_{1}}=\frac{e^{-E_{a}/RT_{2}}}{e^{-E_{a}/RT_{1}}}}

Take both sides (ln)-

\mathrm{\ln \frac{k_{2}}{k_{1}}=\frac{-E_{a}}{RT_{2}}-\frac{-E_{a}}{RT_{1}}}

\mathrm{\ln \frac{k_{2}}{k_{1}}= \frac{-E_a}{R}\left (\frac{1}{T_{2}}-\frac{1}{T_{1}} \right ) }

Put values from the above calculations-

\mathrm{\ln \left (\frac{ln2/5}{ln2/20} \right )=\frac{-E_{a}}{R} \left (\frac{1}{320}-\frac{1}{300} \right )}

\mathrm{\ln4=\frac{-E_{a}}{R} \left (\frac{-20}{320\times 300}\right )}

\mathrm{2\times 2.303\times 0.3010=\frac{-E_{a}}{8.314} \left (\frac{-20}{320\times 300}\right )}

\mathrm{E_{a}= 55.14\ kJ/mol}

Posted by

Sanket Gandhi

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