TS EAMCET · Chemistry · Chemical Kinetics

Calculate the activation energy of a reaction, whose rate constant doubles on raising the temperature from $300 K$ to $600 K .$

A $3.45 kJ / mol$
B $6.90 kJ / mol$
C $9.68 kJ / mol$
D $19.6 kJ / mol$

Verified Solution

Answer & Solution

Correct Answer

(A) $3.45 kJ / mol$

Step-by-step Solution

Detailed explanation

Relation between temperature and activation energy is given by Arrhenius equation. k =Ae-EaRT k2k1 =e-EaRT2+EaRT1ln k2k1=-EaRT2+EaRT1ln 2 =EaR1T1-1T2log 2=Ea2.303R1300-16000.3010=Ea2.303×8.3141600Ea=3.45 kJ/mol

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Calculate the activation energy of a reaction, whose rate constant doubles on raising the temperature from 300 K to 600 K.

Step-by-step Solution

Calculate the activation energy of a reaction, whose rate constant doubles on raising the temperature from $300 K$ to $600 K .$