TS EAMCET · Chemistry · Chemical Kinetics

If the rate constants of a reaction at $500 K$ and $700 K$ are $0.002 s^{- 1}$ and $0.06 s^{- 1}$ , respectively, the value of activation energy is $(R = 8.314 J {mol}^{- 1} K^{- 1}, \log 3 = 0.477)$

A $49.49 kJ {mol}^{- 1}$
B $98.98 kJ {mol}^{- 1}$
C $24.75 kJ {mol}^{- 1}$
D $12.37 kJ {mol}^{- 1}$

Verified Solution

Answer & Solution

Correct Answer

(A) $49.49 kJ {mol}^{- 1}$

Step-by-step Solution

Detailed explanation

We have logK2K1=Ea2.303RT2-T1T1T2 Putting the values, we get, log0.060.002=Ea2.303×8.314700-500700×500 log30=Ea19.147200350000 1.477=Ea19.1472003500 Ea=49.49 kJ mol-1

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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If the rate constants of a reaction at 500 K and 700 K are 0.002 s-1 and 0.06 s-1, respectively, the value of activation energy is R=8.314 J mol-1 K-1, log3=0.477

Step-by-step Solution

If the rate constants of a reaction at $500 K$ and $700 K$ are $0.002 s^{- 1}$ and $0.06 s^{- 1}$ , respectively, the value of activation energy is $(R = 8.314 J {mol}^{- 1} K^{- 1}, \log 3 = 0.477)$