TS EAMCET · Chemistry · Chemical Kinetics

The rate of chemical reaction doubles with every $10^{o} C$ rise in temperature. If the reaction is carried out in the vicinity of $22^{o} C$ , the activation energy of the reaction is (Given $R = 8.3 J K^{- 1} {mol}^{- 1}, \ln 2 = 0.69$ and $\ln 3 = 1.1$ )

A $1.69 kJ {mol}^{- 1}$
B $0.169 kJ {mol}^{- 1}$
C $49.8 kJ {mol}^{- 1}$
D $498 J {mol}^{- 1}$

Verified Solution

Answer & Solution

Correct Answer

(C) $49.8 kJ {mol}^{- 1}$

Step-by-step Solution

Detailed explanation

It is given that rate of a chemical reaction doubled with every rise so So k2k1=2 T1=10+273=283K T2=22+273=295K Now according to activation energy equation , Log k2k1=Ea2.303R [1T1-1T2]Log 2=Ea2.303R [R…

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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The rate of chemical reaction doubles with every 10oC rise in temperature. If the reaction is carried out in the vicinity of 22oC, the activation energy of the reaction is (Given R=8.3 J K-1 mol-1, ln 2=0.69 and ln 3=1.1)

Step-by-step Solution

The rate of chemical reaction doubles with every $10^{o} C$ rise in temperature. If the reaction is carried out in the vicinity of $22^{o} C$ , the activation energy of the reaction is (Given $R = 8.3 J K^{- 1} {mol}^{- 1}, \ln 2 = 0.69$ and $\ln 3 = 1.1$ )