ExamBro
ExamBro
JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

It has been found that for a chemical reaction with rise in temperature by \(9\, K\) the rate constant gets doubled. Assuming a reaction to be occurring at \(300\, K\), the value of activation energy is found to be \(......\,k\,J \,mol ^{-1}\). [nearest integer] (Given \(\ln 10=2.3, R =8.3 JK ^{-1} mol ^{-1}, \log 2=0.30\) )

  1. A \(66\)
  2. B \(12\)
  3. C \(59\)
  4. D \(78\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(59\)

Step-by-step Solution

Detailed explanation

\(\log _{10} \frac{ K _{2}}{ K _{1}}=\frac{ E _{ a }}{2.303 R }\left(\frac{1}{300}-\frac{1}{309}\right)\) \(0.3=\frac{ E _{ a }}{2.303 \times 8.3}\left(\frac{9}{300 \times 309}\right)\) \(E _{ a }=\frac{0.3 \times 2.303 \times 8.3 \times 300 \times 309}{9}\) \(=59065.04\, J\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app