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

The activation energy of one of the reactions in a biochemical process is \(532611\, J\,  mol ^{-1}\). When the temperature falls from \(310 \, K\) to \(300\,  K\), the change in rate constant observed is \(k _{300}= x \times 10^{-3}\,  k _{310}\). The value of \(x\) is \(.....\) [Given: \(\ln 10=2.3\) \(R =8.3\,  J \, K ^{-1}\,  mol ^{-1}\)

  1. A \(1\)
  2. B \(10\)
  3. C \(45\)
  4. D \(985\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1\)

Step-by-step Solution

Detailed explanation

\(\ln \left(\frac{ K _{2}}{ K _{1}}\right)=\frac{ E _{ a }}{ R }\left(\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right)\) \(\operatorname{\ell n}\left(\frac{ K _{2}}{ K _{1}}\right)=\frac{532611}{8.3}\times\left(\frac{10}{310 \times 300}\right)\) where \(K _{2}\) is at \(310 \,K\) and…
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