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: \(\ln 10=2.3\) \(R =8.3\, J \, K ^{-1}\, mol ^{-1}\)
- A \(1\)
- B \(10\)
- C \(45\)
- D \(985\)
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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