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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
Two reactions \(R_1\) and \(R_2\) have identical pre-exponential factors. Activation energy of \(R_1\) exceeds that of \(R_2\) by \(10\, kJ\,mol^{-1}.\) If \(k_1\) and \(k_2\) are rate constants for reactions \(R_1\) and \(R_2\) respectively at \(300 \,K,\) then \(\ln (k_2/k_1)\) is equal to : \((R=8.314\,J\, mol^{-1}\,K^{-1})\)
- A \(8\)
- B \(12\)
- C \(6\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
According to Arrhenius equation \( k=A e^{-E_{2} / R T} \) \( k_{1} =A e^{-E_{\mathrm{A} 1} / R T} \quad (1)\) \(k_{2} =A e^{-E_{22} / R T} \quad (2)\) Dividing Eq. \(( 2 )\) by Eq. \(( 1 )\), we get \(\frac{k_{2}}{k_{1}}=e^{\left(\frac{E_{21}-E_{22}}{R T}\right)}\) Taking…
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