ExamBro
ExamBro
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 : \((R=8.314\,J\, mol^{-1}\,K^{-1})\)

  1. A \(8\)
  2. B \(12\)
  3. C \(6\)
  4. D \(4\)
Verified Solution

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…
From JEE Mains
Explore more questions on app