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

The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at \(300\, K\) is \(1.0 \times 10^{-3} s ^{-1}\) and the activation energy \(E _{ a }=11.488\, kJ\, mol ^{-1},\) the rate constant at \(200\, K\) is ............  \(\quad \times 10^{-5} s ^{-1} .\) (Round of to the Nearest Integer). (Given : \(\left. R =8.314\, J\, mol ^{-1} K ^{-1}\right)\)

  1. A \(10\)
  2. B \(8\)
  3. C \(14\)
  4. D \(16\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(10\)

Step-by-step Solution

Detailed explanation

\(K _{300}=10^{-4} \quad K _{200}=?\) \(E _{ a }=11.488 KJ / mole \quad R =8.314 J / mole - K\) so \(\ell n \left(\frac{ K _{300}}{ K _{200}}\right)=\frac{ E _{ a }}{ R }\left(\frac{1}{200}-\frac{1}{300}\right)\)…
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