For the reaction of H_{2} with I_{2}, the rate constant is 2.5\times 10^{-4}dm^{3}mol^{-1}s^{-1} at 327^{\circ}C and 1.0\; dm^{3}mol^{-1}s^{-1} at 527^{\circ}C. The activation energy for the reaction , in KJmol^{-1} is :

\left ( R=8.314\; J\; K^{-1} mol^{-1}\right )

  • Option 1)

    166

  • Option 2)

    150

  • Option 3)

     72

  • Option 4)

    59

Answers (1)

ln\left ( \frac{P_{1}}{P_{2}} \right )=\frac{Ea}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )

Given T_{1}=327^{\circ}C=600 K

         T_{2}=527^{\circ}C=800 K

          P_{1}=2.5\times 10^{-4}

            P_{2}=1\; atm

 ln\left ( \frac{1}{2.5\times 10^{-4}} \right )=\frac{Ea}{8.314}\times \left ( \frac{1}{600}-\frac{1}{800} \right )

E_{a}=166\; kJ


Option 1)

166

Option 2)

150

Option 3)

 72

Option 4)

59

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