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The activation energy for the reaction ,2HI-----H2+I2 Is 209.5kJ mol-1 at 581K . Calculate the fraction of molecule of reactants having energy equal to or greater then activation energy.?
Most Upvoted Answer
The activation energy for the reaction ,2HI-----H2+I2 Is 209.5kJ mol-1...
K=Ae^(-Ea/RT), here e^(-Ea/RT) represent the molecule of reactants having energy equal to or greater than activation energy. , so , e^(-Ea/RT)=e^ (-209.5×10^3/581×8.314) ,. Solving this you will find number of molecules =1.47 10^-19
Community Answer
The activation energy for the reaction ,2HI-----H2+I2 Is 209.5kJ mol-1...
1.5 × 10^-19
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The activation energy for the reaction ,2HI-----H2+I2 Is 209.5kJ mol-1 at 581K . Calculate the fraction of molecule of reactants having energy equal to or greater then activation energy.?
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