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For a reaction A $\to$ B at 27ºC , 1% reactant molecules possess activation energy, the energy of activation in K.cal/ mole is nearly(R = 2 cal K^-1 mol^-1)

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1.38

0.6

1.2

2.76

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detailed solution

Correct option is D

K=A×e−Ea/KTe−Ea/RT→ fraction of molecular possessing Ea e−Ea/RT=1100−EaRT=−2×2.303Ea=2×300×2.303×21000=4×3×0.2303=2.76K.cal/mole

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For a reaction A →B at 27ºC , 1% reactant molecules possess activation energy, the energy of activation in K.cal/ mole is nearly(R = 2 cal K-1 mol-1)

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For a reaction A $\to$ B at 27ºC , 1% reactant molecules possess activation energy, the energy of activation in K.cal/ mole is nearly(R = 2 cal K^-1 mol^-1)