By Sunil Bhardwaj

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It has been already discussed in the previous video that the collision which involves activated molecules result in to the reaction. Consider a bimolecular reaction and let Z be the number of molecules colliding per second per \({ cm }^{ 3 }\) in a system containing one mole of reactant per \({ dm }^{ 3 }\) and f be the fraction which involves activated molecules i.e. the molecules possessing energy of activation. Then the rate constant k for such bimolecular reaction will be $$ k = Z \times f \qquad ...(1)$$ According to law of distribution of energy, fraction of the molecules in a gas possesing energy greater than a particular value \({ E }_{ a }\) is given by, $$ f = \frac { \triangle N }{ N } = { e }^{ \left( \frac { -{ E }_{ a } }{ RT } \right) }$$ where \(\triangle N\) is number of molecules possesing energy greater than E and N is total number of molecules.

Numerical: The energy of activation for a reaction is \(1.85 \times { 10 }^{ 5 }J{ mol }^{ -1 }\). Calculate fraction of molecules having sufficient energy to react at 556K. (\(R = 8.314 J{ K }^{ -1 }{ mol }^{ -1 }\))

Solution: We have Energy of Activation \(E = 1.85 \times { 10 }^{ 5 } J{ mol }^{ -1 }\)

Temperature \(T = 556K\)

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

Using Maxwell-Boltzman law, $$ f = \frac { \triangle N }{ N } = { { e }^{ { -E }/{ RT } } }$$ $$ f = { { e }^{ { \left( \frac { -\left( 1.85 \times { 10 }^{ 5 } \right) }{ 8.314 \times 556 } \right) } } } = { { e }^{ { -40.021 } } }$$ $$ f = 4.16 \times { 10 }^{ -18 }$$