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

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