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Activation Energy

For the two p...

Question

For the two parallel reactions $A_{1} B$ and $A_{2} C,$ show that the activation E' for the disappearance of A is given in terms of activation energies $E_{1}$ and $E_{2}$ for the two paths by $E^{'} = \frac{k_{1} E_{1} + k_{2} E_{2}}{k_{1} + k_{2}}$

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Solution

Let' s consider that one mole of A is getting converted into B and C.

Now let x moles of A get converted into B and (1-x) moles of A get converted into C.
We know that $\frac{x}{1 - x} = \frac{k_{1}}{k_{2}}$ - - - - - - - 1
Also E'(1) = E $_{1}$ (x)+E $_{2}$ (1-x)---------2

Using equation 1 and equation 2, the required relation can be easily deduced.

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Similar questions

Q. For the two parallel reactions

A_{1} B

A_{2} C

, determine the activation energy

E^{'}

for the disapperance of A is given in terms of activation energies

E_{1}

E_{2}

for the two paths .

Q. A reaction can take place by two paths.

k_{1}

k_{2}

are rate constants for the two paths &

E_{1}

E_{2}

are their respective activation energies.
At temperature

T_{a} : k_{1} > k_{2}, E_{1} < E_{2}

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T_{b}

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k_{1}^{'}

k_{2}^{'}

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k_{1}, k_{2}, k_{1}^{'}

k_{2}^{'}

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A_{1}

B, Activation Energy

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A_{2}

C, Activation Energy

E a_{2}

E a_{2} = 2 E a_{1}

k_{1}

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E_{a_{1}}

E_{a_{2}}

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k_{1}

k_{2}

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Q. A reactant (A) forms two products.

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