A reactant A forms two products -A k1- B- Activation energy Ea1A k2- B- Activation energy Ea2-Assuming same value of pre-expotential factor A for both reactions-

K_1 = k_2 e^E_a_1RT k_1 = Ae^ E_a_1RT , nbsp;k_2 = Ae^ E_a_2RT lnk_1 lnk_2 = E_a_1RT + 2E_a_2RT k_1k_2 = e^E_a_1RT k_1 = k_2 e^E_a_1RT nbsp; ...

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Byju's Answer

Standard XII

Chemistry

Arrhenius Equation

A reactant A ...

Question

A reactant (A) forms two products :
$A_{1} B,$ Activation energy $E a_{1}$
$A_{2} B,$ Activation energy $E_{a_{2}}$
[Assuming same value of pre-expotential factor (A) for both reactions]

k_{1} = k_{2} e^{\frac{E_{a_{1}}}{R T}}

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k_{2} = k_{1} e^{\frac{E_{a_{2}}}{R T}}

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k_{2} = k_{1} e^{\frac{E_{a_{1}}}{R T}}

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k_{1} = 2 k_{2} e^{\frac{E_{a_{2}}}{R T}}

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Solution

The correct option is A $k_{1} = k_{2} e^{\frac{E_{a_{1}}}{R T}}$
$k_{1} = k_{2} e^{\frac{E_{a_{1}}}{R T}}$
$k_{1} = A e^{\frac{- E_{a_{1}}}{R T}}$ , $k_{2} = A e^{\frac{- E_{a_{2}}}{R T}}$
$l n k_{1} - l n k_{2} = \frac{- E_{a_{1}}}{R T} + \frac{2 E_{a_{2}}}{R T}$
$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{1}}}{R T}}$
$k_{1} = k_{2} e^{\frac{E_{a_{1}}}{R T}}$

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Similar questions

Consider figure and mark the correct option.

(a) Activation energy of forward reaction is $E_{1}$ + $E_{2}$ and product is less stable than reactant

(b) Activation energy of forward reaction is $E_{1}$ + $E_{2}$ and product is more stable than reactant

(c) Activation energy of both forward and backward reaction is $E_{1}$ + $E_{2}$ and reactant is more stable than product

(d) Activation energy of backward reaction is $E_{1}$ and product is more stable than reactant

Q. A reactant

(A)

forms two products:

A k_{1} - ---- \to B

, Activation energy

E_{a_{1}}

and Pre-exponential factor

A

A k_{2} - ---- \to C

, Activation energy

E_{a_{2}}

and Pre-exponential factor

A

E_{a_{2}} = 2 E_{a_{1}}

, then

k_{1}

and

k_{2}

will be related as:

A_{1}

B, Activation Energy

E a_{1}

A_{2}

C, Activation Energy

E a_{2}

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

, then

k_{1}

and

k_{2}

are related as:

Q. For the two parallel reactions

A_{1} B

and

A_{2} C

, determine the activation energy

E^{'}

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

E_{1}

and

E_{2}

for the two paths .

Q. A reactant (A) forms two products:

A k_{1} - \to B, Activation energy E a_{1}

A k_{2} - \to C, Activation energy E a_{2}

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

, then

k_{1}

and

k_{2}

are related as:

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A reactant (A) forms two products : Ak1⟶B, Activation energy Ea1 Ak2⟶B, Activation energy Ea2 [Assuming same value of pre-expotential factor (A) for both reactions]

The correct option is A k1=k2eEa1RTk1=k2eEa1RT k1=Ae−Ea1RT , k2=Ae−Ea2RT lnk1−lnk2=−Ea1RT+2Ea2RT k1k2=eEa1RT k1=k2eEa1RT

A reactant (A) forms two products :
$A_{1} B,$ Activation energy $E a_{1}$
$A_{2} B,$ Activation energy $E_{a_{2}}$
[Assuming same value of pre-expotential factor (A) for both reactions]