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Arrhenius Equation

A reactant ...

Question

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 $A$
If $E_{a_{2}} = 2 E_{a_{1}}$ , then $k_{1}$ and $k_{2}$ will be related as:

A

k_{2} = k_{1} e^{- E_{a_{1}} / R T}

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B

k_{2} = k_{1} e^{- E_{a_{2}} / R T}

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C

k_{1} = A k_{2} e^{{- E}_{a_{1}} / R T}

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D

k_{1} = 2 k_{2} e^{{- E}_{a_{2}} / R T}

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Solution

The correct option is B $k_{2} = k_{1} e^{- E_{a_{1}} / R T}$
$k_{2} = A e^{- E_{a_{2}} / R T}$
$k_{1} = A e^{- E_{a_{1}} / R T}$
$∴ \frac{k_{2}}{k_{1}} = e^{(- E_{a_{2}} + E_{a_{1}}) / R T}$
$\frac{k_{2}}{k_{1}} = e^{- E_{a_{1}} / R T}$ Since, $E_{a_{2}} = 2 E_{a_{1}}$
$k_{2} = k_{1} e^{- E_{a_{1}} / R T}$

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Similar questions

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}

k_{1}

k_{2}

are 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}

k_{1}

k_{2}

are related as:

Q. For reaction;

A \to B

, the rate of constant

k_{1} = A_{1} e^{- E a_{1} / R T}

and for the reaction;

P \to Q

, the rate constant

k_{2} = A_{2} e^{- E a_{2} / R T}

A_{1} = 10^{8}, A_{2} = 10^{10}

E a_{1} = 600, E a_{2} = 1200

, then the temperature at which

k_{1} = k_{2}

Q. For a reaction

A ⟶ B

, the rate constant

k_{1} = A_{1} e^{- {E a}_{1} / (R T)}

and for the reaction

X ⟶ Y

, the rate constant

k_{2} = A_{2} e^{- {E a}_{2} / (R T)}

A_{1} = 10^{8}, A_{2} = 10^{10}

{E a}_{1} = 600 c a l / m o l, {E a}_{2} = 1800 c a l / m o l

, then the temperature at which

k_{1} = k_{2}

R = 2 c a l / k m o l

)

Q. For two reactions, activation energies are

E_{a_{1}}

E_{a_{2}}

; rate constants are

k_{1}

k_{2}

at the same temperature. If

k_{1} > k_{2}

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