A reactant A forms two products-A k 1- B- Activation energy EaA k 2- C- Activation energy E a 2If E a 2-2 E a 1- then k 1 and k 2 are related as-A- k 2- k 1 e E a 2- RTD- k 2- k 1 e Ea 1- RT

The correct option is B k_1=Ak_2e^Ea_1RTk_1=A_1e^ Ea_1RT k_2=A_2 e^ Ea_2RT ∴k_1k_2=A_1A_2e^(Ea_2 Ea_1)RT ∴ k_1k_2=A_1A_2e^(2Ea_1 Ea_1)RT nbsp; (∵ Ea_2=2Ea_1 ...

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Byju's Answer

Standard XII

Chemistry

Molecularity of Reaction

A reactant A ...

Question

A reactant (A) forms two products.
$A k_{1} - \to B$ , Activation energy $E a_{1}$
$A k_{2} - \to C$ , Activation energy $E a_{2}$
If $E a_{2} = 2 E a_{1}$ , then $k_{1}$ and $k_{2}$ are related as:

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

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

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Solution

The correct option is D $k_{1} = A k_{2} e^{\frac{E a_{1}}{R T}}$
$k_{1} = A_{1} e^{- \frac{E a_{1}}{R T}}$
$k_{2} = A_{2} e^{- \frac{E a_{2}}{R T}}$
$∴ \frac{k_{1}}{k_{2}} = \frac{A_{1}}{A_{2}} e^{\frac{(E a_{2} - E a_{1})}{R T}}$
$∴$
$\frac{k_{1}}{k_{2}} = \frac{A_{1}}{A_{2}} e^{\frac{(2 E a_{1} - E a_{1})}{R T}}$ $(∵ E a_{2} = 2 E a_{1})$
or,
$k_{1} = k_{2} A e^{\frac{E a_{1}}{R T}}$

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Similar questions

Q. A reactant

(A)

forms two products:

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

, Activation energy

E_{a_{1}}

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. 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}

E_{a_{1}}

and

E_{a_{2}}

; rate constants are

k_{1}

and

k_{2}

at the same temperature. If

k_{1} > k_{2}

then

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}

. If

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

and

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

, then the temperature at which

k_{1} = k_{2}

is:

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A reactant (A) forms two products. Ak1−→B, Activation energy Ea1 Ak2−→C, Activation energy Ea2 If Ea2=2Ea1, then k1 and k2 are related as:

The correct option is D k1=Ak2eEa1RTk1=A1e−Ea1RT k2=A2e−Ea2RT ∴k1k2=A1A2e(Ea2−Ea1)RT ∴ k1k2=A1A2e(2Ea1−Ea1)RT (∵Ea2=2Ea1) or, k1=k2AeEa1RT

A reactant (A) forms two products.
$A k_{1} - \to B$ , Activation energy $E a_{1}$
$A k_{2} - \to C$ , Activation energy $E a_{2}$
If $E a_{2} = 2 E a_{1}$ , then $k_{1}$ and $k_{2}$ are related as: