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Effect of Catalyst

For two react...

Question

For two reactions, activation energies are $E_{a_{1}}$ and $E_{a_{2}}$ ; rate constants are $k_{1}$ and $k_{2}$ at the same temperature. If $k_{1} > k_{2}$ then

A

E_{a_{1}} > E_{a_{2}}

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B

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

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C

E_{a_{1}} < E_{a_{2}}

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D

E_{a_{1}} \geq E_{a_{2}}

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Solution

The correct option is C $E_{a_{1}} < E_{a_{2}}$
For two reactions, activation energies are $E_{a 1}$ and $E_{a 2}$ ; rate constants are $k_{1}$ and $k_{2}$ at the same temperature. If $k_{1} > k_{2}$ then from Arehenius equation,
$k = A e^{\frac{- E a}{R T}}$

$\Rightarrow$ $E_{a_{1}} < E_{a_{2}}$ .
Lower is the activation energy, higher is the rate constant and vice versa.

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Similar questions

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

)

(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}}

k_{1}

k_{2}

will be related as:

Q. If for two reactions

E_{a_{1}} > E_{a_{2}}

T C_{1}

T C_{2}

are temperature coefficient respectively, then which alternative is correct?

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