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Percentage Composition

The rate cons...

Question

The rate constant $(k_{1})$ of one reaction is found to be double that of the rate constant of $(k_{2})$ another reaction. Then the relationship between the corresponding activation energies of two reactions( $E_{1}$ and $E_{2}$ ) can be represented as:

A

E_{1} > E_{2}

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B

E_{1} < E_{2}

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C

E_{1} = E_{2}

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D

E_{1} = 4 E_{2}

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Solution

The correct option is A $E_{1} > E_{2}$

Thus $E_{1} > E_{2}$

Hence, the correct option is $A$

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Similar questions

Q. The rate constant

(K_{1})

of one reaction is found to be double than that of the rate constant of

(K_{2})

another reaction. Then, the relationship between the corresponding activation energies of two reactions

(E_{1}

E_{2})

can be represented as:

Q. The rate constant

K_{1}

of a reaction is found to be double that of a rate constant

K_{2}

of another reaction. The relationship between corresponding activation energies of the two reactions at same temperature

(E_{1}

E_{2})

can be represented as:

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}

Its temperature is raised to

T_{b}

, the rate constants change to

k_{1}^{'}

k_{2}^{'}

. Which relation is correct between

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

k_{2}^{'}

(considering activation energy does not change with temperature)?

E_{1}, E_{2} a n d E_{3}

represents the kinetic energy of an electron, an alpha particle and a proton each having same de-Broglie’s wavelength then

E_{1}

E_{2}

are two events such that

P (E_{1}) = \frac{1}{4}, P (\frac{E_{2}}{E_{1}}) = \frac{1}{2}

P (\frac{E_{1}}{E_{2}}) = \frac{1}{4},

then which of the following is/are correct?

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