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Collision Theory

19.The rate c...

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19.The rate constant varies as logk=10-9.6×10 2/T then the activation energy of the reaction

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Q. The rate constant of a reaction is

1.5 \times 10^{- 3} a t 25^{\circ} C a n d 2.1 \times 10^{- 2} a t 60^{\circ} C

. The activation energy is:

Q. The rate constant of a reaction is

2 \times 10^{- 2} s^{- 1}

300

K

8 \times 10^{- 2} s^{- 1}

340

K

. Calculate the energy of activation of the reaction.

Q. 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}

E_{2}

) can be represented as:

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. Rate constant

k

of a reaction varies with temperature according to the equation

log k = c o n s t a n t - \frac{E_{a}}{2.303 R} \times \frac{1}{T}

E_{a}

is the energy of activation for the reaction. When a graph is plotted for

log k

\frac{1}{T}

a straight line with a slope

- 6670 k

is obtained. The activation energy for this reaction will be:

R = 8.314 J K^{- 1} {m o l}^{- 1})

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