For a given reaction, presence of catalyst reduces the energy of activation by 2 kcal at 27$\xb0$C. The rate of reaction will be increased by:

1. 20 times

2. 14 times

3. 28 times

4. 2 times

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What fraction of a reactant showing first order remains after 40 minute if t_{1/2} is 20 minute?

(a) 1/4

(b) 1/2

(c) 1/8

(d) 1/6

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Rate equation for a second order reaction is:

(a) K = (2.303/t) log {a/(a-x)}

(b) K = (1/t) log {a/(a-x)}

(c) K = (1/t) log {a/a(a-x)}

(d) K = (1/t^{2}) log {a/(a-x)}

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For the reaction 2NO_{2} + F_{2} → 2NO_{2}F, following

mechanism has been provided,

NO_{2} + F_{2} $\stackrel{slow}{\to}$ NO_{2}F+F

NO_{2} + F $\stackrel{fast}{\to}$ NO_{2}F

Thus, rate expression of the above

reaction can be written as:

(a) r = K[NO_{2}]^{2}[F_{2}]

(b) r = K[NO_{2} ][F_{2}]

(c) r = K[NO_{2}]

(d) r = K[F_{2}]

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For the reaction:

[Cu(NH_{3})_{4}]^{2+} + H_{2}O_{$\rightleftharpoons $}[Cu(NH_{3})_{3}H_{2}O]^{2+} + NH_{3}

the net rate of reaction at any time is given by, net rate =

2.0x10^{-4} [Cu(NH_{3})_{4}]^{2+}[H_{2}O] - 3.0x10^{5} [Cu(NH_{3} )_{3} H_{2}0]^{2+}[NH_{3}]

Then correct statement is/are :

(a) rate constant for forward reaction = 2 x 10^{-4}

(b) rate constant for backward reaction = 3 x 10^{5}

(c) equilibrium constant for the reaction = 6.6 x 10^{-10}

(d) all of the above

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Rate constant of reaction can be expressed by Arrhenius equation as,

$K=A{e}^{\raisebox{1ex}{$-{E}_{a}$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.}$

In this equation, *${\mathrm{E}}_{\mathrm{a}}$* represents:

(a) the energy above which all the colliding molecules will react

(b) the energy below which colliding molecules will not react

(c) the total energy of the reacting molecules at a temperature, *T*

(d) the fraction of molecules with energy greater than the activation energy of the reaction

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For a first order reaction A$\to $ Product, the initial concentration of A is 0.1 M and after 40 minute it becomes 0.025 M. Calculate the rate of reaction at reactant concentration of 0.01M:

(a) 3.47x10^{-4} M min^{-1}

(b) 3.47x10^{-5} M min^{-1}

(c) 1.735 x 10^{-6} M min^{-1}

(d) 1.735 x10^{-4} M min^{-1}

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Select the intermediate in the following reaction mechanism:

O_{3}(g) $\rightleftharpoons $ O_{2}(g) +O(g)

O(g) +O_{3}(g) $\to $ 2O_{2}(g)

(a) O_{3}(g)

(b) O(g)

(c) O_{2}(g)

(d) none of these

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A reactant with initial concentration 1.386 mol litre^{-1} showing first order change takes 40 minute to become half. If it shows zero order change taking 20 minute to becomes half under the similar conditions, the ratio, K_{1}/K_{0} for first order and zero order kinetics will be:

(a) 0.5 mol^{-1} litre

(b) 1.0 mol/litre

(c) 1.5 mol/litre

(d) 2.0 mol^{-1} litre

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