The rate constant of are action becomes equal to the pre-exponential factor when

the absolute temperature is zero
the activation energy is infinity
the absolute temperature is infinity
the temperature in Celsius is zero.

Solution

If ‘I’ is the intensity of absorbed light and ‘C’ is the concentration of AB for the photo chemical process,AB + hν ⟶ AB*, the rate of formation of AB* is directly proportional to

Solution

The rate of photo chemical process varies with the intensity of absorption.Since greater the intensity of absorbed light more photons will fall at a point, and further each photon causes one molecule to undergo reaction.
\(H_{2}O \overset{photolysis}{\rightarrow} OH^{-} + H^{+}\)

Diazonium salt decomposes as \(C_{6}H_{5}N_{2}^{+}Cl^{-} \rightarrow C_{6}H_{5}Cl + N_{2}\) At 0°C, the evolution of N2 becomes two times faster when the initial concentration of the salt is doubled. Therefore, it is

a first order reaction
a second order reaction
independent of the initial concentration of the salt
a zero order reaction

Solution

As doubling the initial conc. doubles the rate of reaction, order=1

The decomposition of ammonia on tungsten surface at 500 K follows zero order kinetics. The half-life period of this reaction is 45 minutes when the initial pressure is 4 bar. The half-life period (minutes) of the reaction when the initial pressure is 16 bar at the same temperature is

Solution

The decomposition of N₂O₅ occurs as 2N₂O₅ ⟶ 4NO₂ + O₂ and follows I^st order kinetics,hence:

the reaction is unimolecular
the reaction is bimolecular
t_1/2 ∝ a⁰
None of these

Solution

Rate of a reaction can be expressed by following rate expression Rate = k[A]² [B],if concentration of A is increased by 3 times, and concentration of B is increased by 2 times, how many times rate of reaction increases?

9 times
27 times
18 times
8 times

Solution

For a first order reaction the rate constant is 6.909 min^-1.The time taken for 75% conversion in minutes is

\(\frac{3}{2}log 2\)
\(\frac{2}{3}log 3\)
\(\frac{2}{3}log 2\)
\(\frac{2}{3}log (\frac{3}{4})\)

Solution

The rate equation for the reaction 2A + B ⟶ C is found to be: rate = k[A][B]. The correct statement in relation to this reaction is that the

rate of formation of C is twice the rate of disappearance of A
t_1/2 is a constant
unit of k must be s^–1
value of k is independent of the initial concentrations of A and B

Solution

The velocity constant depends on temperature only. It is independent of concentration of reactants.

The rate law for a reaction between the substances A and B is given by Rate =k[A]ⁿ[B]^m On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

(m + n)
(n – m)
2^{(n – m)}
\(\frac{1}{2^{m-n}}\)

Solution

The rate equation for a reaction,N₂O ⟶ N₂+ 1/2O₂ is Rate = k[N₂O]⁰= k. If the initial concentration of the reactant is a mol Lit^-1, the half-life period of the reaction is

t\(\frac{1}{2}\) = \(\frac{a}{2k}\)
-t\(\frac{1}{2}\) = ka
t\(\frac{1}{2}\) = \(\frac{a}{k}\)
t\(\frac{1}{2}\) = \(\frac{k}{a}\)

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