The time for the half-life of a first-order reaction is 1 hr. The time taken for 87.5% completion of the reaction is-

1. 1 hour

2. 2 hour

3. 3 hour

4. 4 hour

Which order of reaction obeys the relation t_1/2 = 1/Ka?

(a) First

(b) Second

(c) Third

(d) Zero

The chemical reaction, 2O₃ $\to$3O₂ proceeds as follows;

O₃ $\rightleftharpoons$O₂ + O .....(Fast)

O+O₃ $\to$ 2O₂ ....(Slow)

The rate law expression should be:

(a) r = K[O₃]²

(b) r = K[O₃]²[O₂]^-1

(c) r = K[O₃][O₂]

(d) unpredictable

The rate of reaction becomes 2 times for every 10°C rise in temperature. How the rate of reaction will increase when temperature is increased from 30°C to 80°C?

(a) 16

(b) 32

(c) 64

(d) 128

The fraction of a reactant showing first-order reaction remains after 40 minutes if t_1/2 is 20 minutes -

1. 1/4

2. 1/2

c. 1/8

4. 1/6

For the reaction 2NO₂ + F₂ → 2NO₂F, following

mechanism has been provided,

 NO₂ + F₂   $\stackrel{slow}{\to }$  NO₂F+F

NO₂ + F   $\stackrel{fast}{\to }$ NO₂F

Thus, rate expression of the above

reaction can be written as:

(a) r = K[NO₂]²[F₂]

(b) r = K[NO₂ ][F₂]

(c) r = K[NO₂]

(d) r = K[F₂]

For the reaction:

[Cu(NH₃)₄]²⁺ + H₂O_{$\rightleftharpoons$}[Cu(NH₃)₃H₂O]²⁺ + NH₃

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

2.0x10^-4 [Cu(NH₃)₄]²⁺[H₂O] - 3.0x10⁵ [Cu(NH₃ )₃ H₂0]²⁺[NH₃]

Then correct statement is/are :

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

(b) rate constant for backward reaction = 3 x 10⁵

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

(d) all of the above

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, K₁/K₀ for first order and zero order kinetics will be:

(1) 0.5 mol^-1 litre

(2) 1.0 mol/litre

(3) 1.5 mol/litre

(4) 2.0 mol^-1 litre

In a first order reaction, the concentration of the reactant is decreased from 1.0 M to 0.25M in 20 minute. The rate constant of the reaction would be:

(1) 10min^-1

(2) 6.931 min^-1

(3) 0.6931 min^-1

(4) 0.06931 min^-1