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:

1. r = K[O₃]²

2. r = K[O₃]²[O₂]^-1

3. r = K[O₃][O₂]

4. 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?

1. 16

2. 32

3. 64

4. 128

What fraction of a reactant showing first order remains after 40 minute if t_1/2 is 20 minute?

1. 1/4

2. 1/2

3. 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:

1. r = K[NO₂]²[F₂]

2. r = K[NO₂ ][F₂]

3. r = K[NO₂]

4. 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 :

1. rate constant for forward reaction = 2 x 10^-4

2. rate constant for backward reaction = 3 x 10⁵

3. equilibrium constant for the reaction = 6.6 x 10^-10

4. all of the above

Consider the chemical reaction,

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

The rate of this reaction can be expressed in terms of time derivative of concentration of  ${\mathrm{N}}_2 \left(\mathrm{g}\right), {\mathrm{H}}_2\left(\mathrm{g}\right)$ and ${\mathrm{NH}}_3\left(\mathrm{g}\right)$.

The correct relationship amongest the rate expressions is: 

(1) Rate $=\frac{-\mathrm{d}\left[{\mathrm{N}}_2\right]}{dt}=-\frac{1}{3}\frac{ \mathrm{d}\left[{\mathrm{H}}_2\right]}{dt}=\frac{1}{2}\frac{ \mathrm{d}\left[{\mathrm{NH}}_3\right]}{dt}$

(2) Rate $=\frac{-\mathrm{d}\left[{\mathrm{N}}_2\right]}{dt}=-\frac{3 \mathrm{d}\left[{\mathrm{H}}_2\right]}{dt}=\frac{2 \mathrm{d}\left[{\mathrm{NH}}_3\right]}{dt}$

(3) Rate $=\frac{\mathrm{d}\left[{\mathrm{N}}_2\right]}{dt}=\frac{1}{3}\frac{ \mathrm{d}\left[{\mathrm{H}}_2\right]}{dt}=\frac{1}{2}\frac{ \mathrm{d}\left[{\mathrm{NH}}_3\right]}{dt}$

(4) Rate $=\frac{-\mathrm{d}\left[{\mathrm{N}}_2\right]}{dt}=-\frac{ \mathrm{d}\left[{\mathrm{H}}_2\right]}{dt}=\frac{ \mathrm{d}\left[{\mathrm{NH}}_3\right]}{dt}$

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

The rate constant of a first-order reaction is\(4 \times 10^{-3} \mathrm{sec}^{-1}.\) At a reactant concentration of \(0.02~\mathrm{M},\) the rate of reaction would be:

If concentration of reactants is increased by 'X', the rate constant K becomes:

1. e^K/X

2. K/X

3. K

4. X/K