Rate equation for a second order reaction is:

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

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

3. K = $(\frac{1}{{\left(a\right)}_t}-\frac{1}{{\left(a_o\right)}_t})* \frac{1}{t}$

4. K = (1/t²) log {a/(a-x)}

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

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:

1. the energy above which all the colliding molecules will react

2. the energy below which colliding molecules will not react

3. the total energy of the reacting molecules at a temperature, T

4. the fraction of molecules with energy greater than the activation energy of the reaction

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

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:

1. 3.47x10^-4 M min^-1

2. 3.47x10^-5 M min^-1

3. 1.735 x 10^-6 M min^-1

4. 1.735 x10^-4 M min^-1

Select the intermediate in the following reaction mechanism:

O₃(g) $\rightleftharpoons$ O₂(g) +O(g)

O(g) +O₃(g) $\to$ 2O₂(g)

1. O₃(g)

2. O(g)

3. O₂(g)

4. none of these

A reactant with initial concentration 1.386 \(\mathrm{mol} \text { litre }{ }^{-1}\) showing first order change takes 40 minute to become half. If it shows zero order change taking 20 minute to becomes half under 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 following mechanism has been proposed for the reaction of NO with Br₂ to form NOBr:

NO(g) + Br₂(g) $\rightleftharpoons$ NOBr₂(g)

NOBr₂(g) + NO(g) $\to$2NOBr(g)

If the second step is the rate determining step, the order of the reaction with respect to NO(g) will be:

1. 1

2. 0

3. 3

4. 2