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

The rate constant (K) for the reaction 2A +B $\to$ Product was found to be 2.5x10^-5 litre mol^-1 sec^-1 after 15 sec, 2.60 x10^-5 litre mol^-1sec^-1 after 30 sec and 2.55 x10^-5litre mol^-1 sec^-1 after 50 sec. The order of reaction is:

1. 2

2. 3

3. zero

4. 1

A reaction A₂ + B₂ $\to$ 2AB occurs by the following mechanism:

A₂ $\to$ A + A               .....(slow)

A + B₂ $\to$ AB + B       .....(fast)

A + B $\to$ AB               .....(fast)

Its order would be:

1. 3/2

2. 1

3. 0

4. 2

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

1. 20 times

2. 14 times

3. 28 times

4. 2 times

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

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