The gas phase decomposition 2N₂O₅ → 4NO₂ + O₂ follows the first order rate law, K = 7.5 × 10^-3 sec^-1. The initial pressure of N₂O₅${\mathrm{ }}_{}$is 0.1 atm. The time of decomposition of N₂O₅ so that the total pressure becomes 0.15 atm will be -

1.	54 sec	2.	5.4 sec
3.	3.45 sec	4.	34.55 sec

If in the fermentation of sugar in an enzymatic solution that is 0.12 M, the concentration of the sugar is reduced to 0.06 M in 10 h and to 0.03 M in 20 h, the order of the reaction will be:

1. 1                                                   

2. 2

3. 3                                                   

4. 0

In a first-order reaction A $\to$ products, the concentration of the reactant decreases to 6.25 % of its initial value in 80 minutes. The value of the rate constant, if the initial concentration is 0.2 mole/litre, will be:

1. $2.17$ $\times$ ${10}^{-2}$ $mi{n}^{-1}$

2. $3.46$ $\times$ ${10}^{-2}$ $mi{n}^{-1}$

3. $3.46$ $\times$ ${10}^{-3}mi{n}^{-1}$

4. $2.16$ $\times$ ${10}^{-3}$ $mi{n}^{-1}$

A catalyst lowers the activation energy of a reaction from 20 kJ mol^–1 to 10 kJ mol^-1. The temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at 27 ^oC will be:
1. \(-123\ ^{\circ}C\)
2. \(-327\ ^{\circ}C\)
3. \(327\ ^{\circ}C\)
4. \(23\ ^{\circ}C\)

For a certain reaction the variation of the rate constant with temperature is given by the equation

                            $In k_1=In k_0+\left(\frac{In 3}{10}\right)t \left(t\ge 0°C\right)$

The value of the temperature coefficient of the reaction rate is therefore –

(A) 4                                                  

(B) 3

(C) 2                                                   

(D) 10

A catalyst lowers the activation energy of a reaction from $20 kJ mol{e}^{-1} to 10 kJ mol{e}^{-1}$. The temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at $27°C$ is :

(1) $-123°C$                                                     

(2) $327°C$

(3) - 327°C                                                         

(4) $+23°C$

The rate of reaction triples when the temperature changes from \(20{ }^{\circ} \mathrm{C} \text { to } 50^{\circ} \mathrm{C}\). The energy of activation for the reaction will be:

The half-life time for the decomposition of a substance dissolved in ${\mathrm{CCl}}_4$ is 2.5 hours at $\mathrm{30 }°$C. The amount of substance that will be left after 10 hours if the initial weight of the substance is 160 gm is:

Given the following reaction:
N₂O₅   as   N₂O₅   ⇌ 2NO₂ + (1/2)O₂
The values of rate constants for the above reaction are 3.45 × 10^-5 and 6.9 × 10^-3 at 27 ^oC and 67 ^oC respectively. The activation energy for the above reaction is : 

1. $102$ $\times {10}^2$ $J$ $$ $$                                    

2. $488.5$ $kJ$

3. $112$ $J$ $$                                             

4. $112.5$ $kJ$

What is the percentage of the reactant molecules crossing over the energy barrier at 325 K?

Given that $∆{\mathrm{H}}_{325}=0.12$ $\mathrm{kcal},$ ${\mathrm{E}}_{\mathrm{a}\left(\mathrm{b}\right)}=+0.02$ $\mathrm{kcal}$