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:

1.	\(28.81 \mathrm{~kJ} \mathrm{~mol}^{-1} \)	2.	\(38.51 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
3.	\(18.81 \mathrm{~kJ} \mathrm{~mol}^{-1} \)	4.	\(8.31 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

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

1 mole of a gas changes linearly from its initial state (2 atm, 10 lt) to its final state (8 atm, 4 lt). The maximum rate constant is equal to 20 $se{c}^{-1}$ and the value of activation energy is 40 kJ, assuming that the activation energy does not change in this temperature range. The value of the rate constant, at the maximum temperature that the gas can attain, is: 

1. $0.56$ $\times$ ${10}^{-3}$ $se{c}^{-1}$                                          

2. $3.16$ $\times$ ${10}^{-3}$ $se{c}^{-1}$

3. $1.56$ $\times$ ${10}^{-3}se{c}^{-1}$                                          

4. $5.12$ $\times$ ${10}^{-3}$ $se{c}^{-1}$

A first-order reaction was started with a decimolar solution of the reactant. After 8 minutes and 20 seconds, its concentration was found to be M/100. The rate constant of the reaction will be:

1. $4.6$ $\times$ ${10}^{-3}se{c}^{-1}$                                      

2. $16.6$ $\times$ ${10}^{-3}$ $se{c}^{-1}$

3. $24.6$ $\times$ ${10}^{-3}$ $se{c}^{-1}$                                   

4. $40.6$ $\times$ ${10}^{-3}$ $se{c}^{-1}$

87.5% of a radioactive substance disintegrates in 40 minutes. What is the half life of the substance ?

(1) 13.58 min                                                   

(2) 135.8 min

(3) 1358 min                                                    

(4) None of these

A gaseous substance $A{B}_3$ decomposes irreversibly according to the overall equation $A{B}_3\to \frac{1}{2}{A}_2+\frac{3}{2}{B}_2$ Starting with pure $A{B}_3$ the partial pressure of the reactant varies with time for which the data are given below.
Time in hours 0      5.0      15.0       53.0

$P_{A{B}_3}$mm Hg  660   330      165       82.5

What is the order of the reaction ?

(1) 2                                                            

(2) 0.5

(3) 1                                                            

(4) 1.5

The inversion of cane sugar proceeds with half life of 500 minute at pH 5 for any concentration of sugar. However if pH = 6, the half life changes to 50 minute. The rate law expression for the sugar inversion can be written as 

(1) $r = K[sugar{]}^2 [H{]}^6$                                             

(2) $r = K[sugar{]}^1 [H{]}^0$

(3) $r = K[sugar{]}^0 [H^{+}{]}^6$                                          

(4) $r = K[sugar{]}^0 [H^{+}{]}^1$

The decomposition of A into product has value of k as \(4.5 \times 10^3 \mathrm{~s}^{-1} \text { at } 10^{\circ} \mathrm{C}.\) Energy of activation of the reaction is \(60 \mathrm{~kJ}~mol^{-1}.\) The temperature at which value k would become \(1.5\times10^4~s^{-1}\) is :