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 : 

Decomposition on $N{H}_3$on heated tungsten yields the following data :

Initial pressure (mm)      65    105     y    185

Half-life (s)                   290     x     670   820

What are the values of x and y in that order ?

(1) 420 s, 110 mm                                                  

(2) 500 s, 160 mm

(3) 520 s, 170 mm                                                  

(4) 460 s, 150 mm

The half life period of gaseous substance undergoing thermal decomposition was measured for various initial pressure ‘P’ with the following result. 

P(mm)       250     300      400     450

$t_{1/2}$(min)   136    112.5     85      75.5

Calculate the order of reaction.

(A) 2                                                 

(B) 4

(C) 6                                                 

(D) 10

For the reaction 2N₂O₅(g) → 4NO₂(g) + O₂(g)$,$the concentration of $$ $N{O}_2$ increases by 2.4 × 10^-2 mol L^-1

in 6 seconds. The rate of appearance of $N{O}_2$ and the rate of disappearance of $N_2{O}_5$ , respectively, are:

1. 2 x 10^-3 mol L^-1 sec^-1, 4 x 10^-3 mol L^-1 sec^-1

2. 2 x 10^-3 mol L^-1 sec^-1, 1 x 10^-3 mol L^-1 sec^-1

3. 2 x 10^-3 mol L^-1 sec^-1, 2 x 10^-3 mol L^-1 sec^-1

4. 4 x 10^-3 mol L^-1 sec^-1, 2 x 10^-3 mol L^-1 sec^-1

The rate constant of a particular reaction has the dimension of frequency. The order of the reaction is: 

1. Zero.                                                     

2. First.

3. Second.                                                 

4. Fractional.

The reaction of iodomethane with sodium ethoxide proceeds as : $Et{O}^{⊝}+MeI\to EtOMe+ I^{⊝}$

A plot of log $\left\{\frac{\left[MeI\right]}{\left[Et{O}^{⊝}\right]}\right\}$on the Y-axis against 't' on the X-axis gives a straight line with a positive slope. What is the order of the reaction ?

(1) Second                                                

(2) First

(3) Third                                                   

(4) Fractional

For the reaction, C₂H₅I + OH^- → C₂H₅OH + I^- the rate constant was found to have a value of 5.03 × 10^-2 moI^-1 dm³ s^-1 at 289 K and 6.71 mol^-1 dm³ s^-1 at 333 K. 

The rate constant at 305 K will be: 

$1.$ $1.35$ $mo{l}^{-1}$ $d{m}^3$ $s^{-1}$                  

$2.$ $0.35$ $mo{l}^{-1}$ $d{m}^3$ $s^{-1}$

$3.$ $3.15$ $mo{l}^{-1}$ $d{m}^3$ $s^{-1}$                    

$4.$ $7.14$ $mo{l}^{-1}$ $d{m}^3$ $s^{-1}$

A plot of ln rate Vs ln C for the nth order reaction gives 

(1) a straight line with slope n and intercept ln $k_n$

(2) a straight line with slope (n – 1)

(3) a straight line with slope ln $k_n$ and intercept ‘n’

(4) a straight line with slope –n and intercept $k_n$

The first order rate constant for a certain reaction increases from\(1.667 \times 10^{-6} \mathrm{~s}^{-1} \text { at } 727^{\circ} \mathrm{C} \text { to } 1.667 \times 10^{-4} \mathrm{~s}^{-1} \text { at } 1571{ }^{\circ} \mathrm{C}.\) The rate constant at \(1150^{\circ} \mathrm{C}\) is: 
(assume activation energy is constant over the given temperature range)