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Kinetics - Nuclear - Surface - Photochemistry - Chemistry MCQ


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30 Questions MCQ Test Mock Test Series for IIT JAM Chemistry - Kinetics - Nuclear - Surface - Photochemistry

Kinetics - Nuclear - Surface - Photochemistry for Chemistry 2024 is part of Mock Test Series for IIT JAM Chemistry preparation. The Kinetics - Nuclear - Surface - Photochemistry questions and answers have been prepared according to the Chemistry exam syllabus.The Kinetics - Nuclear - Surface - Photochemistry MCQs are made for Chemistry 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Kinetics - Nuclear - Surface - Photochemistry below.
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Kinetics - Nuclear - Surface - Photochemistry - Question 1

Activation energy for decomposition of H2O2 is 76 kJ/mol at 298K and the decomposition is very slow. When a little iodide is added, the activation energy decreases to 57kJ/mol. Hence, rate coefficient increases by a factor of           

Kinetics - Nuclear - Surface - Photochemistry - Question 2

The rate of radioactive disintegration ..........with time:           

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Kinetics - Nuclear - Surface - Photochemistry - Question 3

For opposing reaction, A + B ↔ C + D, Ea = 100 kJ/mol and A = 1010 M-1s-1. The equilibrium concentration of A, B, C, D are 1, 2, 5, 4 M resp., at 700K. The values of k1 and k-1 at this temp are-    

Kinetics - Nuclear - Surface - Photochemistry - Question 4

If 0.001M of a substance quenches the efficiency of fluorescence by 20%, the value of Stern-Volmer constant in M-1 is      

Kinetics - Nuclear - Surface - Photochemistry - Question 5

Decomposition of ammonia on tungsten at 850oC has a rate constant value of 0.10 Torr sec-1. If the initial pressure of ammonia is 100 Torr, the pressure of ammonia at t = 200 sec is     

Kinetics - Nuclear - Surface - Photochemistry - Question 6

The value of the rate constant for the gas phase reaction, 2NO2 + F2 → 2NO2F  is 38 dm3 mol–1 s–1 at 300 K. The order of the reaction is:    

Kinetics - Nuclear - Surface - Photochemistry - Question 7

A reaction follows second order rate law, 

Kinetics - Nuclear - Surface - Photochemistry - Question 8

The specific rate constant of decomposition of a compound is represented by​

The activation energy of decomposition for this compound at 300 K is:           

Kinetics - Nuclear - Surface - Photochemistry - Question 9

The fluorescence life time of a molecule in solution is 10 ns. If the fluorescence quantum yield is 0.1, the rate constant of fluorescence decay is:    

Kinetics - Nuclear - Surface - Photochemistry - Question 10

For the reaction shown below,

The value of k1 is 1 × 10–4 s–1. If the reaction starts from X, the ratio of the concentrations of Y and Z at any given time during the course of the reaction is found to be  The value of k2 is:

Kinetics - Nuclear - Surface - Photochemistry - Question 11

For an adsorbant-adsorbate system obeying the Langmuir adsorption isotherm, K = 0.48 bar-1 and P = 0.16 bar-1. At what pressure will 50% of the surface be covered?             

Detailed Solution for Kinetics - Nuclear - Surface - Photochemistry - Question 11

Given data
b = 0.48 bar-1
p = 0.16 bar-1
Substitute in the corresponding equation

Kinetics - Nuclear - Surface - Photochemistry - Question 12

The reaction, 2NO(g) + O2(g)  → 2NO2(O)  proceeds via the following steps:  
   
The rate of this reaction is equal to    

Kinetics - Nuclear - Surface - Photochemistry - Question 13

The decomposition of N2O5 according to following reaction is first order             
2 N2O5 → 4 NO2(g) + O2 (g). After 30 minutes from the start of decomposition in a closed vessel the total pressure developed is found to be 250 mm of Hg and on complete decomposition the total pressure is 500 mm of Hg. Calculate rate constant of the reaction.    

Detailed Solution for Kinetics - Nuclear - Surface - Photochemistry - Question 13

For the given reaction:
2N2​O5​(g)→4NO2​(g)+O2​(g)

  a                       0                0    -------- initial time (t=0)
 a−X                   2X           X/2  ------- after time (t=30 min)

   0                      2a          a/2     ----- after complete reaction
∵ No. of mole at any time ∝ pressure developed at that time

∴a ∝ P0​ at t=0 -----1
a+(3X/2)∝284.5 at t=30   ------ 2
(5a/2)∝584.5 at t=completely reacted ------ 3

From 1 and 3 equation, we get

∴a ∝ 233.8  ----------------------- 4

Substituting 4 in 2 equation we get

X∝33.8  --------------------------------- 5

Now,
∴ K = 2.303/t ​log a​/(a−X)
∴ K = 2.303/30 ​log233.8​/200

= 5.206 × 10-3min-1

Kinetics - Nuclear - Surface - Photochemistry - Question 14

1 g of 90Sr gets converted to 0.953 g after 2 yr. The half-life of 90Sr, and the amount of 90Sr remaining after 5 yr are:    

Kinetics - Nuclear - Surface - Photochemistry - Question 15

A graph is plotted between log10K and 1/T in which straight line BC has no slope and tan(ᴓ) = -1/2.303 and intercept on y-axis is 5. What is activation energy?    

Kinetics - Nuclear - Surface - Photochemistry - Question 16

In a reaction, → Product, rate is doubled when the concentration of is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (Aand B) are doubled, rate law for the reaction can be written as

Kinetics - Nuclear - Surface - Photochemistry - Question 17

 when the rate of the reaction is equal to the rate constant, the order of the reaction is

Detailed Solution for Kinetics - Nuclear - Surface - Photochemistry - Question 17

Ans: a 

Explanation: According to rate law the rate of reaction is given by the expression: 

rate = k[conc.] x 

Whereas, x is the overall order of the reaction. To make the rate of reaction equal to the rate constant (k), the order of the reaction (x) should be zero. 

Thus, the rate of reaction is equal to rate constant, the order of the reaction is '0'.

Kinetics - Nuclear - Surface - Photochemistry - Question 18

The inversion of cane sugar proceeds with half-life of 500 minutes 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:

Kinetics - Nuclear - Surface - Photochemistry - Question 19

The reaction, A(g) + 2B(g) → C(g) + D(g) is an elementary process. In an experiment, the initial partial pressure of A and B are PA = 0.60 and PB = 0.80 atm. When PC = 0.2 atm the rate of reaction relative to the initial rate is:    

Kinetics - Nuclear - Surface - Photochemistry - Question 20

For an enzyme catalyzed reaction, a Lineweaver-Burk plot gave the slope value = 40 s and intercept = 4 (mmoldm-3s-1)-1. If initial concentration of enzyme is 2.5× 10-9 moldm-3, catalytic efficiency is 10x. Calculate the value of ‘x’.           

Kinetics - Nuclear - Surface - Photochemistry - Question 21

At 273 K, N2 is adsorbed on mica surface. A plot of 1/V vs 1/P (V in m3 and P in Torr) gives a straight line with slope equal to 2.0 × 10–5 torr m–3 and an intercept Vm equal to 4 × 10–8 m3. The adsorption coefficient and number of molecules of N2 forming monolayer, resp., are:

Kinetics - Nuclear - Surface - Photochemistry - Question 22

Physiosorbed particles undergo desorption at 270C with an activation energy of 16.628 kJ/mol. Assuming first order process and a frequency factor of 1012 Hz, the average residence time (in sec) of particles on the surface is

Kinetics - Nuclear - Surface - Photochemistry - Question 23

A monoatomic gas, X absorbed on a surface, Langmuir, A plot of the fraction of surface coverage, q against the concentration of the gas [X], for very low concentration of the gas, is described by the equation: 

Kinetics - Nuclear - Surface - Photochemistry - Question 24

Which of the following transformations is not correct?

Kinetics - Nuclear - Surface - Photochemistry - Question 25

Which ones of the following reactions are artificial radioactive decay?

Kinetics - Nuclear - Surface - Photochemistry - Question 26

A, B, C and D elements form compounds AC, A2D and BD. If AC and A2D are radioactive and BD is not radioactive compound, find the following radioactive compounds-           
(I) A2                                       (II) A2C                          (III) C2D                        (IV) BC           

Kinetics - Nuclear - Surface - Photochemistry - Question 27

A substance absorbs 2.0 x 1016 quanta or radiations per second and 0.002 mole of it reacts in 1200 seconds. What is the quantum yield or the reaction (N = 6.02 x 1023)?           

Kinetics - Nuclear - Surface - Photochemistry - Question 28

Which regions of the light radiations of the visible ultraviolet lying between – wavelength are chiefly concerned in bringing about photochemical reactions?           

Kinetics - Nuclear - Surface - Photochemistry - Question 29

A closed vessel with rigid walls contain 1 mol of 92U238 and 1 mol of air at 298K. Considering complete decay of 92U238 to 82U206, the ratio of the final pressure to the initial pressure of the system at 298K is____?           

Kinetics - Nuclear - Surface - Photochemistry - Question 30

The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. To which group, element X belongs in periodic table? 29Cu63 + 1H1 → X + 2 1H1 + α + 6 0n1    

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