Chapter 10

A Complete Resource Book in Chemistry for JEE Main ยท 142 exercises

Problem 1

The unit of specific reaction rate constant for a firstorder (if the concentration is expressed in molarity) would be (a) \(\mathrm{s}^{-1}\) (b) mole \(\mathrm{s}^{-1}\) (c) mole \(\mathrm{L}^{-1} \mathrm{~s}^{-1}\) (d) mole \(\mathrm{L}^{-1}\)

4 step solution

Problem 4

\(3 \mathrm{~A} \longrightarrow \mathrm{B}+\mathrm{C}\) It would be a zero order reaction when (a) the rate of reaction is proportional to square of concentration of \(\mathrm{A}\) (b) the rate of reaction remains same at any concentration of \(\mathrm{A}\) (c) the rate remains unchanged at any concentration of \(\mathrm{B}\) and \(\mathrm{C}\) (d) the rate of reaction doubles if concentration of is increased to double

6 step solution

Problem 5

The rate of reaction depends upon (a) molar concentration (b) atomic mass (c) equivalent mass (d) none of these

5 step solution

Problem 6

For a first-order reaction, the half-life period is independent of (a) initial concentration (b) cube root of initial concentration (c) first power of final concentration (d) square root of final concentration

3 step solution

Problem 8

Activation energy of a chemical reaction can be determined by (a) evaluating rate constant at standard temperature (b) evaluating velocities of reaction at two different temperatures (c) evaluating rate constants at two different tempera tures (d) changing concentration of reactants

3 step solution

Problem 9

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(6.2 \times 10^{-4} \mathrm{~s}^{-1} .\) The half-life for this decomposition is (a) \(1177.7 \mathrm{~s}\) (b) \(1117.7\) (c) \(6.077 \mathrm{~s}\) (d) \(110.77\)

5 step solution

Problem 11

If the rate of the reaction is equal to the rate constant, the order of the reaction is (a) 3 (b) 0 (c) 1 (d) 2

6 step solution

Problem 12

Which of the following best explains the effects of a catalyst on the rate of a reversible reaction? (a) It decreases the rate of the reverse reaction (b) It increases the kinetic energy of the reacting molecules (c) It moves the equilibrium position to the right (d) It provides a new reaction path with a lower activation energy

2 step solution

Problem 14

For a chemical reaction \(\mathrm{A} \longrightarrow \mathrm{B}\), the rate of reaction doubles when the concentration of A is in creased four times. The order of reaction for \(\mathrm{A}\) is (a) zero (b) one (c) two (d) half

5 step solution

Problem 15

The unit of second-order reaction rate constant is (a) \(\mathrm{L}^{-1} \cdot \mathrm{mol}^{-1} \mathrm{ds}^{-1}\) (b) \(\mathrm{L}^{2} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}\) (c) \(\mathrm{L} \cdot \mathrm{mol}^{-1} \mathrm{~s}^{-1}\) (d) \(\mathrm{s}^{-1}\)

4 step solution

Problem 18

Among which of the following factor the specific reaction rate of a first- order reaction depends on (a) temperature (b) concentration of reactant (c) pressure (d) volume

3 step solution

Problem 19

The molecularity of a reaction is (a) always two (b) same as its order (c) different than the other (d) may be same or different as compared to order

5 step solution

Problem 21

Which of the following is incorrect about order of reaction? (a) it is calculated experimentally (b) it is sum of powers of concentration in rate law expression (c) the order of reaction cannot be fractional (d) there is not necessarily a connection between order and stoichiometry of a reaction.

5 step solution

Problem 23

If \(\mathrm{T}\) is the intensity of absorbed light and ' \(\mathrm{C}\) is the concentration of \(\mathrm{AB}\) for the photochemical process \(\mathrm{AB}+\mathrm{hv} \longrightarrow \mathrm{AB} \bullet\), the rate of formation of \(\mathrm{AB}\) ' is directly proportional to (a) \(\mathrm{C}\) (b) I (c) \(\mathrm{I}^{2}\) (d) C.I

4 step solution

Problem 26

The equation for the rate constant is \(k=\mathrm{Ae}^{-\mathrm{E} \nu \mathrm{RT}} . \mathrm{A}\) chemical reaction will proceed more rapidly if there is a decrease in (a) \(k\) (b) \(\mathrm{A}\) (c) \(\mathrm{E}\) (d) \(\mathrm{T}\)

5 step solution

Problem 27

The rate law has the form; rate \(=k[\mathrm{~A}][\mathrm{B}]^{3 / 2}\), can the reaction be an elementary process? (a) yes (b) no (c) may be yes or no (d) cannot be predicted

3 step solution

Problem 28

For a second-order reaction, \(2 \mathrm{~A} \longrightarrow\) Product, a straight line is obtained if we plot (a) concentration vs time (b) log (conc.) vs time (c) log (conc.) vs time \(^{1}\) (d) (concc')vs time"

5 step solution

Problem 29

For an endothermic reaction, where \(\Delta \mathrm{H}\) represents the enthalpy of the reaction in \(\mathrm{kJ} / \mathrm{mole}\), the minimum value for the energy of activation will be (a) less than \(\Delta \mathrm{H}\) (b) zero (c) more than \(\Delta \mathrm{H}\) (d) equal to \(\Delta \mathrm{H}\)

3 step solution

Problem 31

The rate constant of a reaction depends on (a) extent of reaction (b) time of reaction (c) temperature (d) initial concentration of the reactants

4 step solution

Problem 32

The function of catalyst in chemical reaction is to (a) increase the product (b) decrease the product (c) accelerate the rate of reaction (d) increase the reactants

3 step solution

Problem 34

According to the collision theory of reaction rates, an increase of the temperature at which the reaction oc curs will inturn increase the rate of the reaction. This caused due to (a) greater number of molecules are having the activation energy (threshold energy) (b) greater velocity of reaction molecules (c) greater number of collisions (d) none of these

6 step solution

Problem 35

For a chemical reaction which can never be a fractional number. (a) order (b) half-life (c) molecularity (d) rate constant

6 step solution

Problem 36

Which of the following is correct for a first order reaction? \(\left(k=\right.\) rate constant \(t_{1 / 2}=\) half-life \()\) (a) \(t_{1 / 2}=0.693 \times \mathrm{k}\) (b) \(\mathrm{k} \cdot \mathrm{t}_{1 / 2}=1 / 0.693\) (c) \(\mathrm{k} \cdot \mathrm{t}_{1 / 2}=0.693\) (d) \(6.93 \times k \times t_{1 / 2}=1\)

7 step solution

Problem 37

Which of the following relation is correct for a first order reaction? \((k=\) rate constant; \(\mathrm{r}=\) rate of reaction; \(\mathrm{C}=\) conc, of reactant) (a) \(k=\mathrm{r} \times \mathrm{C}^{2}\) (b) \(k=\mathrm{r} \mathrm{x}\) (c) \(k=\mathrm{C} / r\) (d) \(k=\mathrm{r} / \mathrm{C}\)

3 step solution

Problem 38

If the rate law of a reaction \(\mathrm{nA} \longrightarrow \mathrm{B}\) is expressed as Rate \(=-\frac{1}{n} \frac{d[A]}{d t}=+\frac{d[B]}{d t}=k[A]^{x}\) The unit of the rate constant will be (a) \(\mathrm{mol}^{\mathrm{x}} / \mathrm{L}^{\mathrm{x}} / \mathrm{s}\) (b) \(\mathrm{L}^{\mathrm{x}} / \mathrm{mol}^{\mathrm{x}} \mathrm{s}\) (c) \(\operatorname{mol}^{(1-x)} / L^{(x-1)} \cdot S^{-1}\) (d) \(m o l^{(x-1)} / L^{(1-x)} \cdot S^{-1}\)

4 step solution

Problem 40

Rate constant of a reaction \((k)\) is \(175 \mathrm{~L}^{2} \mathrm{~mol}^{-2} \mathrm{sec}^{-1}\). What is the order of reaction? (a) first (b) second (c) third (d) zero

3 step solution

Problem 41

A catalyst is a substance which (a) supplies energy to the reaction (b) increases the equilibrium concentration of the product (c) changes the equilibrium constant of the reaction (d) shortens the time to each equilibrium

5 step solution

Problem 42

The rate of reaction was found to be equal to its rate constant at any concentration of the reactant. The order of the reaction is (a) zero-order (b) first-order (c) second-order (d) third-order

3 step solution

Problem 45

A catalyst increases rate of reaction by (a) decreasing enthalpy (b) decreasing activation energy (c) decreasing internal energy (d) increasing activation energy

4 step solution

Problem 50

Which of the following statement is correct? (a) A plot of \(\log k_{0}\) vs \(1 / \mathrm{t}\) is linear (b) A plot of log [X] vs time is linear for a first-order reaction, \(\mathrm{X} \longrightarrow \mathrm{P}\) (c) A plot of log \(\mathrm{P}\) vs \(1 / \mathrm{t}\) is linear at constant volume (d) A plot of P vs \(1 / \mathrm{V}\) is linear at constant pressure

3 step solution

Problem 51

For a zero-order reaction, the plot of concentration vs time is linear with (a) +ve slope and zero intercept (b) -ve slope and zero intercept (c) +ve slope and non-zero intercept (d) -ve slope and non-zero intercept

5 step solution

Problem 54

A reaction rate is given by \(k=1.5 \times 10^{15} \exp (-25000 / \mathrm{RT}) \mathrm{s}^{-1}\) it means that (a) half-life period of the reaction will be smaller at high temperature (b) log vs \(\mathrm{T}\) will give a straight line (c) half-life of the reaction will be smaller at lower temperature (d) log vs \(1 / \mathrm{T}\) will give a straight line having a slope of \(-25000\)

5 step solution

Problem 57

For the reaction, \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+1 / 2 \mathrm{O}_{2} .\) Given values are \(-\frac{\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{dt}}=k_{1}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) \(\frac{\mathrm{d}\left[\mathrm{NO}_{2}\right]}{\mathrm{dt}}=k_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) \(\frac{\mathrm{d}\left[\mathrm{O}_{2}\right]}{\mathrm{dt}}=k_{3}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) Now the relation in between \(k_{1} k_{2}\) and \(k_{3}\) is (a) \(k_{1}=k_{2}=k_{3}\) (b) \(3 k_{1}=k_{2}=2 k_{3}\) (c) \(2 k_{1}=4 k_{2}=k_{3}\) (d) \(2 k_{1}=k_{2}=4 k_{3}\)

4 step solution

Problem 58

\(k\) for a zero-order reaction is \(2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). If the concentration of the reactant after \(25 \mathrm{~s}\) is \(0.5 \mathrm{M}\), the initial concentration must have been (a) \(0.5 \mathrm{M}\) (b) \(1.25 \mathrm{M}\) (c) \(12.5 \mathrm{M}\) (d) \(1.0 \mathrm{M}\)

4 step solution

Problem 60

For a first-order reaction, \(\mathrm{t}_{0.75}\) is \(138.6\) seconds. Its specific reaction rate constant (in sec \(^{-1}\) ) is (a) \(10^{-2}\) (b) \(10^{-4}\) (c) \(10^{-5}\) (d) \(10^{-6}\)

4 step solution

Problem 61

The rate of a first-order reaction is \(1.5 \times 10^{2} \mathrm{~mol} \mathrm{~L}^{-1}\) \(\min ^{-1}\) at \(0.5 \mathrm{M}\) concentration of the reactant. The halflife of the reaction is (a) \(0.383\) hour (b) \(23.1 \mathrm{~min}\) (c) \(8.73 \mathrm{~min}\) (d) \(7.53 \mathrm{~min}\)

6 step solution

Problem 62

For the first-order reaction half-life is \(14 \mathrm{~s}\). The time required for the initial concentration to reduce to \(1 / 8\) th of its value is (a) \(21 \mathrm{~s}\) (b) \(32 \mathrm{~s}\) (c) \(42 \mathrm{~s}\) (d) \(14^{2} \mathrm{~s}\)

5 step solution

Problem 65

Under the same reactions conditions, initial concentration of \(1.386 \mathrm{~mol} \mathrm{dm}^{-3}\) of a substance becomes half in \(40 \mathrm{sec}\) and 20 sec through \(1^{\text {st }}\) order and zero order kinetics respectively. Ratio \(\left[\frac{\mathrm{K}_{1}}{\mathrm{~K}_{0}}\right]\) of the rate constants for first order \(\left(\mathrm{k}_{1}\right)\) and zero order \(\left(\mathrm{k}_{0}\right)\) of the reactions is (a) \(0.5 \mathrm{~mol}^{-1} \mathrm{dm}^{-3}\) (b) \(1 \mathrm{~mol} \mathrm{dm}^{-3}\) (c) \(1.5 \mathrm{~mol} \mathrm{dm}^{-3}\) (d) \(2 \mathrm{~mol}^{-1} \mathrm{dm}^{3}\)

5 step solution

Problem 66

The half-life of a substance in a first-order reaction is 15 minutes. The rate constant is (a) \(2.46 \times 10^{2} \mathrm{~min}^{-1}\) (b) \(4.62 \times 10^{-2} \mathrm{~min}^{-1}\) (c) \(3 \times 10^{-5} \mathrm{~min}^{-1}\) (d) \(3 \times 10^{-9} \mathrm{~min}^{-1}\)

5 step solution

Problem 69

The rate constant of first-order reaction is \(3 \times 10^{-6}\) per second. The initial concentration is \(0.10 \mathrm{M}\). The initial rate is (a) \(3 \times 10^{-7} \mathrm{Ms}^{-1}\) (b) \(3 \times 10^{-4} \mathrm{Ms}^{-1}\) (c) \(3 \times 10^{-5} \mathrm{Ms}^{-1}\) (d) \(3 \times 10^{-6} \mathrm{Ms}^{-1}\)

4 step solution

Problem 70

Consider the following statements: (1) rate of a process is directly proportional to its free energy change (2) the order of an elementary reaction step can be determined by examining the stoichiometry (3) the first-order reaction describe exponential time course. Of the statements (a) 1 and 2 are correct (b) 1 and 3 are correct (c) 2 and 3 are correct (d) 1,2 and 3 are correct

4 step solution

Problem 71

In a first-order reaction \(\mathrm{A} \longrightarrow \mathrm{P}\), the ratio of \(\mathrm{a} /(\mathrm{a}-\mathrm{x})\) was found to be 8 after 60 minutes. If the concentration is \(0.1 \mathrm{M}\) then the rate of reaction in moles of A reacted per minutes is (a) \(2.226 \times 10^{-3} \mathrm{~mol}\) litre \(^{-1} \mathrm{~min}^{-1}\) (b) \(3.466 \times 10^{-3} \mathrm{~mol}\) litre \(^{-1} \mathrm{~min}^{-1}\) (c) \(4.455 \times 10^{-3} \mathrm{~mol}\) litre \(^{-1} \mathrm{~min}^{-1}\) (d) \(5.532 \times 10^{-3} \mathrm{~mol}\) litre \(^{-1} \mathrm{~min}^{-1}\)

6 step solution

Problem 73

In the following reaction, how is the rate of appear ance of the underlined product related to the rate of disappearance of the underlined reactant? \(\mathrm{BrO}_{3}^{-}(\mathrm{aq})+5 \mathrm{Br}^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow 3 \mathrm{Br}_{2}(1)\) \(+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (a) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{5}{3} \frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}}\) (b) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\) (c) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=\frac{\mathrm{d}[\mathrm{Br}-]}{\mathrm{dt}}\) (d) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{3}{5} \frac{\mathrm{d}\left[\mathrm{Br}^{-}\right]}{\mathrm{dt}}\)

4 step solution

Problem 75

The reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) follows first order kinetics. The time taken for \(0.8\) mole of \(\mathrm{A}\) to produce \(0.6\) mole of is 1 hour. What is the time taken for conversion of \(0.9\) mole of A to produce \(0.675\) mole of \(\mathrm{B} ?\) (a) 2 hour (b) 1 hour (c) \(0.5\) hour (d) \(0.25\) hour

4 step solution

Problem 79

A substance reacts according to first-order kinetics. The rate constant for the reaction is \(1 \times\) \(10^{-2} \mathrm{sec}^{1} .\) Its initial concentration is IM. Its initial rate is (a) \(2 \times 10^{2} \mathrm{Ms}^{-1}\) (b) \(1 \times 10^{2} \mathrm{Ms}^{-1}\) (c) \(1 \times 10^{-2} \mathrm{Ms}^{-1}\) (d) \(2 \times 10^{-2} \mathrm{Ms}^{-1}\)

5 step solution

Problem 80

What will be the initial rate of a reaction if its con stant is \(10^{-3} \mathrm{~min}^{-1}\) and the concentration of reactant is \(0.2 \mathrm{~mol} \mathrm{dm}^{3} ?\) (a) \(0.02 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~min}^{-1}\) (b) \(0.002 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~min}^{-1}\) (c) \(0.0002 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~min}^{-1}\) (d) \(2 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~min}^{-1}\)

5 step solution

Problem 81

The rate of a certain hypothetical reaction \(\mathrm{A}+\mathrm{B}+\mathrm{C} \longrightarrow\) Products is given by \(\mathrm{r}=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=k[\mathrm{~A}]^{1 / 2}[\mathrm{~B}]^{1 / 3}[\mathrm{C}]^{1 / 4}\) The order of the reaction is (a) \(13 / 12\) (b) \(13 / 14\) (c) \(12 / 13\) (d) \(13 / 11\)

6 step solution

Problem 82

The decay constant of \({ }_{6} \mathrm{C}^{14}\) is \(2.31 \times 10^{-4}\) year \(^{-1} .\) Its half life is (a) \(2 \times 10^{3}\) yrs (b) \(2.5 \times 10^{3} \mathrm{yrs}\) (c) \(3 \times 10^{3} \mathrm{yrs}\) (d) \(3.5 \times 10^{3}\) yrs

4 step solution

Problem 83

A first-order reaction is \(50 \%\) completed in 30 minutes at \(27^{\circ} \mathrm{C}\). Its rate constant is (a) \(2.31 \times 10^{-2} \mathrm{~min}^{-1}\) (b) \(3.21 \times 10^{-2} \mathrm{~min}^{-1}\) (c) \(4.75 \times 10^{-2} \mathrm{~min}^{1}\) (d) \(1.33 \times 10^{-3} \mathrm{~min}^{-1}\)

5 step solution

Problem 84

The rate constant of a first-order reaction is \(6 \times 10^{-3}\) \(\mathrm{s}^{-1} .\) If the initial concentration is \(0.10 \mathrm{M}\), the initial rate of reaction is (a) \(6 \times 10^{-3} \mathrm{Ms}^{-1}\) (b) \(6 \times 10^{-1} \mathrm{Ms}^{-1}\) (c) \(6 \times 10^{-6} \mathrm{Ms}^{-1}\) (d) \(6 \times 10^{-8} \mathrm{Ms}^{-1}\)

6 step solution

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