Chapter 10
A Complete Resource Book in Chemistry for JEE Main · 159 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 2
The temperature coefficient of most of the reactions lies between (a) 1 and 3 (b) 2 and 3 (c) 1 and 2 (d) 2 and 4
3 step solution
Problem 3
The activation energy for a simple chemical reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) is \(\mathrm{E}_{a}\) in forward direction. The activation energy for reverse reaction (a) can be less than or more than \(\mathrm{E}_{-}\) (b) is always double of \(E_{\text {a }}\) (c) is negative of \(\mathrm{E}_{a}\) (d) is always less than \(\mathrm{E}_{\mathrm{a}}\)
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
5 step solution
Problem 5
The rate of reaction depends upon (a) molar concentration (b) atomic mass (c) equivalent mass (d) none of these
7 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
4 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
4 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) \(110.77\)
3 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
4 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
6 step solution
Problem 14
For a chemical reaction \(\mathrm{A} \longrightarrow \mathrm{B}\), the rate of reaction doubles when the concentration of \(\mathrm{A}\) is in creased four times. The order of reaction for \(\mathrm{A}\) is (a) zero (b) one (c) two (d) half
7 step solution
Problem 15
The unit of second-order reaction rate constant is (a) \(\mathrm{L}^{-1}, \mathrm{~mol}^{-1} \mathrm{~d} \mathrm{~s}^{-1}\) (b) \(\mathrm{L}^{2} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}\) (c) \(\mathrm{L} \cdot \mathrm{mol}^{-1} \mathrm{~s}^{-1}\) (d) \(\mathrm{s}^{-1}\)
5 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
4 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
4 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.
6 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 \(\mathrm{AB}^{\prime}\) is directly proportional to (a) \(\vec{C}\) (b) I (c) \(\mathrm{I}^{2}\) (d) C.I
4 step solution
Problem 24
For a first-order reaction, (a) The degree of dissociation is equal to \(\left(1-\mathrm{e}^{\mathrm{kt}}\right)\) (b) The pre-exponential factor in the Arrhenius equation has the dimensions of time \(t^{-1}\). (c) The time taken for the completion of \(75 \%\) reaction is thrice the t \(1 / 2\) of the reaction. (d) both (a) and (b)
5 step solution
Problem 25
The rate law for the reaction \(\mathrm{RCl}+\mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{ROH}+\mathrm{NaCl}\) is given by Rate \(=k[\mathrm{RCl}]\). The rate of the reaction will be (a) doubled on doubling the concentration of sodium hydroxide (b) halved on reducing the concentration of alkyl halide to one half (c) decreased on increasing the temperature of reaction (d) unaffected by increasing the temperature of the reaction.
4 step solution
Problem 26
The equation for the rate constant is \(k=\) Ae \(^{-E a R T} . A\) 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
4 step solution
Problem 29
For an endothermic reaction, where \(\Delta \mathrm{H}\) represents the enthalpy of the reaction in \(\mathrm{kJ} /\) 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
4 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
4 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
5 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) \(t_{1 / 2}=0.693 \times k\) (b) \(\mathrm{k} \cdot \mathrm{t}_{1 / 2}=1 / 0.693\) (c) \(\mathrm{k} \cdot \mathrm{t}_{1 / 2}=0.693\) (d) \(6.93 \times k \times t_{1 / 2}=1\)
3 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) \(k=r x\) (c) \(k=\mathrm{C} / r\) (d) \(k=\mathrm{r} / \mathrm{C}\)
5 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) \(\mathrm{L}^{\mathrm{x}} / \mathrm{mol}^{\mathrm{t}} \mathrm{s}\) (c) \(m o l^{(1-x)} / L^{(x-1)} \cdot S^{-1}\) (d) \(\operatorname{mol}^{(x-1)} / L^{(1-x)} \cdot S^{-1}\)
5 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
4 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
5 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
3 step solution
Problem 47
The rate constant is given by the equation \(\mathrm{K}=\mathrm{P} \mathrm{Ze}^{-\mathrm{E} / \mathrm{kT}}\). Which factor should register a decrease for the reaction to proceed more rapidly (a) \(\mathrm{T}\) (b) \(\mathrm{Z}\) (c) \(\mathrm{E}\) (d) \(\mathrm{P}\)
4 step solution
Problem 50
Which of the following statement is correct? (a) A plot of \(\log k_{\mathrm{p}}\) vs \(1 / \mathrm{t}\) is linear (b) A plot of \(\log [\mathrm{X}]\) vs time is linear for a first-order reaction, \(\mathrm{X} \longrightarrow \mathrm{P}\) (c) A plot of \(\log \mathrm{P}_{\mathrm{vs}} 1 / \mathrm{t}\) is linear at constant volume (d) A plot of \(\mathrm{P}\) vs \(1 / \mathrm{V}\) is linear at constant pressure
4 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) tve 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 \(-25000\)
4 step solution
Problem 56
A substance undergoes first-order decomposition, it follows two parallel reactions \(k_{1}=1.26 \times 10^{-4} \mathrm{~s}^{-1}\) and \(k_{2}=3.8 \times 10^{-5} \mathrm{~s}^{-1}\) The percentage distribution of \(\mathrm{Y}\) and \(\mathrm{Z}\) are (a) \(80 \% \mathrm{Y}\) and \(20 \% \mathrm{Z}\) (b) \(72.83 \% \mathrm{Y}\) and \(32.71 \% \mathrm{Z}\) (c) \(76.83 \% \mathrm{Y}\) and \(23.17 \% \mathrm{Z}\) (d) \(62.4 \% \mathrm{Y}\) and \(90.5 \% \mathrm{Z}\)
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) \(2 k_{1}=k_{2}=4 k_{3}\)
5 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) \(1.0 \mathrm{M}\)
5 step solution
Problem 59
\(75 \%\) of a first-order reaction was completed in 32 min. When was \(50 \%\) of the reaction completed? (a) \(24 \mathrm{~min}\) (b) \(16 \mathrm{~min}\) (c) \(8 \mathrm{~min}\) (d) \(64 \mathrm{~min}\)
4 step solution
Problem 60
For a first-order reaction, \(t_{0.75}\) 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) \(7.53 \mathrm{~min}\)
5 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) \(14^{2} \mathrm{~s}\)
4 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) \(2 \mathrm{~mol}^{-1} \mathrm{dm}^{3}\)
6 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) \(3 \times 10^{-9} \mathrm{~min}^{-1}\)
4 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) \(4.455 \times 10^{-3}\) mol litre \(^{-1} \min ^{-1}\) (d) \(5.532 \times 10^{-3}\) mol litre \(^{-1} \min ^{-1}\)
7 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) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{5}{3} \frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\) (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) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{3}{5} \frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\)
5 step solution
Problem 74
The experimental data for the reaction \(2 \mathrm{~A}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB}\) is $$ \begin{array}{llll} \hline \text { Exp. } & {[\mathrm{A}]} & {\left[\mathrm{B}_{2}\right]} & \text { Rate }\left(\mathrm{Ms}^{-1}\right) \\ \hline 1 . & 0.50 \mathrm{M} & 0.50 \mathrm{M} & 1.6 \times 10^{-4} \\ 2 . & 0.50 \mathrm{M} & 1.00 \mathrm{M} & 3.2 \times 10^{-4} \\ 3 . & 1.00 \mathrm{M} & 1.00 \mathrm{M} & 3.2 \times 10^{-4} \\ \hline \end{array} $$ the rate equation for the above data is (a) rate \(=\mathrm{k}\left[\mathrm{B}_{2}\right]\) (b) rate \(=k\left[\mathrm{~B}_{2}\right]^{2}\) (c) rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]^{2}\) (d) rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}]\)
5 step solution