Chapter 10
A Complete Resource Book in Chemistry for JEE Main · 142 exercises
Problem 85
A graph plotted between concentration of reactant, consumed at any time \((x)\) and time ' \(\mathrm{t}\) ' is found to be a straight line passing through the origin. The reaction is of (a) first-order (b) zero-order (c) third-order (d) second-order
3 step solution
Problem 87
The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at \(25^{\circ} \mathrm{C}\) are \(3.0 \times 10^{-4} \mathrm{~s}^{-1}, 104.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(6 \times 10^{14} \mathrm{~s}^{-1}\) respectively. The value of the rate constant as \(\mathrm{T} \longrightarrow \infty\) is (a) \(2.0 \times 10^{18} \mathrm{~s}^{-1}\) (b) \(6.0 \times 10^{14} \mathrm{~s}^{-1}\) (c) infinity (d) \(3.6 \times 10^{30} \mathrm{~s}^{-1}\)
4 step solution
Problem 88
The experimental rate law for a reaction \(2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow\) Product, is \(\mathrm{V} \alpha \mathrm{C}_{\mathrm{A}} \mathrm{C}_{\mathrm{B}}^{-2} .\) If the concentration of both \(\mathrm{A}\) and are doubled the rate of reaction increases by a factor of (a) \(\sqrt{2}\) (b) 2 (c) \(2 . \sqrt{2}\) (d) 4
6 step solution
Problem 89
The rate equation for a chemical reaction is Rate of reaction \(=[\mathrm{X}][\mathrm{Y}]\) Consider the following statements in this regard (1) The order of reaction is one (2) The molecularity of reaction is two (3) The rate constant depends upon temperature Of these statements (a) 1 and 3 are correct (b) 1 and 2 are correct (c) 2 and 3 are correct (d) 1,2 and 3 are correct
5 step solution
Problem 90
In a zero-order reaction, \(47.5 \%\) of the reactant remains at the end of \(2.5\) hours. The amount of reactant consumed in one hour is (a) \(11.0 \%\) (b) \(33.0 \%\) (c) \(42.0 \%\) (d) \(21.0 \%\)
4 step solution
Problem 91
During the decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) to give oxygen, \(48 \mathrm{~g} \mathrm{O}_{2}\) is formed per minute at a certain point of time. The rate of formation of water at this point is (a) \(0.75 \mathrm{~mol} \min ^{1}\) (b) \(1.5 \mathrm{~mol} \mathrm{~min}^{-1}\) (c) \(2.25 \mathrm{~mol} \mathrm{~min}^{-1}\) (d) \(3.0 \mathrm{~mol} \mathrm{~min}^{-1}\)
3 step solution
Problem 92
The rate constant of a first-order reaction, \(\mathrm{A} \longrightarrow\) products, is \(60 \times 10^{-4} \mathrm{~s}^{-1} .\) Its rate at \([\mathrm{A}]=\) \(0.01 \mathrm{~mol} \mathrm{~L}^{-1}\) would be (a) \(60 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) (b) \(36 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) (c) \(60 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) (d) \(36 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\)
4 step solution
Problem 93
In a first-order reaction the concentration of reactant decreases from \(800 \mathrm{~mol} / \mathrm{dm}^{3}\) to \(50 \mathrm{~mol} / \mathrm{dm}^{3}\) in \(2 \times 10^{4}\) sec. The rate constant of reaction in \(\mathrm{sec}^{-1}\) is (a) \(2 \times 10^{4}\) (b) \(3.45 \times 10^{-5}\) (c) \(1.386 \times 10^{-4}\) (d) \(2 \times 10^{-4}\)
7 step solution
Problem 94
The activation energies of two reactions with rate constants \(k\), and \(k_{2}\), are \(E_{a 1}\) and \(E_{a 2}\) respectively. If \(E_{a 1}\) \(<\mathrm{E}_{\mathrm{a} 2}\), when the temperature is increased from \(\mathrm{T}\), to \(\mathrm{T}_{2}\) the rate constants are \(k_{1}^{\prime}\) and \(k_{2}^{\prime} .\) Which one of the following statements is correct? (a) \(\mathbf{k}_{1}^{\prime}=\mathrm{k}_{2}\) (b) \(\frac{\mathrm{k}_{1}}{\mathrm{k}_{1}}<\frac{\mathrm{k}_{2}}{\mathrm{k}_{2}}\) (c) \(\frac{\mathrm{k}_{1}}{\mathrm{k}_{1}}>\frac{\mathrm{k}_{2}}{\mathrm{k}_{2}}\) (d) \(\frac{k_{1}}{k_{1}^{\prime}}=\frac{k_{2}}{k_{2}}\)
4 step solution
Problem 95
In a chemical reaction two reactants take part. The rate of reaction is directly proportional to the concentration of one of them and inversely proportional to the concentration of the other. The order of reaction is (a) 0 (b) 1 (c) 2 (d) 4
4 step solution
Problem 96
Which of the following statements is correct? (1) order of a reaction can be known from experimental results and not from the stoichiometry of a reaction. (2) molecularity a reaction refers to (i) each of the elementary steps in (an overall mechanism of) a complex reaction or (ii) a single step reaction. (3) overall molecularity of a reaction may be determined in a manner similar to overall order of reaction. (4) overall order of a reaction \(\mathrm{A}^{\mathrm{m}}+\mathrm{B}^{n} \longrightarrow \mathrm{AB}_{\mathrm{x}}\) is \(\mathrm{m}+\mathrm{n} .\) Select the correct answer using the following codes: (a) 2 and 3 (b) 1,3 and 4 (c) 2,3 and 4 (d) 1,2 and 3
5 step solution
Problem 97
The half-life of a chemical reaction at a particular concentration is \(50 \mathrm{~min}\), when the concentration of reactants is doubled, the half-life becomes \(100 \mathrm{~min}\). Find the order. (a) zero (b) first (c) second (d) third
3 step solution
Problem 98
The rate constant of first-order reaction is \(10^{-2} \mathrm{~min}^{-1}\). The half-life period of reaction is (a) \(693 \mathrm{~min}\) (b) \(69.3 \mathrm{~min}\) (c) \(6.93 \mathrm{~min}\) (d) \(0.693 \mathrm{~min}\)
4 step solution
Problem 100
If the half life period of a radioactive isotope is \(10 \mathrm{~s}\), then its average life will be (a) \(14.4 \mathrm{~s}\) (b) \(1.44 \mathrm{~s}\) (c) \(0.144 \mathrm{~s}\) (d) \(2.44 \mathrm{~s}\)
3 step solution
Problem 101
In the first-order reaction, half of the reaction is com pleted in 100 seconds. The time for \(99 \%\) reaction to occur will be (a) \(664.64 \mathrm{~s}\) (b) \(646.6 \mathrm{~s}\) (c) \(660.9 \mathrm{~s}\) (d) \(654.5 \mathrm{~s}\)
6 step solution
Problem 102
For a certain reaction, the activation energy is zero. What is the value of rate constant at \(300 \mathrm{~K}\), if \(\mathrm{K}=1.6\) \(\times 10^{6} \mathrm{~s}^{-1}\) at \(280 \mathrm{~K} ?\) (a) \(1.6 \times 10^{6} \mathrm{~s}^{-1}\) (b) zero (c) \(4.8 \times 10^{8} \mathrm{~s}^{-1}\) (d) \(3.2 \times 10^{12} \mathrm{~s}^{-1}\)
3 step solution
Problem 103
When the temperature of a reaction increases from \(27^{\circ} \mathrm{C}\) to \(37^{\circ} \mathrm{C}\), the rate increases by \(2.5\) times, the activation energy in the temperature range is (a) \(70.8 \mathrm{~kJ}\) (b) \(7.08 \mathrm{~kJ}\) (c) \(35.8 \mathrm{~kJ}\) (d) \(14.85 \mathrm{~kJ}\)
6 step solution
Problem 106
Consider the chemical reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) The rate of this reaction can be expressed in terms of time derivatives of concentration of \(\mathrm{N}_{2}(\mathrm{~g}), \mathrm{H}_{2}(\mathrm{~g})\) or \(\mathrm{NH}_{3}(\mathrm{~g})\). Identify the correct relationship amongst the rate expressions. (a) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (b) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (c) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (d) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\)
4 step solution
Problem 108
Consider the following reaction $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ The rate of this reaction in terms of \(\mathrm{N}_{2}\) at \(\mathrm{T}\) is \(-\mathrm{d}\left[\mathrm{N}_{2}\right] /\) \(\mathrm{dt}=0.02 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) What is the value of \(\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}\) (in units of \(\left.\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) at the same temperature? (a) \(0.02\) (b) 50 (c) \(0.06\) (d) \(0.04\)
3 step solution
Problem 110
Which one of the following equations is correct for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) ?\) (a) \(3 \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}=2 \frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\) (b) \(2 \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}=-3 \frac{\mathrm{d}\left[\mathrm{H}_{3}\right]}{\mathrm{dt}}\) (c) \(2 \frac{\mathrm{d}\left[\mathrm{NH}_{1}\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\) (d) \(3 \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}=-2 \frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\)
5 step solution
Problem 112
Observe the following reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow \mathrm{C}\) the rate of formation of is \(2.2 \times 10^{-3}\) mol. \(L^{-1}\). What is the value of \(-\mathrm{d}[\mathrm{A}] / \mathrm{dt}\left(\right.\) in \(\left.\mathrm{mol} \mathrm{L}^{-1} \mathrm{~min}^{-1}\right)\) ? (a) \(2.2 \times 10^{-3}\) (b) \(1.1 \times 10^{-3}\) (c) \(4.4 \times 10^{-3}\) (d) \(5.5 \times 10^{-3}\)
4 step solution
Problem 113
\(3 \mathrm{~A} \longrightarrow 2 \mathrm{~B}\), rate of reaction \(+\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\) is equal to (a) \(-\frac{3}{2} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) (b) \(-\frac{2}{3} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) (c) \(-\frac{1}{3} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) (d) \(+2 \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\)
5 step solution
Problem 114
For a reaction \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D}\) if the concentration of \(\mathrm{A}\) is doubled without altering the concentration of \(\mathrm{B}\), the rate gets doubled. If the concentration of is increased by nine times without altering the concentration of \(\mathrm{A}\), the rate gets tripled. The order of the reaction is (a) 2 (b) 1 (c) \(3 / 2\) (d) \(4 / 3\)
4 step solution
Problem 115
For the reaction \(\mathrm{A} \longrightarrow\) Products, it is found that the rate of reaction increases by a factor of \(6.25\), when the concentration of \(\mathrm{A}\) is increased by a factor of \(2.5 .\) The order of reaction with respect to \(\mathrm{A}\) is (a) \(0.5\) (b) 1 (c) 2 (d) 3
5 step solution
Problem 117
The basic theory of Arrhenius equation is that (1) activation energy and pre-exponential factors are always temperature independent (2) the number of effective collisions is proportional to the number of molecule above a certain thresh old energy. (3) as the temperature increases, the number of molecules with energies exceeding the threshold energy increases. (4) the rate constant in a function of temperature (a) 2,3 and 4 (b) 1,2 and 3 (c) 2 and 3 (d) 1 and 3
7 step solution
Problem 119
For a \(1^{\text {st }}\) order reaction \(\mathrm{A} \longrightarrow \mathrm{P}\), the temperature (T) dependent rate constant (K) was found to follow the equation \(\log \mathrm{k}=-(2000) \frac{1}{\mathrm{~T}}+6\) The pre- exponential factor A and activation energy Ea are respectively? (a) \(1 \times 10^{\circ} \mathrm{S}^{-1}\) and \(9.2 \mathrm{~kJ} / \mathrm{M}\) (b) \(1 \times 10^{6} \mathrm{~S}^{-1}\) and \(38.3 \mathrm{~kJ} / \mathrm{M}\) (c) \(1 \times 10^{6} \mathrm{~S}^{-1}\) and \(16.6 \mathrm{~kJ} / \mathrm{M}\) (d) \(6 \mathrm{~S}^{-1}\) and \(16.6 \mathrm{~kJ} / \mathrm{M}\)
5 step solution
Problem 120
If for a certain first-order reaction, the initial rate is \(0.65 \% \mathrm{~min}^{-1}\), the half-life in one hour is (a) \(17.76\) (b) \(27.14\) (c) \(1.776\) (d) \(11.66\)
4 step solution
Problem 121
The reaction \(\mathrm{X} \longrightarrow\) Product follows first-order kinetics, in 40 minutes, the concentration of \(X\) changes from \(0.1 \mathrm{M}\) to \(0.025 \mathrm{M}\), then the rate of reaction when concentration of \(\mathrm{X}\) is \(0.01 \mathrm{M}\) is? (a) \(3.47 \times 10^{-5} \mathrm{M} / \mathrm{min}\) (b) \(1.73 \times 10^{-4} \mathrm{M} / \mathrm{min}\) (c) \(1.73 \times 10^{-5} \mathrm{M} / \mathrm{min}\) (d) \(3.47 \times 10^{-4} \mathrm{M} / \mathrm{min}\)
5 step solution
Problem 122
The following data are obtained from the decomposition of a gaseous compound Initial pressure in arm \(\begin{array}{lll}1.6 & 0.8 & 0.4\end{array}\) Time for \(50 \%\) reaction in \(\min 80\) \(113 \quad 160\) The order of the reaction is (a) \(0.5\) (b) \(1.0\) (c) \(1.5\) (d) \(2.0\)
4 step solution
Problem 126
The rate constant of a reaction is \(1.5 \times 10^{7} \mathrm{~s}^{-1}\) at \(50^{\circ} \mathrm{C}\) and \(4.5 \times 10^{7} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). What is the value of activation energy? (a) \(2.2 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\) (b) \(2300 \mathrm{~J} \mathrm{~mol}^{-1}\) (c) \(2.2 \times 10^{4} \mathrm{~J} \mathrm{~mol}^{-4}\) (d) \(220 \mathrm{~J} \mathrm{~mol}^{-1}\)
6 step solution
Problem 127
The activation energies of two reactions are \(18 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) and \(4.0 \mathrm{~kJ} \mathrm{~mol}^{-4}\) respectively. Assuming the pre-exponential factor to be the same for both reactions, the ratio of their rate constants at \(27^{\circ} \mathrm{C}\) is (a) \(3.656 \times 10^{-3}\) (b) \(3.624 \times 10^{-6}\) (c) \(36.52 \times 10^{-8}\) (d) \(4.656 \times 10^{-4}\)
6 step solution
Problem 129
At \(380^{\circ} \mathrm{C}\), half-life period for the first-order decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(360 \mathrm{~min}\). The energy of activation of the reaction is \(200 \mathrm{~kJ} \mathrm{~mol}^{1}\). Calculate the time required for \(75 \%\) decomposition at \(450^{\circ} \mathrm{C}\) if half-life for decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(10.17 \mathrm{~min}\) at \(450^{\circ} \mathrm{C}\). (a) \(20.4 \mathrm{~min}\) (b) \(408 \mathrm{~min}\) (c) \(10.2 \mathrm{~min}\) (d) none of these
5 step solution
Problem 130
A gaseous compound decomposes on heating as per the following equation: \(\mathrm{A}(\mathrm{g}) \longrightarrow B(\mathrm{~g})+2 \mathrm{C}(\mathrm{g}) .\) After 5 minutes and 20 seconds, the pressure increases by \(96 \mathrm{~mm} \mathrm{Hg}\). If the rate constant for this first order reaction is \(5.2 \times 10^{-4} \mathrm{~s}^{-1}\), the initial pressure of \(\mathrm{A}\) is (a) \(226 \mathrm{~mm} \mathrm{Hg}\) (b) \(37.6 \mathrm{~mm} \mathrm{Hg}\) (c) \(616 \mathrm{~mm} \mathrm{Hg}\) (d) \(313 \mathrm{~mm} \mathrm{Hg}\)
7 step solution
Problem 132
For the reaction, \(\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{H}_{2} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}, \Delta \mathrm{E}^{0}=-30\) kcal. If the reaction is reversible and if the activation energy for the forward reaction is \(28.0 \mathrm{kcal}\) and its drops to \(10.5 \mathrm{kcal}\) in the presence of a catalyst, the activation energies for the uncatalyzed and catalysed reverse reaction are respectively (in kcal) (a) \(58,40.5\) (b) \(-58,-40.5\) (c) \(40.5,58\) (d) \(58.0,-58.0\)
4 step solution
Problem 133
For the thermal rearrangement of vinyl allyl ether (g) to allyl acetaldehyde (g) at \(175^{\circ} \mathrm{C},=5 \times 10^{11}\) exp \((-128000 / \mathrm{RT})\), where is in \(\mathrm{s}^{-1}\) and activation energy in \(\mathrm{J} \mathrm{mol}^{-1}\). The enthalpy of activation in \(\mathrm{kJ} \mathrm{mol}^{-1}\) is (a) \(224.3\) (b) \(142.3\) (c) \(12.43\) (d) \(124.3\)
4 step solution
Problem 134
Which of the following statements are correct about half-life period? (1) time required for \(99.9 \%\) completion of a reaction is 100 times the half-life period (2) time required for \(75 \%\) completion of a first-order reaction is double the half-life of the reaction (3) average life \(=1.44\) times the half-life for firstorder reaction
4 step solution
Problem 135
When concentrations of the reactants is increased sixteen times, the rate becomes two times. The reaction is of (a) \(1 / 4\) order (b) fourth-order (c) third-order (d) \(1 / 8\) order
5 step solution
Problem 137
For the reaction a \(\mathrm{A} \longrightarrow \mathrm{xP}\) when \([\mathrm{A}]=2.2 \mathrm{mM}\) the rate was found to be \(2.4 \mathrm{~m} \mathrm{M} \mathrm{s}^{-1}\) On reducing concentration of \(\mathrm{A}\) to half, the rate changes to \(0.6 \mathrm{~m} \mathrm{M} \mathrm{s}^{-1}\). The order of reaction with respect to \(\mathrm{A}\) is (a) \(1.5\) (b) \(2.0\) (c) \(2.5\) (d) \(3.0\)
4 step solution
Problem 139
For a zero order reaction, the plot of concentration versus time is linear with (a) positive slope with zero intercept (b) positive slope with non-zero intercept (c) negative slope with non-zero intercept (d) parallel to time axis.
6 step solution
Problem 140
A certain reaction proceeds in a sequence of three elementary steps with the rate constants \(\mathrm{k}_{1}, \mathrm{k}_{2}\) and \(\mathrm{k}_{3} .\) If the observed rate constant of the expressed as \(\mathrm{k}\) (obs) \(=\mathrm{k}(\mathrm{obs})=\left[\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}\right]^{1 / 2} \mathrm{k}_{3}\), the observed energy of activa- tion of the reaction is (a) \(\frac{\mathrm{E}_{3}+\mathrm{E}_{1}}{2}\) (b) \(\frac{1}{2}\left[\frac{E_{1}}{E_{2}}\right]+E_{3}\) (c) \(\mathrm{E}_{3}+\frac{1}{2}\left[\mathrm{E}_{1}-\mathrm{E}_{2}\right]\) (d) \(\mathrm{E}_{3}\left[\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}\right]^{12}\)
5 step solution
Problem 141
Decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is a first order reaction. Rate constant for decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(0.022 \mathrm{hr}\). A solution of \(\mathrm{H}_{2} \mathrm{O}_{2}\) labelled as ' \(\mathrm{x}\) ' volume was left open. Due to this some \(\mathrm{H}_{2} \mathrm{O}_{2}\) decomposed. To determine new volume strength after \(6 \mathrm{hr}, 10 \mathrm{ml}\) of this solution was diluted to \(100 \mathrm{ml} .10 \mathrm{ml}\) of this diluted solution was titrated against \(25 \mathrm{ml}\) of \(0.025 \mathrm{M} \mathrm{KMnO}_{4}\) acidified solution. The value of ' \(x\) ' is (a) 20 volume (b) 10 volume (c) 15 volume (d) none of these
4 step solution
Problem 142
Reaction \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}\) is completed according to the following mechanism. \(\mathrm{A}_{2} \rightleftharpoons 2 \mathrm{~A}\) \(\mathrm{A}+\mathrm{B}_{2} \rightarrow \mathrm{AB}+\mathrm{B} \quad\) (slow step) \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB}\) The order of reaction is (a) 1 (b) \(3 / 2\) (c) \(1 / 2\) (d) 2
5 step solution
Problem 143
For the following reaction \(\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\) The initial pressure was \(\mathrm{P}_{0}\) while pressure after time ' \(\mathrm{t}\) ' was \(\mathrm{P}_{1}\). The rate constant \(\mathrm{k}\) will be (a) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{0}-\mathrm{P}_{\mathrm{t}}}\) (b) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{2 \mathrm{P}_{0}-\mathrm{P}_{\mathrm{1}}}\) (c) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{1}}\) (d) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{0}-2 \mathrm{P}_{\mathrm{t}}}\)
5 step solution
Problem 144
A first order reaction is carried out with an initial concentration of \(10 \mathrm{~mol}\) per litre and \(80 \%\) of the reactant changes into the product in \(10 \mathrm{sec}\). Now if the same reaction is carried out with an initial concentration of 5 mol per litre the percentage of the reactant changing to the produce in \(10 \mathrm{sec}\) is (a) 160 (b) 80 (c) 50 (d) 40
5 step solution
Problem 145
The rate low for the hydrolysis of thioacetamide, \(\mathrm{CH}_{3}
\mathrm{CSNH}_{2}\)
4 step solution
Problem 146
A follows parallel path Ist order reactions giving \(\mathrm{B}\) and C as shown: If initial concentration of \(\mathrm{A}\) is \(0.25 \mathrm{M}\), calculate the concentration of \(\mathrm{C}\) after 5 hour of reaction. Given, \(\lambda_{1}=1.5 \times 10^{-5} \mathrm{~s}^{-1}, \lambda_{2}=5 \times 10^{-6} \mathrm{~s}^{-1}\) (a) \(7.55 \times 10^{-3} \mathrm{M}\) (b) \(1.89 \times 10^{-2} \mathrm{M}\) (c) \(5.53 \times 10^{-3} \mathrm{M}\) (d) \(3.51 \times 10^{-3} \mathrm{M}\)
7 step solution
Problem 147
Two substance 'A' and 'B' are present such that \(\left[\mathrm{A}_{0}\right]=\) \(4\left[\mathrm{~B}_{0}\right]\), and half-life of ' \(^{4} \mathrm{~A}\) ' is 5 minutes and that of ' \(\mathrm{B}\) ' is 15 minute. If they start decaying at the same time following first order, how much time later will the concentration of both of them would be same. (a) \(10 \mathrm{~min}\) (b) \(12 \mathrm{~min}\) (c) \(5 \mathrm{~min}\) (d) \(15 \mathrm{~min}\)
5 step solution
Problem 148
For a reaction, \(\mathrm{A} \rightarrow \mathrm{B}+\mathrm{C}\), it was found that at the end of 10 min from the start the total optical, rotation of the system was \(50^{\circ} \mathrm{C}\) and when the reaction is complete it was \(100^{\circ}\). Assuming that only \(\mathrm{B}\) and \(\mathrm{C}\) are optically active and dextro rotator, the rate constant of this first order reaction would be (a) \(6.9 \mathrm{~min}^{-1}\) (b) \(0.069 \mathrm{~min}^{-1}\) (c) \(6.9 \times 10^{-2} \mathrm{~min}^{-1}\) (d) \(0.69 \mathrm{~min}^{-1}\)
6 step solution
Problem 149
Two reactants (A) and (B) separately show two chemical reactions. Both reactions are carried out with same initial concentration of each reactant. Reactant (A) follows first order kinetics whereas reactant (b) follows second order kinetics. If both have same life period. The ratio of their at the start of reaction is (a) \(2.303 \log 2\) (b) \(\log 2\) (c) \(\frac{2.303}{\log 2}\) (d) \(\frac{\log 2}{2.303}\)
5 step solution
Problem 151
For a first order reaction \(\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\), the total pressure of \(\mathrm{A}+\mathrm{B}+\mathrm{C}\) at time ' \(\mathrm{t}^{\prime}\) and \(\infty\) are \(\mathrm{P}_{2}\) and \(\mathrm{P}_{3}\) respectively. The constant \(\mathrm{k}\) of the reaction is (a) \(\frac{1}{t} \ln \frac{P_{3}}{2\left(P_{3}-P_{2}\right)}\) (b) \(\frac{1}{t} \ln \frac{2 P_{3}}{P_{3}-P_{2}}\) (c) \(\frac{1}{\mathrm{t}} \ln \frac{\mathrm{P}_{3}}{\mathrm{P}_{3}-\mathrm{P}_{2}}\) (d) \(\frac{1}{t} \ln \frac{P_{3}}{2 P_{3}-P_{2}}\)
8 step solution