Chapter 10
A Complete Resource Book in Chemistry for JEE Main · 112 exercises
Problem 82
The decay constant of \(C^{14}\) is \(2.31 \times 10^{-4}\) year \(^{-1}\). Its half life is (a) \(2 \times 10^{3} \mathrm{yrs}\) (b) \(2.5 \times 10^{3} \mathrm{yrs}\) (c) \(3 \times 10^{3} y r s\) (d) \(3.5 \times 10^{3} \mathrm{yrs}\)
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) \(1.33 \times 10^{-3} \mathrm{~min}^{-1}\)
4 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
Problem 85
A graph plotted between concentration of reactant, consumed at any time \((\mathrm{x})\) 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
4 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) \(3.6 \times 10^{30} \mathrm{~s}^{-1}\)
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) \(0.75 \mathrm{~mol} \mathrm{~min}^{1}\) (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) \(36 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\)
5 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 sec \({ }^{-1}\) is (a) \(2 \times 10^{4}\) (b) \(3.45 \times 10^{-5}\) (c) \(1.386 \times 10^{-4}\) (d) \(2 \times 10^{4}\)
6 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 \(\mathrm{A}^{\mathrm{m}}+\mathrm{B}^{\mathrm{n}} \longrightarrow \mathrm{AB}_{\mathrm{x}}\) 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
6 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) \(2.44 \mathrm{~s}\)
5 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) \(654.5 \mathrm{~s}\)
5 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 \(\mathrm{K}=1.6\) \(\times 10^{\circ} \mathrm{s}^{-1}\) at \(280 \mathrm{~K} ?\) (a) \(1.6 \times 10^{6} \mathrm{~s}^{-1}\) (b) zero (c) \(4.8 \times 10^{4} \mathrm{~s}^{-4}\) (d) \(3.2 \times 10^{12} \mathrm{~s}^{-1}\)
3 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 \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\)
5 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) \(0.04\)
5 step solution
Problem 109
The rate constant of a reaction at temperature 200 is 10 times less than the rate constant at \(400 \mathrm{~K}\). What is the activation energy \(\left(\mathrm{E}_{\alpha}\right)\) of the reaction? \((\mathrm{R}=\) gas constant) (a) \(1842.4 \mathrm{R}\) (b) \(921.2 \mathrm{R}\) (c) \(460.0 \mathrm{R}\) (d) \(230.3 \mathrm{R}\)
7 step solution
Problem 111
The time taken for the completion of \(90 \%\) of a first-order reaction is 't' min. What is the time (in seconds) taken for the completion of \(99 \%\) of the reaction? (a) \(2 t\) (b) \(t / 30\) (c) \(120 t\) (d) \(60 \mathrm{t}\)
5 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) \(-\frac{1}{3} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) \((\mathrm{d})+2 \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\)
3 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 \(B\), 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) \(4 / 3\)
5 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
4 step solution
Problem 116
For a gaseous reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{AB}\) this following rate data were obtained at \(300 \mathrm{~K}\). Table \(10.4\) \begin{tabular}{llll} \hline Expt & Concentration & Rate of disappearance \\ & {\([\mathrm{A}]\)} & {\(\left[\mathrm{B}_{2}\right]\)} & of \(\mathrm{B}_{2}\left(\mathrm{~mol} \mathrm{~L} \min ^{-1}\right)\) \\ \hline \(1 .\) & \(0.015\) & \(0.15\) & \(1.8 \times 10^{-2}\) \\ \(2 .\) & \(0.09\) & \(0.15\) & \(1.08 \times 10^{-2}\) \\ \(3 .\) & \(0.015\) & \(0.45\) & \(5.4 \times 10^{-2}\) \\ \hline \end{tabular} What is the rate law? (a) \(\mathrm{r}=k[\mathrm{~A}]\left[\mathrm{B}_{2}\right]\) (b) \(r=[\mathrm{A}]^{2}\left[\mathrm{~B}_{2}\right]^{1}\) (c) \(r=k[A]\left[B_{2}\right]^{2}\) (d) \(\mathrm{r}=k\left[\mathrm{~B}_{2}\right]\)
4 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
4 step solution
Problem 118
The slope of the line for the graph of \(\log k\) vs \(1 / T\) for the reaction, \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+1 / 2 \mathrm{O}_{2}\) is \(-5000\). Calculate the energy of activation of the reaction. (a) \(95.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(9.57 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(957 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(0.957 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\)
5 step solution
Problem 119
For a \(1^{\text {* }}\) order reaction \(\mathrm{A} \longrightarrow \mathrm{P}\), the temperature (T) dependent rate constant \((\mathrm{K})\) was found to follow the equation \(\log \mathrm{k}=-(2000) \frac{1}{\mathrm{~T}}+6\) The pre- exponential factor \(\mathrm{A}\) and activation energy Ea are respectively? (a) \(1 \times 10^{6} \mathrm{~S}^{-1}\) 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 \(16.6 \mathrm{~kJ} / \mathrm{M}\)
4 step solution
Problem 121
The reaction \(\mathrm{X} \longrightarrow\) Product follows first-order kinetics, in 40 minutes, the concentration of \(\mathrm{X}\) 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 \(\quad 1.6 \quad 0.8 \quad 0.4\) Time for \(50 \%\) reaction in min \(80 \quad 113 \quad 160\) The order of the reaction is (a) \(0.5\) (b) \(1.0\) (c) \(1.5\) (d) \(2.0\)
6 step solution
Problem 123
The following data pertains to the reaction between \(\mathrm{A}\) and \(\underline{B}\) \begin{tabular}{llll} \multicolumn{4}{c} { Table \(10.5\)} \\ \hline S. No. & {\([\mathrm{A}] \mathrm{mol} \mathrm{L}^{-1}\)} & {\([\mathrm{~B}] \mathrm{mol} \mathrm{L}^{-1}\)} & Rate mol \(\mathrm{L}^{-1} \mathrm{~S}^{-1}\) \\ \hline 1. & \(1 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(2 \times 10^{-4}\) \\ 2\. & \(2 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(4 \times 10^{-4}\) \\ 3\. & \(2 \times 10^{-2}\) & \(4 \times 10^{-2}\) & \(8 \times 10^{-4}\) \\ \hline \end{tabular} Which of the following inferences are drawn from the above data? (1) rate constant of the reaction is \(10^{-4}\) (2) rate law of the reaction is \([\mathrm{A}][\mathrm{B}]\) (3) rate of reaction increases four times by doubling the concentration of each reactant. Select the correct answer the codes given below: (a) 1 and 3 (b) 2 and 3 (c) land 2 (d) 1,2 and 3
5 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) \(313 \mathrm{~mm} \mathrm{Hg}\)
6 step solution
Problem 131
The data given below is for the reaction of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) to form \(\mathrm{NOCl}\) at 295 \begin{tabular}{lll} \multicolumn{2}{c} { Table \(10.6\)} \\ \hline [CI_] & [NO] & Initial rate \(\left(\mathrm{molL}^{-4} \mathrm{~s}^{-1}\right)\) \\ \hline \(0.05\) & \(0.05\) & \(1 \times 10^{-3}\) \\ \(0.15\) & \(0.05\) & \(3 \times 10^{-3}\) \\ \(0.05\) & \(0.15\) & \(9 \times 10^{-3}\) \\ \hline \end{tabular} What is the rate law? (a) \(\mathrm{r}=k[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right]\) (b) \(\mathrm{r}=k\left[\mathrm{Cl}_{2}\right]^{\mathrm{2}}[\mathrm{NO}]^{2}\) (c) \(\mathrm{r}=k\left[\mathrm{Cl}_{2}\right]^{2}[\mathrm{NO}]\) (d) \(\mathrm{r}=k\left[\mathrm{Cl}_{2}\right]^{1}\)
5 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) it is proportional to initial concentration for zeroth-order (a) 1,2 and 3 (b) 2,3 and 4 (c) 2 and 3 (d) 3 and 4
6 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 \(\mathrm{a} \mathrm{A} \longrightarrow \mathrm{xP}\) 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) \(3.0\)
5 step solution
Problem 138
If the initial concentration of reactant in certain reaction is doubled, the half life period of the reaction is also doubled. The order of reaction is (a) zero (b) first (c) second (d) \(1.5\)
3 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.
4 step solution
Problem 142
Reaction \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}\) is completed according to the following mechanism. \(A_{2}=2 A\) \(\mathrm{A}+\mathrm{B}_{2} \rightarrow \mathrm{AB}+\mathrm{B} \quad\) (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 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 sec is (a) 160 (b) 80 (c) 50 (d) 40
5 step solution
Problem 145
The rate low for the hydrolysis of thioacetamide, \(\mathrm{CH}_{3}
\mathrm{CSNH}_{2}\),
5 step solution
Problem 146
A follows parallel path Ist order reactions giving 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) \(3.51 \times 10^{-3} \mathrm{M}\)
6 step solution
Problem 147
Two substance ' \(\mathrm{A}^{\prime}\) and ' \(\mathrm{B}^{\prime}\) are present such that \(\left[\mathrm{A}_{0}\right]=\) \(4\left[\mathrm{~B}_{0}\right]\), and half-life of ' \(\mathrm{A}^{\prime}\) 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) \(15 \mathrm{~min}\)
7 step solution
Problem 148
For a reaction, \(\mathrm{A} \rightarrow \mathrm{B}+\mathrm{C}\), it was found that at the end of \(10 \mathrm{~min}\) 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) \(0.69 \mathrm{~min}^{-1}\)
9 step solution
Problem 150
The following data pertains to the reaction between \(\mathrm{A}\) and \(\mathrm{B}\) : \begin{tabular}{llll} \hline S. No. & {\([\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}\)} & {\([\mathrm{~B}] / \mathrm{mol} \mathrm{L}^{-1}\)} & Rate \(/ \mathrm{mol} \mathrm{L}^{-1} \mathrm{t}^{-1}\) \\ \hline 1 & \(1 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(2 \times 10^{-4}\) \\ II & \(2 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(4 \times 10^{-4}\) \\ \hline III & \(2 \times 10^{-2}\) & \(4 \times 10^{-2}\) & \(8 \times 10^{-4}\) \\ \hline \end{tabular} Which of the following inferences are drawn from the above data? 1\. Rate law of the reaction is \(\mathrm{k}[\mathrm{A}][\mathrm{B}]\). 2\. Rate constant of the reaction is \(10^{-4}\). 3\. Rate of reaction increases four times by doubling the concentration of each reactant. Select the correct answer from the codes given below: (a) 1 and 2 (b) 1 and 3 (c) 1,2 and 3 (d) 3 alone
7 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) \(\frac{1}{t} \ln \frac{2 P_{3}}{P_{1}-P_{2}}\) (c) \(\frac{1}{t} \ln \frac{P_{3}}{P_{3}-P_{2}}\) (d) \(\frac{1}{t} \ln \frac{P_{3}}{2 P_{3}-P_{2}}\)
6 step solution
Problem 152
The energy of activation and specific rate constant for a first order reaction at \(25^{\circ} \mathrm{C}\) are \(100 \mathrm{~kJ} / \mathrm{mol}\) and \(3.46\) \(\times 10^{-5} \mathrm{sec}^{-1}\) respectively. Determine the temperature at which half life of reaction is 2 hour. \(2 \mathrm{~N}_{2} \mathrm{O}_{3} \rightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2}\) \(\begin{array}{ll} \left.\text { (in } \mathrm{CCl}_{4}\right) & \left.\text { (in } \mathrm{CCl}_{4}\right)\end{array}\) (a) \(300 \mathrm{~K}\) (b) \(302 \mathrm{~K}\) (c) \(304 \mathrm{~K}\) (d) \(306 \mathrm{~K}\)
7 step solution
Problem 153
A redox reaction is carried out at \(127^{\circ} \mathrm{C}\). If the same reaction is carried out in presence of a catalyst at the same temperature, the rate of reaction is doubled. To what extent is the energy barrier lowered by the catalyst? [Use \(\mathrm{R}=2\) cal \(\mathrm{mol}^{-1} \mathrm{~K}^{-1}\) and \(\left.\log 2=0.301\right]\) (a) \(455 \mathrm{cal}\) (b) \(231 \mathrm{cal}\) (c) \(693 \mathrm{cal}\) (d) \(554 \mathrm{cal}\)
6 step solution
Problem 154
The rate constant, the activation energy and Arrhenius parameter of a chemical reaction at \(300 \mathrm{~K}\) are \(\mathrm{K}, \mathrm{Ea}\) and \(\mathrm{A}\) respectively. The value of rate constant at \(\mathrm{T} \rightarrow\) \(\infty\) is (a) \(\mathrm{A}\) (b) \(\mathrm{Ea}\) (c) \(\mathrm{Ea} \times \mathrm{A}\) (d) \(\mathrm{A}-\mathrm{Ea}\)
4 step solution
Problem 155
An aqueous solution of sugar undergoes acid catalysed hydrolysis. \(50 \mathrm{~g}\) sugar in \(125 \mathrm{~mL}\) water rotates the plane of plane polarized light by \(+13.1^{\circ}\) at \(\mathrm{t}=0 .\) After complete hydrolysis, it shows a rotation of \(-3.75^{\circ} .\) The percentage hydrolysis of sugar at time ' \(t\) ' in the same solution having a rotation of \(5^{\circ}\) is (a) \(42 \%\) (b) \(58 \%\) (c) \(48 \%\) (d) \(55 \%\)
5 step solution