Chapter 10
A Complete Resource Book in Chemistry for JEE Main · 159 exercises
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 \(\mathrm{A}\) to produce \(0.675 \mathrm{~mole}\) of B? (a) 2 hour (b) 1 hour (c) \(0.5\) hour (d) \(0.25\) hour
5 step solution
Problem 76
Which of the following are the examples of pseudo-unimolecular reactions? (1) acid catalyzed hydrolysis of an ester (2) inversion of cane sugar (3) decomposition of ozone (4) decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) Select the correct answers using the codes given below: (a) 1 and 2 (b) 1 and 3 (c) 2,3 and 4 (d) 1,2 and 4
6 step solution
Problem 77
For the reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 3 \mathrm{C}+\mathrm{D}\) which of the following does not express the reaction rate? (a) \(\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}\) (b) \(-\frac{\mathrm{d}[\mathrm{A}]}{2 \mathrm{dt}}\) (c) \(\frac{\mathrm{d}[\mathrm{C}]}{3 \mathrm{dt}}\) (d) \(-\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\)
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) \(2 \times 10^{-2} \mathrm{Ms}^{-1}\)
4 step solution
Problem 81
The rate of a certain hypothetical reaction \(\mathrm{A}+\mathrm{B}+\mathrm{C} \longrightarrow\) Products is given by \(\mathrm{r}=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=k[\mathrm{~A}]^{1 / 2}[\mathrm{~B}]^{1 / 3}[\mathrm{C}]^{\mathrm{L} / 4}\) The order of the reaction is (a) \(13 / 12\) (b) \(13 / 14\) (c) \(12 / 13\) (d) \(13 / 11\)
7 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) \(3.5 \times 10^{3} \mathrm{yrs}\)
5 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) \(6 \times 10^{-8} \mathrm{Ms}^{-1}\)
4 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
3 step solution
Problem 86
From the following data for the reaction between \(\mathrm{A}\) and \(\mathrm{B}\) $$ \begin{array}{cccc} \hline[\mathrm{A}] & {[\mathrm{B}]} & \text { Initial rate } & \left(\mathrm{mol} \mathrm{L}^{-1} \mathbf{s}^{-1}\right) \\ \mathrm{molL}^{-1} & \mathrm{molL}^{-1} & \mathbf{3 0 0} \mathrm{K} & \mathbf{3 2 0} \mathrm{K} \\ \hline 2.5 \times 10^{-4} & 3.0 \times 10^{-5} & 5.0 \times 10^{-4} & 2.0 \times 10^{-3} \\ 5.0 \times 10^{-4} & 6.0 \times 10^{-5} & 4.0 \times 10^{-3} & \- \\ 1.0 \times 10^{-3} & 6.0 \times 10^{-5} & 1.6 \times 10^{-2} & \- \\ \hline \end{array} $$ Calculate the rate equation (a) \(\mathbf{r}=k[\mathrm{~B}]^{1}\) (b) \(\mathrm{r}=k[\mathrm{~A}]^{2}\) (c) \(\mathbf{r}=k[\mathrm{~A}]^{2}[\mathrm{~B}]^{1}\) (d) \(\mathrm{r}=\mathrm{A}[\mathrm{A}][\mathrm{B}]\)
5 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) \(3.6 \times 10^{30} \mathrm{~s}^{-1}\)
3 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
5 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 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) \(2 \times 10^{-4}\)
5 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
4 step solution
Problem 98
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
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) \(654.5 \mathrm{~s}\)
4 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 105
Consider the following reaction, \(\mathrm{A} \longrightarrow\) products This reaction is completed in 100 minutes. The rate constant of this reaction at \(\mathrm{t}_{1}=10 \mathrm{~min}\) is \(10^{-2} \mathrm{~min}^{-1}\). What is the rate constant (in \(\min ^{-1}\) ) at \(\mathrm{t}_{2}=20 \mathrm{~min} ?\) (a) \(2 \times 10^{-2}\) (b) \(10^{-2}\) (c) \(5 \times 10^{-3}\) (d) \(0.1\)
4 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 107
The rate constant for the reaction, \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) is \(3.0 \times 10^{-5} \mathrm{~s}^{1} .\) If the rate is \(2.40 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{1}\), then the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (in \(\mathrm{mol} \mathrm{L}^{4}\) ) is (a) \(1.4\) (b) \(1.2\) (c) \(0.04\) (d) \(0.8\)
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) \(0.04\)
4 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 (E) of the reaction? \((\mathrm{R}=\) gas constant \()\) (a) \(1842.4 \mathrm{R}\) (b) \(921.2 \mathrm{R}\) (c) \(460.0 \mathrm{R}\) (d) \(230.3 \mathrm{R}\)
6 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) \(2 \frac{\mathrm{d}\left[\mathrm{NH}_{2}\right]}{\mathrm{dt}}=-3 \frac{\mathrm{d}\left[\mathrm{H}_{3}\right]}{\mathrm{dt}}\) (c) \(2 \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\) (d) \(3 \frac{\mathrm{d}\left[\mathrm{NH}_{2}\right]}{\mathrm{dt}}=-2 \frac{\mathrm{d}\left[\mathrm{H}_{3}\right]}{\mathrm{dt}}\)
4 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) \(2 \mathrm{t}\) (b) \(\mathrm{t} / 30\) (c) \(120 \mathrm{t}\) (d) \(60 \mathrm{t}\)
6 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) \(+2 \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\)
4 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
6 step solution
Problem 116
For a gaseous reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{AB}\) this following rate data were obtained at \(300 \mathrm{~K}\). $$ \begin{array}{llll} \hline \text { Expt } & \text { Concentration } & \text { Rate of disappearance } \\ & {[\mathbf{A}]} & {\left[\mathrm{B}_{2}\right]} & \text { of } \mathrm{B}_{2}\left(\mathrm{~mol} \mathrm{~L} \mathbf{~ m i n}^{-1}\right) \\ \hline 1 . & 0.015 & 0.15 & 1.8 \times 10^{-3} \\ 2 . & 0.09 & 0.15 & 1.08 \times 10^{-2} \\ 3 . & 0.015 & 0.45 & 5.4 \times 10^{-3} \\ \hline \end{array} $$ 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) \(\mathrm{r}=k[\mathrm{~A}]\left[\mathrm{B}_{2}\right]^{2}\) (d) \(\mathrm{r}=k\left[\mathrm{~B}_{2}\right]\)
4 step solution
Problem 117
The basic theory of Arrhenius equation is that 12 (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 \(1 / \mathrm{T}\) 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) \(0.957 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\)
4 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) \(1 \times 10^{6} \mathrm{~S}^{-1}\) 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 \(16.6 \mathrm{~kJ} / \mathrm{M}\)
6 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 \(X\) 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) \(3.47 \times 10^{-4} \mathrm{M} / \mathrm{min}\)
4 step solution
Problem 122
The following data are obtained from the decomposition of a gaseous compound Initial pressure in arm \(\begin{array}{lll}1.6 & 0.8 & 0.4\end{array}\) \(\begin{array}{lll}\text { Time for } 50 \% \text { reaction in } \min & 80 & 113 & 160\end{array}\) 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 \(\mathrm{B}\) $$ \begin{array}{llll} \hline \text { S. No. } & {[\mathrm{A}] \mathrm{mol} \mathrm{L}^{-1}} & {[\mathrm{~B}] \mathrm{mol} \mathrm{L}^{-1}} & \begin{array}{l} \text { Rate mol } \\ \mathrm{L}-\mathbf{S}^{-1} \end{array} \\ \hline \text { 1. } & 1 \times 10^{-2} & 2 \times 10^{-2} & 2 \times 10^{-4} \\\ 2 . & 2 \times 10^{-2} & 2 \times 10^{-2} & 4 \times 10^{-4} \\ 3 . & 2 \times 10^{-2} & 4 \times 10^{-2} & 8 \times 10^{-8} \\ \hline \end{array} $$ 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
6 step solution
Problem 125
Log \(1.5 \times 10^{7}=\log \mathrm{A}-\left(\frac{2}{2.303 \mathrm{R}}\right)\) In the above equa- tion, what is the value of Arrhenius factor? (a) \(3 \times 10^{6}\) (b) \(6.3 \times 10^{9}\) (c) \(5.42 \times 10^{13}\) (d) \(5.42 \times 10^{10}\)
4 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
6 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 $$ \begin{array}{lll} \hline\left[\mathrm{Cl}_{2}\right] & {[\mathrm{NO}]} & \text { Initial rate }\left(\mathrm{mol}_{\mathbf{L}}^{-1} \mathbf{s}^{-1}\right) \\ \hline 0.05 & 0.05 & 1 \times 10^{-3} \\ 0.15 & 0.05 & 3 \times 10^{-3} \\ 0.05 & 0.15 & 9 \times 10^{-3} \\ \hline \end{array} $$ What is the rate law? (a) \(\mathrm{r}=k[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right]\) (b) \(\mathrm{r}=k\left[\mathrm{Cl}_{2}\right]^{1}[\mathrm{NO}]^{2}\) (c) \(\mathrm{r}=k\left[\mathrm{Cl}_{2}\right]^{2}[\mathrm{NO}]\) (d) \(\mathrm{r}=k\left[\mathrm{Cl}_{2}\right]^{1}\)
4 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 \(28.0\) kcal and its drops to \(10.5\) 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 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
6 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) \(3.0\)
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) \(1.5\)
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.
4 step solution