Chapter 10
A Complete Resource Book in Chemistry for JEE Main · 159 exercises
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) \(=\mathrm{k}(\mathrm{obs})=\left[\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}\right]^{1 / 2} \cdot \mathrm{k}_{3}\), the observed energy of activa- tion of the reaction is (a) \(\frac{E_{3}+E_{1}}{2}\) (b) \(\frac{1}{2}\left[\frac{E_{1}}{E_{2}}\right]+E_{3}\)
6 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
3 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 \(\mathrm{P}_{t}\). The rate constant \(\mathrm{k}\) will be (a) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{0}-\mathrm{P}_{1}}\) (b) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{2 \mathrm{P}_{0}-\mathrm{P}_{1}}\) (c) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{\mathrm{t}}}\) (d) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{\mathrm{P}_{0}}{\mathrm{P}_{0}-2 \mathrm{P}_{\mathrm{t}}}\)
6 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
4 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 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) \(3.51 \times 10^{-3} \mathrm{M}\)
9 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 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) \(0.69 \mathrm{~min}^{-1}\)
4 step solution
Problem 150
The following data pertains to the reaction between \(\mathrm{A}\) and \(\mathrm{B}\) : $$ \begin{array}{llll} \hline \text { S. No. } & {[\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}} & {[\mathrm{~B}] / \mathrm{mol} \mathrm{L}^{-1}} & \text { Rate } / \mathrm{mol} \mathrm{L}^{-1} \mathbf{t}^{-1} \\ \hline \text { I } & 1 \times 10^{-2} & 2 \times 10^{-2} & 2 \times 10^{-4} \\\ \text { II } & 2 \times 10^{-2} & 2 \times 10^{-2} & 4 \times 10^{-4} \\ \text { III } & 2 \times 10^{-2} & 4 \times 10^{-2} & 8 \times 10^{-4} \\ \hline \end{array} $$ 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
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 ' \(\mathrm{t}\) ' 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) \(\frac{1}{t} \ln \frac{P_{3}}{P_{3}-P_{2}}\) (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. $$ \begin{aligned} &2 \mathrm{~N}_{2} \mathrm{O}_{5} \quad \rightarrow \quad 2 \mathrm{~N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2} \\ &\text { (in } \left.\mathrm{CCl}_{4}\right) \quad\left(\text { in } \mathrm{CCl}_{4}\right) \end{aligned} $$ (a) \(300 \mathrm{~K}\) (b) \(302 \mathrm{~K}\) (c) \(304 \mathrm{~K}\) (d) \(306 \mathrm{~K}\)
6 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 \(\mathrm{R}=2 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\) and \(\log 2=0.301\) ] (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 A respectively. The value of rate constant at \(\mathrm{T} \rightarrow\) \(\infty\) is (a) \(\mathrm{A}\) (b) \(\mathrm{Ea}\) (c) \(\mathrm{EaxA}\) (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 ' \(\mathrm{t}\) ' in the same solution having a rotation of \(5^{\circ}\) is (a) \(42 \%\) (b) \(58 \%\) (c) \(48 \%\) (d) \(55 \%\)
4 step solution
Problem 157
Hydrogenation of vegetable ghee at \(27^{\circ} \mathrm{C}\) reduces the pressure of \(\mathrm{H}_{2}\) from \(3 \mathrm{~atm}\) to \(2.18 \mathrm{~atm}\) in 40 minutes. The rate of reaction in terms of molarity per second is \(\left(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) \(1.357 \times 10^{-6}\) (b) \(1.537 \times 10^{-5}\) (c) \(1.375 \times 10^{-5}\) (d) \(6.250 \times 10^{-4}\)
6 step solution
Problem 159
The half life \(\left(\mathrm{t}_{1}\right)\) of the first order reaction and half life \(\left(t_{2}\right)\) of the second order reaction are in the ratio \(2: 1\). Hence the ratio of the rates of the above first and second order reactions at the start is (a) \(1: 0.4365\) (b) \(0.3465: 1\) (c) \(2: 1\) (d) \(1: 2\)
5 step solution
Problem 162
For a zero order reaction, the concentration of reactant after \(10 \mathrm{~s}\) is \(0.2 \mathrm{~mol} / /\). If \(\mathrm{k}_{0}\) is \(2 \times 10^{-2}\) mol \(l^{-1} \mathrm{~s}^{-1}\), the initial concentration of reactant is (a) \(0.6 \mathrm{~mol} / l\) (b) \(0.4 \mathrm{~mol} / /\) (c) \(0.8 \mathrm{~mol} / l\) (d) \(1 \mathrm{~mol} / l\)
5 step solution
Problem 164
Consider the following statements (a) The rate of a process is always proportional to its free energy change. (b) The molecularity of an elementary chemical reaction step can be determined by examining its stoichiometry. (c) The first order reactions follow an exponential time course. (d) Energy of activation is inversely proportional to temperature. The correct statement (s) is/are (a) \(1,2,3\) (b) \(1,2,3,4\) (c) 2 and 3 (d) 1 and 3
5 step solution
Problem 165
The hypothetical reaction, \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}\) follows the following mechanism: \(\mathrm{A}_{2}=\mathrm{A}+\mathrm{A}\) (fast) \(\mathrm{A}+\mathrm{B}_{2} \longrightarrow \mathrm{AB}+\mathrm{B}\) (slow) \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{AB} \quad\) (fast) The order of the overall reaction is: (a) 1 (b) 2 (c) \(3 / 2\) (d) 0
5 step solution
Problem 168
The chemical kinetics of the reaction \(\mathrm{a} \mathrm{A}+\mathrm{bB} \rightarrow\) \(\mathrm{C}\) at \(298 \mathrm{~K}\) were followed. The initial rates were recorded rates were recorded under different initial conditions and are summarized as follows. $$ \begin{array}{lll} \hline \begin{array}{l} \text { Initial conc. } \\ {[\mathrm{A}]_{0}(\mathrm{~mol} / \mathrm{L})} \end{array} & \begin{array}{l} \text { Initial conc. } \\ {[\mathrm{B}]_{0}(\mathrm{~mol} / \mathrm{L})} \end{array} & \begin{array}{l} \text { Initial rate } \\ \text { (mol } / \mathrm{L} \mathrm{s}) \end{array} \\ \hline 0.1 & 0.1 & 2.4 \times 10^{-3} \\ 0.2 & 0.1 & 4.8 \times 10^{-3} \\ 0.4 & 0.1 & 9.7 \times 10^{-3} \\ 0.1 & 0.2 & 9.6 \times 10^{-3} \\ 0.1 & 0.4 & 3.8 \times 10^{-2} \\ \hline \end{array} $$ Which of the following statements is incorrect? (a) The rate constant \(\mathrm{k}\) is governed by the activation energy of the reaction (b) Reaction rate \(=\mathrm{k}[\mathrm{A}][\mathrm{B}]^{2}\) (c) In the chemical equation of \(\mathrm{aA}+\mathrm{bB} \rightarrow \mathrm{C}\), a is 0 and \(\mathrm{b}\) is 3 . (d) The overall order of reaction is third order.
5 step solution
Problem 169
If a is the initial concentration of reactant and \((a-x) \quad\) (b) \(x=a\left(1-e^{-k_{1}}\right)\) \(\begin{aligned}&\text { is the remaining concentration after time 't' in a first } \\\&\text { order reaction of rate constant } k_{1} \text {, then which of the } & \text { (c) } t_{1 / 2}=\frac{1.414}{k_{1}}\end{aligned}\) following relations is /are correct? (a) \(k_{1}=\frac{2.303}{t} \log \left(\frac{\mathrm{a}}{\mathrm{a}-x}\right)\) (d) \(\mathrm{t}_{\mathrm{av}}=\frac{1}{k_{1}}\)
5 step solution
Problem 171
Identify the correct statements: (a) The order of an elementary reaction is equal to its molecularity (b) The order of a reaction can be zero (c) For second order reaction, order of reaction \(=2 \times\) molecularity. (d) The order of inversion of cane sugar is 2 .
4 step solution
Problem 174
A radioactive element, \(\mathrm{X}\), decays by the sequence and with half lives, given below: $$ \begin{aligned} &\mathrm{X}\left(\mathrm{t}_{1 / 2}=30 \mathrm{~min}\right) \\ &\mathrm{Y}\left(\mathrm{t}_{1 / 2}=2 \text { days }\right) \quad \stackrel{\lambda_{1}}{\longrightarrow_{2}} \longrightarrow \mathrm{Y}+\alpha \\\ &\cline { 1 } \longrightarrow \mathrm{Z}+2 \beta \end{aligned} $$ Which of the following statement(s) is/are incorrect? (a) Atomic numbers of \(\mathrm{X}\) and \(\mathrm{Z}\) are same (b) Disintegration constant \(\lambda_{2}>\lambda_{1}\) (c) The mass number of \(\mathrm{Y}\) is greater than that of \(\mathrm{X}\). (d) \(\mathrm{Y}\) and \(\mathrm{Z}\) are isotopes.
3 step solution
Problem 175
In a hypothetical reaction \(\mathrm{X} \rightarrow \mathrm{Y}\), the activation energy for the forward and backward reaction is 15 and \(9 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) respectively. The potential energy of \(\mathrm{X}\) is \(10 \mathrm{~kJ}\) \(\mathrm{mol}^{-1} .\) Identify the correct statement(s). (a) The threshold energy of the reaction is \(25 \mathrm{~kJ}\). (b) The potential energy \(\mathrm{f} \mathrm{Y}\) is \(16 \mathrm{~kJ}\) (c) Heat of reaction is \(6 \mathrm{~kJ}\). (d) The reaction is endothermic.
5 step solution
Problem 176
When sucrose is hydrolysed, it produces dextrorotatory "glucose" and laevorotatory "fructose" whereas the reactant itself is a dextrorotatory compound. The above a conversion process follows first order kinetics. Similarly, an optically active compound A is hydrolysed as follows. $$ \mathrm{A}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{+}}{\longrightarrow} 2 \mathrm{~B}+\mathrm{C} $$ The value of rate constant of the above reaction at \(27^{\circ} \mathrm{C}\) is (a) \(1.94 \times 10^{-3} \mathrm{~min}^{-1}\) (b) \(1.37 \times 10^{-6} \mathrm{~min}^{-1}\) (c) \(6.13 \times 10^{-3} \mathrm{~min}^{-1}\) (d) \(5.57 \times 10^{-3} \mathrm{~min}^{-1}\) The observed rotation of compound \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are \(60^{\circ}\), \(50^{\circ}\) and \(-80^{\circ}\) per mole respectively. The angles of rotation after 40 minutes and after the completion of reaction were \(26^{\circ}\) and \(10^{\circ}\) respectively. At \(27^{\circ} \mathrm{C}\) activation energy for conversion is \(27 \mathrm{~kJ} \mathrm{~mol}^{-1} .\) (Use: \(\log 1.25=0.0969, \log 14.97=\) 1.175)
8 step solution
Problem 177
When sucrose is hydrolysed, it produces dextrorotatory "glucose" and laevorotatory "fructose" whereas the reactant itself is a dextrorotatory compound. The above a conversion process follows first order kinetics. Similarly, an optically active compound A is hydrolysed as follows. $$ \mathrm{A}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{+}}{\longrightarrow} 2 \mathrm{~B}+\mathrm{C} $$ The observed rotation of compound \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are \(60^{\circ}\), \(50^{\circ}\) and \(-80^{\circ}\) per mole respectively. The angles of rotation after 40 minutes and after the completion of reaction were \(26^{\circ}\) and \(10^{\circ}\) respectively. At \(27^{\circ} \mathrm{C}\) activation energy for conversion is \(27 \mathrm{~kJ} \mathrm{~mol}^{-1} .\) (Use: \(\log 1.25=0.0969, \log 14.97=\) 1.175) The value of \(t_{1 / 2}\) for the above process at \(127^{\circ} \mathrm{C}\) is (a) \(1.2\) hours (b) \(8.4\) minutes (c) \(1.5\) hours (d) \(5.68\) minutes
5 step solution
Problem 178
When sucrose is hydrolysed, it produces dextrorotatory "glucose" and laevorotatory "fructose" whereas the reactant itself is a dextrorotatory compound. The above a conversion process follows first order kinetics. Similarly, an optically active compound A is hydrolysed as follows. $$ \mathrm{A}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{+}}{\longrightarrow} 2 \mathrm{~B}+\mathrm{C} $$ The observed rotation of compound \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are \(60^{\circ}\), \(50^{\circ}\) and \(-80^{\circ}\) per mole respectively. The angles of rotation after 40 minutes and after the completion of reaction were \(26^{\circ}\) and \(10^{\circ}\) respectively. At \(27^{\circ} \mathrm{C}\) activation energy for conversion is \(27 \mathrm{~kJ} \mathrm{~mol}^{-1} .\) (Use: \(\log 1.25=0.0969, \log 14.97=\) 1.175) At what temperature half life of above process is equal to \(31.1\) minutes? (a) \(490 \mathrm{~K}\) (b) \(345 \mathrm{~K}\) (c) \(72 \mathrm{~K}\) (d) \(280 \mathrm{~K}\)
7 step solution
Problem 180
For a reaction if effective rate constant \(\mathrm{k}\) ' is given by \(\mathrm{k}^{\prime}=\frac{2 \mathrm{k}_{2}}{\mathrm{k}_{3}}\left(\frac{\mathrm{k}_{1}}{\mathrm{k}_{5}}\right)^{1 / 5}\) where \(\mathrm{k}_{1}, \mathrm{k}_{2}, \mathrm{k}_{3}, \mathrm{k}_{5}\) are rate constants for different steps of the reaction. The effective frequency factor (a) \(\mathrm{A}^{\prime}=\frac{\mathrm{A}_{2}}{\mathrm{~A}_{3}}\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{5}}\right)^{2}\) (b) \(\mathrm{A}^{\prime}=\frac{2 \mathrm{~A}_{2}}{\mathrm{~A}_{3}}\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{5}}\right)^{1 / 5}\) (c) \(A^{\prime}=A_{1} \times A_{2} \times A_{3} \times A_{4} \times A_{5}\) (d) \(\mathrm{A}^{\prime}=\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\left(\frac{\mathrm{A}_{3}}{\mathrm{~A}_{5}}\right)^{1 / 5}\)
4 step solution
Problem 183
Match the following $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \begin{array}{l} \text { (a) Fraction of } \\ \text { effective } \\ \text { collisions } \end{array} & \text { (p) } \frac{1}{\mathrm{k} \cdot \mathrm{A}_{0}} \\ \text { (b) Half-life of } & \\ \begin{array}{l} \text { second order } \\ \text { reaction } \end{array} & \text { (q) } \log [\mathrm{A}]-\frac{-\mathrm{kt}}{2.303}+\log \left[\mathrm{A}_{0}\right] \\ \text { (c) Integrated rate } \\ \begin{array}{ll} \text { law for first } \\ \text { order reaction } \end{array} & \text { (r) } \frac{\mathrm{A}_{0}}{2 \mathrm{k}} \\ \end{array} $$
5 step solution
Problem 184
Match the following $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-II } \\ \hline \begin{array}{ll} \text { (a) } t_{1 / 2}=\frac{0.693}{\mathrm{k}} & \text { (p) Zero order } \\\ \text { (b) } t_{1 / 2}-\frac{\mathrm{a}}{2 \mathrm{k}} & \text { (q) First order } \\ \text { (c) } \mathrm{t}=\frac{1}{\mathrm{k}} & \text { (r) Average life } \\ \text { (d) } \mathrm{t}_{\frac{1}{2}}=\frac{1}{\mathrm{ak}} & \text { (s) Second order } \\ & \end{array} \\ \hline \end{array} $$
4 step solution
Problem 186
Assertion: For a reaction \(\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})\) $$ \begin{aligned} -\mathrm{r}_{\mathrm{A}} &=2.5 \mathrm{P}_{\mathrm{A}} \text { at } 400 \mathrm{~K} \\ -\mathrm{r}_{\mathrm{A}} &=2.5 \mathrm{P}_{\mathrm{A}} \text { at } 600 \mathrm{~K} \end{aligned} $$ Activation energy is \(4125 \mathrm{~J} / \mathrm{mol}\) Reason: Since for any reaction values of rate constant at two different temperatures is same therefore activation energy of the reaction is zero.
5 step solution
Problem 187
Assertion: For a first order reaction, A (g) \(\rightarrow\) Product the time required to reduce successively the concentration of reactant by a constant fraction is always same. Reason: At any instant, the rate of a first order reaction is given by \(\mathrm{k}[\mathrm{A}]\).
4 step solution
Problem 189
The \(\mathrm{t}_{1 / 2}\) for the decomposition of \(\mathrm{CH}_{3} \mathrm{CHO}\) at constant temperature and at initial pressure of \(340 \mathrm{~mm}\) and 170 \(\mathrm{mm}\) of \(\mathrm{Hg}\) were 410 and \(820 \mathrm{~s}\) respectively. The order of the reaction is
4 step solution
Problem 190
The rate of a first order reaction at \(20 \mathrm{~min}\) is \(0.55 \mathrm{~mol}\) \(\mathrm{L}^{-1} \min ^{-1}\) and \(0.055 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) at 40 min after initi- ation. Find half life of the reaction in minutes.
4 step solution
Problem 191
For the reaction \(\mathrm{A} \rightarrow\) Product, the following data were got. \([\mathbf{A}] \quad \frac{\mathbf{d}[\mathbf{A}]}{\mathbf{d t}}\) (i) 1 1 (ii) 2 8 (iii) 3 27 The order of the reaction is
5 step solution
Problem 192
In gaseous reactions important for understanding the upper atmosphere, \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{O}\) react bimolecularly to form two OH radicals. \(\Delta \mathrm{H}\) for this reaction is \(72 \mathrm{~kJ}\) at \(500 \mathrm{~K}\) and \(\mathrm{Ea}=77 \mathrm{~kJ} \mathrm{~mol}^{-1}\), then \(\mathrm{E}_{9}\) for the bimolecu- lar recombination of \(2 \mathrm{OH}\) radicals to form \(\mathrm{H}_{2} \mathrm{O} \& \mathrm{O}\) at \(500 \mathrm{~K}\) is
3 step solution
Problem 194
Two first order reactions \(\mathrm{A}\) and \(\mathrm{B}\) have the same frequency factor and activation energy of \(\mathrm{A}\) exceeds that of \(\mathrm{B}\) by \(10.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\). If \(\mathrm{A}\) is \(30 \%\) complete in \(60 \mathrm{~min}\). at \(100^{\circ} \mathrm{C}\), how long will the \(\mathrm{B}\) take for \(70 \%\) decomposition. [Given : \(\log 3=0.4771, \log 7=0.845\), antilog \(0.4633=2.9]\)
6 step solution
Problem 195
The half life period and initial concentration of a reaction are as follows. What is order of reaction? $$ \begin{array}{llll} \hline \text { Intitial cocentration } & 350 & 540 & 158 \\ \hline \mathbf{t}_{1 / 2} & 425 & 275 & 941 \\ \hline \end{array} $$
4 step solution
Problem 198
A reaction, \(\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{C}(\mathrm{g})+\mathrm{D}(\mathrm{g})\) was studied using an initial concentration of B which was \(1.5\) times that of \(\mathrm{A}\). But the equilibrium concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) were found to be equal. What is the value of \(K_{p}\) for above equilibrium?
7 step solution
Problem 199
Units of rate constant for the first and zero-order reactions in terms of molarity \(\mathrm{M}\), units are respectively? (a) \(\mathrm{s}^{-1}, \mathrm{Ms}^{-1}\) (b) \(\mathrm{s}^{-1}, \mathrm{M}\) (c) \(\mathrm{M} \mathrm{s}^{-1}, \mathrm{~s}^{-1}\) (d) \(\mathrm{M}, \mathrm{s}^{-1}\)
4 step solution
Problem 200
For a reaction \(\mathrm{A}+2 \mathrm{~B} \longrightarrow \mathrm{C}\), rate is given by \(+\mathrm{d}[\mathrm{C}] / \mathrm{dt}=k[\mathrm{~A}][\mathrm{B}]\), hence the order of the reaction is (a) 3 (b) 2 (c) 1 (d) 0
5 step solution
Problem 201
Consider the following two reactions \(\mathrm{A} \longrightarrow\) Product \(-\mathrm{d}[\mathrm{A}] / \mathrm{dt}=k_{1}[\mathrm{~A}]^{0}\) \(\mathrm{B} \longrightarrow\) Product \(-\mathrm{d}[\mathrm{B}] / \mathrm{dt}=\mathrm{k}_{2}[\mathrm{~B}]\) \(k_{1}\), and \(k_{2}\) are expressed in terms of molarity (mol \(\mathrm{L}^{-1}\) ) and time \(\left(\mathrm{s}^{-1}\right)\) as (a) \(\mathrm{s}^{-1}, \mathrm{M} \mathrm{s}^{-1} \mathrm{~L}^{-1}\) (b) \(\mathrm{Ms}^{-1}, \mathrm{M} \mathrm{s}^{-1}\) (c) \(\mathrm{s}^{-1} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) (d) \(\mathrm{M} \mathrm{s}^{-1}, \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
4 step solution
Problem 204
For the reaction system \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})\) volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to \(\mathrm{O}_{2}\) and second order with respect to \(\mathrm{NO}_{2}\), the rate of reaction will (a) diminish to one fourth of its initial value (b) diminish to one eighth of its initial value (c) increase to eight times of its initial value (d) increase to four times of its initial value
4 step solution
Problem 206
In a first-order reaction, the concentration of the reactant, decreases from \(0.8 \mathrm{M}\) to \(0.4 \mathrm{M}\) in 15 minutes. The time taken for the concentration to change from \(0.1 \mathrm{M}\) to \(0.025 \mathrm{M}\) is (a) 30 minutes (b) 60 minutes (c) \(7.5\) minutes (d) 15 minutes
5 step solution
Problem 208
Consider an endothermic reaction \(\mathrm{X} \longrightarrow \mathrm{Y}\) with the activation energies \(\mathrm{E}_{b}\) and \(\mathrm{E}_{\mathrm{f}}\) for the backward and forward reactions respectively. In general, (a) \(\mathrm{E}_{b}<\mathrm{E}_{f}\) (b) \(E_{h}>E_{f}\) (c) \(\mathrm{E}_{b}-\mathrm{E}_{f}\) (d) there is no definite relation between \(\mathrm{E}_{\mathrm{b}}\) and \(\mathrm{E}_{\mathrm{t}}\)
3 step solution
Problem 209
\(\mathrm{t}_{1}\) can be taken as the time taken for the concentration of a reactant to drop to \(\frac{3}{4}\) of its initial value. If the rate constant for a first order reaction is then \(\mathrm{t}_{1 / 4}\) can be written as (a) \(0.1 / k\) (b) \(0.29 / k\) (c) \(0.69 / k\) (d) \(0.75 / k\)
6 step solution
Problem 210
A reaction involving two different reactants can never be? (a) Unimoleculur reaction (b) Ist order reaction (c) IInd order reaction (d) Bimoleculur reaction
3 step solution
Problem 211
The following mechanism has been proposed for the reaction of \(\mathrm{NO}\) with \(\mathrm{Br}_{2}\) to form \(\mathrm{NO} \mathrm{Br}\) \(\mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NOBr}_{2}(\mathrm{~g})\) \(\mathrm{NOBr}_{2}(\mathrm{~g})+\mathrm{NO}(\mathrm{g}) \longrightarrow 2 \mathrm{NOBr}(\mathrm{g})\) If the second step is the rate determining step, the order of the reaction with respect to \(\mathrm{NO}(\mathrm{g})\) is (a) 1 (b) 0 (c) 3 (d) 2
3 step solution
Problem 212
A reaction was found to be second-order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will (a) remain unchanged (b) triple (c) increase by a factor of 4 (d) double
5 step solution
Problem 213
Rate of a reaction can be expressed by Arrhenius 2 equation as, \(k=\mathrm{Ae}^{\mathrm{E} / \mathrm{RT}}\) In this equation, E represents (a) the energy above which all the colliding molecules will react (b) the energy below which colliding molecules will not react (c) the total energy of the reacting molecules at a tem perature, T (d) the fraction of molecules with energy greater than the activation energy of the reaction
4 step solution
Problem 214
The energies of activation for forward and reverse reactions for \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}\) are \(180 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(200 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by \(100 \mathrm{~kJ} \mathrm{~mol}^{-1}\). The enthalpy change of the reaction \(\left(\mathrm{A}_{2}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB}\right)\) in the presence of catalyst will be (in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) \(\quad\) (a) 120 (b) 280 (c) 20 (d) 300
3 step solution