Problem 29
A solution of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}\) yields by decomposition at \(45^{\circ} \mathrm{C}, 4.8 \mathrm{ml}\) of \(\mathrm{O}_{2}\), 20 min after the start of the experiment and \(9.6 \mathrm{ml}\) of \(\mathrm{O}_{2}\) after a very long time. The decomposition obeys first-order kinetics. What volume of \(\mathrm{O}_{2}\) would have evolved, 40 min after the start? (a) \(7.2 \mathrm{ml}\) (b) \(2.4 \mathrm{ml}\) (c) \(9.6 \mathrm{ml}\) (d) \(6.0 \mathrm{ml}\)
Problem 31
A first-order reaction: \(\mathrm{A}(\mathrm{g}) \rightarrow n \mathrm{~B}(\mathrm{~g})\) is started with 'A'. The reaction takes place at constant temperature and pressure. If the initial pressure was \(P_{0}\) and the rate constant of reaction is ' \(K\), then at any time, \(t\), the total pressure of the reaction system will be (a) \(P_{0}\left[n+(1-n) e^{-k t}\right]\) (b) \(P_{0}(1-n) e^{-k t}\) (c) \(P_{0} \cdot n \cdot e^{-k t}\) (d) \(P_{0}\left[n-(1-n) e^{-k t}\right.\)
Problem 32
For a reaction of order \(n\), the integrated form of the rate equation is: \((n-1) \cdot K \cdot t\) \(=\left(C_{0}\right)^{1-n}-(C)^{1-n}\), where \(C_{0}\) and \(C\) are the values of the reactant concentration at the start and after time ' \(t\) '. What is the relationship between \(t_{3 / 4}\) and \(t_{1 / 2}\), where \(t_{3 / 4}\) is the time required for \(C\) to become \(C_{0} / 4\). (a) \(t_{3 / 4}=t_{1 / 2} \cdot\left[2^{n-1}+1\right]\) (b) \(t_{3 / 4}=t_{1 / 2}\left[2^{n-1}-1\right]\) (c) \(t_{3 / 4}=t_{1 / 2}\left[2^{n+1}-1\right]\) (d) \(t_{3 / 4}=t_{1 / 2} \cdot\left[2^{n+1}+1\right]\)
Problem 32
The rate law for a reaction between the substances \(\mathrm{A}\) and \(\mathrm{B}\) is given by rate \(=\) \(K[\mathrm{~A}]^{n}[\mathrm{~B}]^{m} .\) On doubling the concentration of \(\mathrm{A}\) and halving the concentration of \(\mathrm{B}\), the ratio of the new rate to the earlier rate of the reaction will be as (a) \(1 / 2^{m+n}\) (b) \((m+n)\) (c) \((n-m)\) (d) \(2^{(n-m)}\)
Problem 34
For a reaction \(2 \mathrm{~A}+\mathrm{B}+3 \mathrm{C} \rightarrow \mathrm{D}+3 \mathrm{E}\), the following date is obtained: $$ \begin{array}{ccccc} \hline \text { Reaction } & \multicolumn{2}{c} {\text { Concentration in }} & \text { Initial rate of } \\ & \multicolumn{2}{c} {\text { mole per litre }} & \text { formation of } \\ & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D}\left(\text { torr } \mathbf{s}^{-1}\right) \\ \hline 1 & 0.01 & 0.01 & 0.01 & 2.5 \times 10^{-4} \\ 2 & 0.02 & 0.01 & 0.01 & 1.0 \times 10^{-3} \\ 3 & 0.01 & 0.02 & 0.01 & 2.5 \times 10^{-4} \\ 4 & 0.01 & 0.02 & 0.02 & 5.0 \times 10^{-4} \\ \hline \end{array} $$ The order with respect to \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are, respectively, (a) \(0,1,2\) (b) \(2,0,1\) (c) \(1,0,2\) (d) \(2,1,1\)
Problem 35
In the gas phase, two butadiene molecules can dimerizes to give larger molecules according to the reaction: \(2 \mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{~g})\) \(\rightarrow \mathrm{C}_{8} \mathrm{H}_{12}(\mathrm{~g})\). The rate law for this reac- tion is, \(r=K\left[\mathrm{C}_{4} \mathrm{H}_{6}\right]^{2}\) with \(K=6.1 \times 10^{-2}\) \(1 \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\) at the temperature of reaction. The rate of formation of \(\mathrm{C}_{8} \mathrm{H}_{12}\), when the concentration of \(\mathrm{C}_{4} \mathrm{H}_{6}\) is \(0.02 \mathrm{M}\), is (a) \(2.44 \times 10^{-5} \mathrm{Ms}^{-1}\) (b) \(1.22 \times 10^{-5} \mathrm{Ms}^{-1}\) (c) \(1.22 \times 10^{-3} \mathrm{Ms}^{-1}\) (d) \(2.44 \times 10^{-6} \mathrm{Ms}^{-1}\)
Problem 35
When the concentration of 'A' is 0.1 M, it decomposes to give ' \(\mathrm{X}\) ' by a firstorder process with a rate constant of \(6.93 \times 10^{-2} \mathrm{~min}^{-1}\). The reactant 'A', in the presence of catalyst, gives ' \(\mathrm{Y}\) ' by a secondorder mechanism with the rate constant of \(0.2 \mathrm{~min}^{-1} \mathrm{M}^{-1} .\) In order to make half-life of both the processes, same, one should start the second-order reaction with an initial concentration of 'A' equal to (a) \(0.01 \mathrm{M}\) (b) \(2.0 \mathrm{M}\) (c) \(1.0 \mathrm{M}\) (d) \(0.5 \mathrm{M}\)
Problem 36
The half-life periods of two first-order reactions are in the ratio \(3: 2\). If \(t_{1}\) is the time required for \(25 \%\) completion of the first reaction and \(t_{2}\) is the time required for \(75 \%\) completion of the second reaction, then the ratio, \(t_{1}: t_{2}\), is \((\log 3=0.48\), \(\log 2=0.3\) ) (a) \(3: 10\) (b) \(12: 25\) (c) \(3: 5\) (d) \(3: 2\)
Problem 41
A complex reaction: \(2 \mathrm{X}+\mathrm{Y} \rightarrow \mathrm{Z}\), takes
place in two steps
$$
\begin{array}{l}
\mathrm{X}+\mathrm{Y} \stackrel{K_{1}}{\longrightarrow} 2 \mathrm{~W} \\
\mathrm{X}+2 \mathrm{~W} \stackrel{K_{2}}{\longrightarrow} \mathrm{Z}
\end{array}
$$
If \(K_{1}
Problem 45
The suggested mechanism for the reaction: \(\mathrm{CHCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CCl}_{4}(\mathrm{~g})\) \(+\mathrm{HCl}(\mathrm{g})\), is $$ \mathrm{Cl}_{2} \underset{K_{2}}{\stackrel{K_{1}}{\leftrightarrows}} 2 \mathrm{C} \mathrm{C} \text { (fast) } $$ \(\mathrm{CHCl}_{3}+\mathrm{Ci} \stackrel{K_{3}}{\longrightarrow} \mathrm{HCl}+\dot{\mathrm{C}} \mathrm{Cl}_{3}(\mathrm{slow})\) $$ \dot{\mathrm{C}} \mathrm{Cl}_{3}+\mathrm{C} \mathrm{\textrm{l }} \stackrel{K_{4}}{\longrightarrow} \mathrm{CCl}_{4} \text { (fast) } $$ The experimental rate law consistent with the mechanism is (a) rate \(=K_{3}\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]\) (b) rate \(=K_{4}\left[\mathrm{CCl}_{3}\right][\mathrm{Cl}]\) (c) rate \(=K_{\text {eq }}\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]\) (d) rate \(=K_{3} K_{\text {eq }}^{1 / 2}\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]^{1 / 2}\)
