Chapter 18

43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry · 89 exercises

Problem 2

A flask contains a mixture of compounds \(\mathrm{A}\) and \(\mathrm{B}\). Both compounds decompose by first-order kinetics. The half-lives for \(\mathrm{A}\) and \(\mathrm{B}\) are 300 \(\mathrm{s}\) and \(180 \mathrm{~s}\), respectively. If the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are equal initially, the time required for the concentration of \(A\) to be four times that of \(\mathrm{B}\) (in s) is: (Use \(\ln 2=0.693\) ) [Main Sep. \(\mathbf{0 5}, \mathbf{2 0 2 0}\) (I)](a) 180 (b) 900 (c) 300 (d) 120

5 step solution

Problem 2

For following reactions: \(\mathrm{A} \stackrel{700 \mathrm{~K}}{\longrightarrow}\) Product \(\mathrm{A} \frac{500 \mathrm{~K}}{\text { catalyst }}{\longrightarrow}\) Product it was found that the \(\mathrm{E}_{\mathrm{a}}\) is decrease by \(30 \mathrm{~kJ} / \mathrm{mol}\) in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same): (a) \(75 \mathrm{~kJ} / \mathrm{mol}\) (b) \(105 \mathrm{~kJ} / \mathrm{mol}\) (c) \(135 \mathrm{~kJ} / \mathrm{mol}\) (d) \(198 \mathrm{~kJ} / \mathrm{mol}\)

5 step solution

Problem 2

A positron is emitted from \({ }_{11}^{23} \mathrm{Na}\). The ratio of the atomic mass and atomic number of the resulting nuclide is (a) \(22 / 10\) (b) \(22 / 11\) (c) \(23 / 10\) (d) \(23 / 12\)

4 step solution

Problem 3

It is true that:(a) A second order reaction is always a multistep reaction (b) A zero order reaction is a multistep reaction (c) A first order reaction is always a single step reaction (d) A zero order reaction is a single step reaction

5 step solution

Problem 3

The rate of a certain biochemical reaction at physiological temperature \((T)\) occurs \(10^{6}\) times faster with enzyme than without. The change in the activation energy upon adding enzyme is: (a) \(-6(2.303) \mathrm{RT}\) (b) \(-6 \mathrm{RT}\) (c) \(+6(2.303) \mathrm{RT}\) (d) \(+6 \mathrm{RT}\)

7 step solution

Problem 3

\({ }^{23} \mathrm{Na}\) is the more stable isotope of \(\mathrm{Na}\). Find out the process by which \({ }_{1}^{24}\) Na can undergo radioactive decay (a) \(\beta\) - emission (b) \(\alpha\) emission (c) \(\beta^{+}\)emission (d) \(\mathrm{K}\) electron capture

5 step solution

Problem 4

For the reaction \(2 \mathrm{~A}+3 \mathrm{~B}+\frac{3}{2} \mathrm{C} \rightarrow 3 \mathrm{P}\), which statement is correct ? [Main Sep. \(\mathbf{0 3}, \mathbf{2 0 2 0}\) (II)](a) \(\frac{\mathrm{dn}_{\mathrm{A}}}{\mathrm{dt}}=\frac{3}{2} \frac{\mathrm{dn}_{\mathrm{B}}}{\mathrm{dt}}=\frac{3}{4} \frac{\mathrm{dn}_{\mathrm{C}}}{\mathrm{dt}}\) (b) \(\frac{\mathrm{dn}_{\mathrm{A}}}{\mathrm{dt}}=\frac{\mathrm{dn}_{\mathrm{B}}}{\mathrm{dt}}=\frac{\mathrm{d} n_{\mathrm{C}}}{\mathrm{dt}}\) (c) \(\frac{\mathrm{dn}_{\mathrm{A}}}{\mathrm{dt}}=\frac{2}{3} \frac{\mathrm{dn}_{\mathrm{B}}}{\mathrm{dt}}=\frac{4}{3} \frac{\mathrm{dn}_{\mathrm{C}}}{\mathrm{dt}}\) (d) \(\frac{\mathrm{dn}_{\mathrm{A}}}{\mathrm{dt}}=\frac{2}{3} \frac{\mathrm{dn}_{\mathrm{B}}}{\mathrm{dt}}=\frac{3}{4} \frac{\mathrm{dn}_{\mathrm{C}}}{\mathrm{dt}}\)

4 step solution

Problem 4

The number of neutrons accompanying the formation of \({ }_{54}^{139} \mathrm{Xe}\) and \({ }_{38}^{94} \mathrm{Sr}\) from the absorption of a slow neutron by \({ }_{92}^{235} \mathrm{U}\), followed by nuclear fission is, (a) 0 (b) 2 (c) 1 (d) 3

4 step solution

Problem 5

\begin{tabular}{|c|c|c|c|} \hline Experiment & \([\mathrm{A}] /\) \(\mathrm{molL}^{-1}\) & \([\mathrm{B}] /\) \(\mathrm{molL}^{-1}\) & Initial rate/ \(\mathrm{moL}^{-1} \mathrm{~min}^{-1}\) \\\ \hline I & \(0.1\) & \(0.1\) & \(6.00 \times 10^{-3}\) \\ \hline II & \(0.1\) & \(0.2\) & \(2.40 \times 10^{-2}\) \\ \hline II & \(0.2\) & \(0.1\) & \(1.20 \times 10^{-2}\) \\ \hline IV & \(\mathrm{X}\) & \(0.2\) & \(7.20 \times 10^{-2}\) \\ \hline V & \(0.3\) & \(\mathrm{Y}\) & \(2.88 \times 10^{-1}\) \\ \hline \end{tabular} \(\mathrm{X}\) and \(\mathrm{Y}\) in the given table are respectively: (a) \(0.4,0.4\) (b) \(0.4,0.3\) (c) \(0.3,0.4\) (d) \(0.3,0.3\)The results given in the below table were obtained during kinetic studies of the following reaction: \(2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}\) [Main Sep. 02, 2020 (II)]

7 step solution

Problem 5

For the reaction of \(\mathrm{H}_{2}\) with \(\mathrm{I}_{2}\), the rate constant is \(2.5 \times 10^{-4} \mathrm{dm}^{3} \mathrm{~mol}^{-1} \mathrm{~s}^{-}\) \({ }^{1}\) at \(327{ }^{\circ} \mathrm{C}\) and \(1.0 \mathrm{dm}^{3} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\) at \(527^{\circ} \mathrm{C}\). The activation energy for the reaction, in \(\mathrm{kJ} \mathrm{mol}^{-1}\) is : \(\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\) (a) 166 (b) 150 (c) 72 (d) 59

6 step solution

Problem 6

The half-life period of a radioactive element is 140 days. After 560 days, one gram of the element will reduced to : (a) \(\frac{1}{2} \mathrm{~g}\) (b) \(\frac{1}{4} \mathrm{~g}\) (c) \(\frac{1}{8} \mathrm{~g}\) (d) \(\frac{1}{16} \mathrm{~g}\)

4 step solution

Problem 8

\(\mathrm{NO}_{2}\) required for a reaction is produced by the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}\) as per the equation, \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\)The initial concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(3.00 \mathrm{~mol} \mathrm{~L}^{-1}\) and it is \(2.75 \mathrm{~mol} \mathrm{~L}^{-1}\) after 30 minutes. The rate of formation of \(\mathrm{NO}_{2}\) is : [Main April 12, \(\mathbf{2 0 1 9} \mathbf{( I I )}]\)(a) \(4.167 \times 10^{3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{1}\) (b) \(1.667 \times 10^{2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{1}\) (c) \(8.333 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) (d) \(2.083 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\)

5 step solution

Problem 8

Two reactions \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) have identical pre- exponential factors. Activation energy of \(\mathrm{R}_{1}\) exceeds that of \(\mathrm{R}_{2}\) by \(10 \mathrm{~kJ} \mathrm{~mol}^{-1}\). If \(k_{1}\) and \(k_{2}\) are rate constants for reactions \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) respectively at \(300 \mathrm{~K}\), then \(\ln \left(k_{2} / k_{1}\right)\) is equal to : \(\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) 8 (b) 12 (c) 6 (d) 4

7 step solution

Problem 8

If uranium (mass number 238 and atomic number 92) emits an \(\alpha\) particle, the product has mass no. and atomic no. (a) 236 and 92 (b) 234 and 90 (c) 238 and 90 (d) 236 and 90

4 step solution

Problem 9

The rate of a reaction quadruples when the temperature changes from 300 to \(310 \mathrm{~K}\). The activation energy of this reaction is : (Assume activation energy and pre-exponential factor are independent of temperature; \(\left.\ln 2=0.693 ; \mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) \(107.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(53.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(26.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(214.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

6 step solution

Problem 10

The rate coefficient \((k)\) for a particular reactions is \(1.3 \times 10^{-4} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\), and \(1.3 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(150{ }^{\circ} \mathrm{C}\). What is the energy of activation \(\left(E_{\mathrm{a}}\right)\) (in \(\mathrm{kJ}\) ) for this reaction? \((\mathrm{R}=\) molar gas constant \(=\) \(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) ) (a) 16 (b) 60 (c) 99 (d) 132

7 step solution

Problem 10

The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element \(X\) as shown below. To which group, element \(X\) belongs in the periodic table? [2012] \({ }_{29}^{63} \mathrm{Cu}+{ }_{1}^{1} \mathrm{H} \rightarrow 6_{0}^{1} n+{ }_{2}^{4} \alpha+2{ }_{1}^{1} \mathrm{H}+X\)

4 step solution

Problem 11

The rate of a reaction doubles when its temperature changes from \(300 \mathrm{~K}\) to \(310 \mathrm{~K}\). Activation energy of the reaction will be : \((\mathrm{R}=\) \(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) and \(\log 2=0.301\) ) (a) \(53.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(48.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(58.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(60.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

8 step solution

Problem 11

The number of neutrons emitted when \({ }_{92}^{235} \mathrm{U}\) undergoes controlled nuclear fission to \({ }_{54}^{142} \mathrm{Xe}\) and \({ }_{38}^{90} \mathrm{Sr}\) is

5 step solution

Problem 12

The reaction \(2 \mathrm{X} \square \mathrm{B}\) is a zeroth order reaction. If the initial concentration of \(X\) is \(0.2 \mathrm{M}\), the half-life is \(6 \mathrm{~h}\). When the initial concentration of \(X\) is \(0.5 \mathrm{M}\), the time required to reach its final concentration of \(0.2 \mathrm{M}\) will be : [Main Jan. 11, 2019 (II)](a) \(9.0 \mathrm{~h}\) (b) \(12.0 \mathrm{~h}\) (c) \(18.0 \mathrm{~h}\) (d) \(7.2 \mathrm{~h}\)

5 step solution

Problem 12

The reaction \(X \rightarrow Y\) is an exothermic reaction. Activation energy of the reaction for \(\mathrm{X}\) into \(\mathrm{Y}\) is \(150 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Enthalpy of reaction is 135 \(\mathrm{kJ} \mathrm{mol}^{-1}\). The activation energy for the reverse reaction, \(\mathrm{Y} \rightarrow \mathrm{X}\) will be : (a) \(280 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(285 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(270 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(15 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

4 step solution

Problem 12

The total number of \(\alpha\) and \(\beta\) particles emitted in the nuclear reaction \({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{82}^{214} \mathrm{~Pb}\) is

5 step solution

Problem 13

For an elementary chemical reaction, \(\mathrm{A}_{2} \frac{\mathrm{k}_{\mathrm{l}}}{\mathrm{k}_{-1}} 2 \mathrm{~A}\),the expression for \(\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\) is:(a) \(\mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]-\mathrm{k}_{-1}[\mathrm{~A}]^{2}\) (b) \(2 \mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]-\mathrm{k}_{-1}[\mathrm{~A}]^{2}\) (c) \(\mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]+\mathrm{k}_{-1}[\mathrm{~A}]^{2}\) (d) \(2 \mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]-2 \mathrm{k}_{-1}[\mathrm{~A}]^{2}\)

3 step solution

Problem 13

\({ }_{92}^{238} \mathrm{U}\) is known to undergo radioactive decay to form \({ }_{82}^{206} \mathrm{~Pb}\) by emitting alpha and beta particles. A rock initially contained \(68 \times 10^{-6}\) g of \({ }_{92}^{238} \mathrm{U}\). If the number of alpha particles that it would emit during its radioactive decay of \({ }_{92}^{238} \mathrm{U}\) to \({ }_{82}^{206} \mathrm{~Pb}\) in three half- lives is \(Z \times 10^{18}\), then what is the value of \(Z\) ?

6 step solution

Problem 14

At \(518{ }^{\circ} \mathrm{C}\), the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was \(1.00\) Torr \(\mathrm{s}^{-1}\) when \(5 \%\) had reacted and \(0.5\) Torr \(\mathrm{s}^{-1}\) when \(33 \%\) had reacted. The order of the reaction is : [Main 2018] (a) 2 (b) 3 (c) 1 (d) 0

8 step solution

Problem 14

For a first order reaction \(A \rightarrow P\), the temperature ( \(T\) ) dependent rate constant \((k)\) was found to follow the equation \(\log k=-(2000)\) \(\frac{1}{T}+6.0 .\) The pre-exponential factor \(A\) and the activation energy \(E_{a}\), respectively, are (a) \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) and \(9.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(6.0 \mathrm{~s}^{-1}\) and \(16.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) and \(16.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) and \(38.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

6 step solution

Problem 14

One of the hazards of nuclear explosion is the generation of \(90 \mathrm{Sr}\) and its subsequent incorporation in bones. This nuclide has a half-life of 28.1 years. Suppose one microgram was absorbed by a new-born child, how much \({ }^{90} \mathrm{Sr}\) will remain in his bones after 20 years?

7 step solution

Problem 15

If \(50 \%\) of a reaction occurs in 100 seconds and \(75 \%\) of the reaction occurs in 200 seconds, the order of this reaction is: [Main Online April 16, 2018] (a) 2 (b) 3 (c) Zero (d) 1

4 step solution

Problem 15

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 \(6.0\) \(\times 10^{14} \mathrm{~s}^{-1}\) respectively. The value of the rate constant as \(T \rightarrow \infty\) 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 16

Decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) follows a first order reaction. In fifty minutes the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) decreases from \(0.5\) to \(0.125 \mathrm{M}\) in one such decomposition. When the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) reaches \(0.05 \mathrm{M}\), the rate of formation of \(\mathrm{O}_{2}\) will be: [Main 2016] (a) \(2.66 \mathrm{~L} \mathrm{~min}^{-1}\) at STP (b) \(1.34 \times 10^{-2} \mathrm{~mol} \mathrm{~min}^{-1}\) (c) \(6.96 \times 10^{-2} \mathrm{~mol} \min ^{-1}\) (d) \(6.93 \times 10^{-4} \mathrm{~mol} \mathrm{~min}^{-1}\)

5 step solution

Problem 16

A catalyst is a substance which (a) increases the equilibrium concentration of the product (b) changes the equilibrium constant of the reaction (c) shortens the time to reach equilibrium (d) supplies energy to the reaction

3 step solution

Problem 17

The rate law for the reaction below is given by the expression \(\mathrm{k}[\mathrm{A}]\) [B] \(\mathrm{A}+\mathrm{B} \rightarrow\) Product If the concentration of \(\mathrm{B}\) is increased from \(0.1\) to \(0.3 \mathrm{~mole}\), keeping the value of \(A\) at \(0.1\) mole, the rate constant will be: [Main Online April 10, 2016] (a) \(3 k\) (b) \(9 k\) (c) \(k / 3\) (d) \(k\)

5 step solution

Problem 17

The rate of a reaction decreased by \(3.555\) times when the temperature was changed from \(40^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\). The activation energy (in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) of the reaction is \(.\) Take; \(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) In \(\overline{3.555}=1.268\)

6 step solution

Problem 18

Higher order \((>3)\) reactions are rare due to : [Main 2015] (a) shifting of equilibrium towards reactants due to elastic collisions (b) loss of active species on collision (c) low probability of simultaneous collision of all the reacting species (d) increase in entropy and activation energy as more molecules are involved

4 step solution

Problem 18

The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from \(27^{\circ} \mathrm{C}\) to \(42^{\circ} \mathrm{C}\). Its energy of activation in \(\mathrm{J} / \mathrm{mol}\) is (Take \(\left.\ln 5=1.6094 ; \mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\)

4 step solution

Problem 19

\(\mathrm{A}+2 \mathrm{~B} \rightarrow \mathrm{C}\), the rate equation for this reaction is given as Rate \(=\mathrm{k}[\mathrm{A}]\) [B] [Main Online April 11, 2015] If the concentration of \(\mathrm{A}\) is kept the same but that of \(\mathrm{B}\) is doubled what will happen to the rate itself? (a) halved (b) the same (c) doubled (d) quadrupled

2 step solution

Problem 20

For the non-stoichiometric reaction \(2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}\), the following kinetic data were obtained in three separate experiments, all at \(298 \mathrm{~K}\). \begin{tabular}{|c|c|c|} \hline Initial Concentration (A) & Initial Concentration (B) & Initial rate of formation of C \(\left(\mathbf{m o l} \mathbf{L}^{-1} \mathbf{s}^{-1} \mathbf{)}\right.\) \\ \hline \(0.1 \mathrm{M}\) & \(0.1 \mathrm{M}\) & \(1.2 \times 10^{-3}\) \\ \hline \(0.1 \mathrm{M}\) & \(0.2 \mathrm{M}\) & \(1.2 \times 10^{-3}\) \\ \hline \(0.2 \mathrm{M}\) & \(0.1 \mathrm{M}\) & \(2.4 \times 10^{-3}\) \\ \hline \end{tabular}The rate law for the formation of \(\mathrm{C}\) is: [Main 2014] (a) \(\frac{d \mathrm{c}}{d t}=k[\mathrm{~A}][\mathrm{B}]\) (b) \(\frac{d \mathrm{c}}{d t}=k[\mathrm{~A}]^{2}[\mathrm{~B}]\)(c) \(\frac{d \mathrm{c}}{d t}=k[\mathrm{~A}][\mathrm{B}]^{2}\) (d) \(\frac{d \mathrm{c}}{d t}=k[\mathrm{~A}]\)

4 step solution

Problem 21

The half-life period of a first order reaction is 15 minutes. The amount of substance left after one hour will be: [Main Online April 9, 2014] (a) \(\frac{1}{4}\) of the original amount (b) \(\frac{1}{8}\) of the original amount (c) \(\frac{1}{16}\) of the original amount (d) \(\frac{1}{32}\) of the original amount

5 step solution

Problem 21

A hydrogenation reaction is carried out at \(500 \mathrm{~K}\). If same reaction is carried out in the presence of a catalyst at the same rate, the temperature required is \(400 \mathrm{~K}\). Calculate the activation energy of the reaction if the catalyst lowers the activation barrier by \(20 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

3 step solution

Problem 21

In the nuclear transmutation $$ { }_{4}^{9} \mathrm{Be}+X \rightarrow{ }_{4}^{8} \mathrm{Be}+Y $$ \((\mathrm{X}, \mathrm{Y})\) is (are) (a) \((\gamma, \mathrm{n})\) (b) (p, D) (c) (n, D) (d) \((\gamma, \mathrm{p})\)

4 step solution

Problem 22

For the elementary reaction \(\boldsymbol{M} \rightarrow \boldsymbol{N}\), the rate of disappearance of \(\boldsymbol{M}\) increases by a factor of 8 upon doubling the concentration of \(\boldsymbol{M}\). The order of the reaction with respect to \(\boldsymbol{M}\) is [Adv. 2014](a) 4 (b) 3 (c) 2 (d) 1

3 step solution

Problem 22

At \(380^{\circ} \mathrm{C}\), the 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 \(450^{\circ} \mathrm{C}\).

6 step solution

Problem 22

Decrease in atomic number is observed during (a) alpha emission (b) beta emission (c) positron emission (d) electron capture.

5 step solution

Problem 23

The rate constant of a zero order reaction is \(2.0 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). If the concentration of the reactant after 25 seconds is \(0.5 \mathrm{M}\). What is the initial concentration? [Main Online April 23, 2013](a) \(0.5 \mathrm{M}\) (b) \(1.25 \mathrm{M}\) (c) \(12.5 \mathrm{M}\) (d) \(1.0 \mathrm{M}\)

4 step solution

Problem 23

In the Arrhenius equation for a certain reaction, the value of \(A\) and \(E_{a}\) (activation energy) are \(4 \times 10^{13} \mathrm{sec}^{-1}\) and \(98.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. If the reaction is of first order, at what temperature will its half-life period be ten minutes?

5 step solution

Problem 24

Why do we use the carbon dating to calculate the age of the fossil? \([2006-5 \mathrm{M},-2]\) (a) Rate of exchange of carbon between atmosphere and living is slower than decay of \(\mathrm{C}^{14}\) (b) It is not appropriate to use \(\mathrm{C}^{14}\) dating to determine age (c) Rate of exchange of \(\mathrm{C}^{14}\) between atmosphere and living organism is so fast that an equilibrium is set up between the intake of \(\mathrm{C}^{14}\) by organism and its exponential decay (d) none of the above

3 step solution

Problem 25

Under the same reaction conditions, initial concentration of \(1.386 \mathrm{~mol}\) \(\mathrm{dm}^{-3}\) of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio \(\left(k_{1} / k_{0}\right)\) of the rate constant for first order \(\left(k_{1}\right)\) and zero order \(\left(k_{0}\right)\) of the reaction is \(-\) [2008](a) \(0.5 \mathrm{~mol}^{1} \mathrm{dm}^{3}\) (b) \(1.0 \mathrm{~mol} \mathrm{dm}^{3}\) (c) \(1.5 \mathrm{~mol} \mathrm{dm}^{-3}\) (d) \(2.0 \mathrm{~mol}^{-1} \mathrm{dm}^{3}\)

6 step solution

Problem 26

Consider a reaction \(a G+b H \rightarrow\) Products. When concentration of both the reactants \(G\) and \(H\) is doubled, the rate increases by eight times. However, when concentration of \(G\) is doubled keeping the concentration of \(H\) fixed, the rate is doubled. The overall order of the reaction is [2007] (a) 0 (b) 1 (c) 2 (d) 3

5 step solution

Problem 26

A nuclear explosion has taken place leading to increase in concentration of \(C^{14}\) in nearby areas. \(C^{14}\) concentration is \(C_{1}\) in nearby areas and \(C_{2}\) in areas far away. If the age of the fossil is determined to be \(T_{1}\) and \(T_{2}\) at the respective places then (a) The age of the fossil will increase at the place where explosion has taken place and \(T_{1}-T_{2}=\frac{1}{\lambda} \ln \frac{C_{1}}{C_{2}}\) (b) The age of the fossil will decrease at the place where explosion has taken place and \(T_{1}-T_{2}=\frac{1}{\lambda} \ln \frac{C_{1}}{C_{2}}\) (c) The age of fossil will be determined to be same (d) \(\frac{T_{1}}{T_{2}}=\frac{C_{1}}{C_{2}}\)

5 step solution

Problem 27

Statement-1 : The plot of atomic number (y-axis) versus number of neutrons (x-axis) for stable nuclei shows a curvature towards \(\mathrm{x}\)-axis from the line of \(45^{\circ}\) slope as the atomic number is increased. Statement-2 : Proton-proton electrostatic repulsions begin to overcome attractive forces involving protons and neutrons in heavier nuclides. (a) Statement- 1 is True, Statement-2 is True; Statement- 2 is a correct explanation for Statement-1. (b) Statement-1 is True, Statement-2 is True; Statement- 2 is NOT a correct explanation for Statement-1 (c) Statement- 1 is True, Statement- 2 is False (d) Statement- 1 is False, Statement- 2 is True

4 step solution

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