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

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