Physical Chemistry

CSIR NET Chemical Science — Chemical Kinetics previous year questions

16 questions · 2011–2019 · all with verified answer · last checked 2026-11-11

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Q1 · Dec 2019 · 2 marks

The oxidation of NO to \mathrm{NO_2} occurs via the mechanism given below.

\mathrm{2NO \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} N_2O_2}; \mathrm{N_2O_2 + O_2 \xrightarrow{k_2} 2NO_2}

\frac{d[\mathrm{NO_2}]}{dt} in the presence of large excess of \mathrm{O_2} can be written as

(a) 2k_1(\mathrm{NO})^2
(b) 2k_1k_2(\mathrm{NO})^2(\mathrm{O_2})
(c) \frac{k_1}{k_2}(\mathrm{NO})^2
(d) 2k_2(\mathrm{NO})^2

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Answer: (a)

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Q2 · Dec 2019 · 2 marks

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Answer: (a)

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Q3 · June 2016 · 2 marks

Experimentally determined rate law for the chemical reaction, \mathrm{2NO_2F \rightarrow 2NO_2 + F_2} is R = k[\mathrm{NO_2F}] The rate determining step consistent with the rate law is

(a) \mathrm{2NO_2F \rightarrow 2NO_2 + F_2}
(b) \mathrm{NO_2F + F \rightarrow NO_2 + F_2}
(c) \mathrm{NO_2F \rightarrow NO_2 + F}
(d) \mathrm{NO_2 + F \rightarrow NO_2F}

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Answer: (c)

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Q4 · Dec 2016 · 2 marks

According to the transition state theory, the plot with slope equal to \frac{\Delta H^{\ddagger}}{R} is

(a) \ln k vs. T
(b) \ln\frac{k}{T} vs T
(c) \ln\frac{k}{T} vs \frac{1}{T}
(d) \ln k vs. \frac{1}{T}

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Answer: (c)

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Q5 · June 2015 · 2 marks

In a photochemical reaction, radicals formed according to the equation

\begin{array}{l} \mathrm{C}_{4} \mathrm{H}_{10}+\mathrm{hv} \sqrt{ } 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{C}_{2} \mathrm{H}_{5}+\mathrm{C}_{2} \mathrm{H}_{5} \xrightarrow{k_{2}} \mathrm{C}_{2} \mathrm{H}_{6}+\mathrm{C}_{2} \mathrm{H}_{4} \end{array}

If I is the intensity of light absorbed, the rate of the overall reaction is proportional to-

(a) I
(b) \mathrm{I}^{1 / 2}
(c) \mathrm{I}\left[\mathrm{C}_{4} \mathrm{H}_{10}\right]
(d) \mathrm{I}^{1 / 2}\left[\mathrm{C}_{4} \mathrm{H}_{10}\right]^{1 / 2}

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Answer: (d)

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Q6 · Dec 2014 · 2 marks

For an enzyme-substrate reaction, a plot between \frac{1}{v} and \frac{1}{[S O]} yields a slope of 40 s . If the enzyme concentration is \mathbf{2 . 5} \mu \mathbf{M}, then the catalytic efficiency of the enzyme is (1) 40 \mathrm{Lmol}^{-\mathbf{1}} \mathrm{s}^{-\mathbf{1}} (2) \mathbf{1 0}^{-\mathbf{4}} \mathbf{~ L m o l}^{-\mathbf{1}} \mathbf{s}^{-\mathbf{1}} (3) \mathbf{1 0}^{\mathbf{7}} \mathbf{~ L m o l}^{\boldsymbol{-} \mathbf{1}} \mathbf{s}^{\mathbf{- 1}} (4) \mathbf{1 0}^{\mathbf{4}} \mathbf{~ L m o l}^{\boldsymbol{-} \mathbf{1}} \mathbf{s}^{\boldsymbol{-} \mathbf{1}} 38. For a gaseous reaction, \mathbf{2 N O}(\mathbf{g})+\mathbf{C l}_{\mathbf{2}}(\mathbf{g})-\mathbf{N o n}-linear T.S.-2NOCl, the pre-exponential factor in the rate constant is proportional to

(a) \mathbf{T}^{\mathbf{1} / \mathbf{2}}
(b) \mathrm{T}^{-1 / 2}
(c) \mathrm{T}^{-5 / 2}
(d) \mathrm{T}^{-7 / 2}

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Answer: (d)

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Q7 · Dec 2014 · 2 marks

The rate constant for a reaction A^{1+}+B^{n+1} \rightarrow P is measured in two different aqueous solutions of ionic strengths \mathbf{0 . 0 1 ~ M} and \mathbf{0 . 0 4 ~ M}. If \boldsymbol{\operatorname { l o g }}_{\mathbf{K 0 . 0 1}} \frac{K 0.04}{\mathbf{K}}=\mathbf{0 . 3}, the charge n on B is closest to

(a) 1
(b) 2
(c) 3
(d) 6

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Answer: (c)

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Q8 · Dec 2013 · 2 marks

A first order gaseous reaction is 25 \% complete in 30 minutes at 227^{\circ} \mathrm{C} and in 10 minutes at 237^{\circ} \mathrm{C}. The activation energy of the reaction is closest to \left(\mathrm{R}=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)

(a) 27 \mathrm{kcal} \mathrm{mol}^{-\mathbf{1}}
(b) \mathbf{1 1 0 ~ k c a l ~ m o l}^{\boldsymbol{-} \mathbf{1}} (3) 55 \mathrm{kcal} \mathrm{mol}^{-\mathbf{1}} (4) 5.5 \mathrm{kcal} \mathrm{mol}^{-1}

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Answer: (c)

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Q9 · Dec 2013 · 2 marks

For a first order reaction \mathrm{A} \rightarrow products, the plot of \mathrm{In}\left(\frac{[A]_{t}}{[A]_{0}}\right) v s time, where [\mathrm{A}]_{0} and [\mathrm{A}]_{t} to cocentration at time =0 and t respectively, is

(a) a straight line with a positive slope passing through origin
(b) a straight line with a negative slope passing through origin.
(c) an exponential curve asymptotic to the time axis
(d) a curve asymptotic to the \operatorname{In}\left(\frac{[A]_{t}}{[A]_{0}}\right) axis.

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Answer: (b)

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Q10 · Dec 2012 · 2 marks

The decomposition of \mathrm{NH}_3 on Mo surface follows Langmuir-Hinshelwood mechanism. The decomposition was carried out at low pressures. The initial pressure of \mathrm{NH}_3 was 10^{-2} torr. The pressure of \mathrm{NH}_3 was reduced to 10^{-4} torr in 10 minutes. The rate constant of decomposition of \mathrm{NH}_3 is:

(a) 9.9 \times 10^{-4}\ \mathrm{torr\ min^{-1}}
(b) 0.4606\ \mathrm{min^{-1}}
(c) 9.9 \times 10^{-3}\ \mathrm{torr\ min^{-1}}
(d) 0.693\ \mathrm{min^{-1}}

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Answer: (b)

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Q11 · Dec 2012 · 2 marks

The rate equation for the reaction 2 \mathrm{AB}+\mathrm{B}_{2} \rightarrow 2 \mathrm{~A} \mathrm{~B}_{2} is given by

\text { rate }=k[A B]\left[B_{2}\right]

A possible mechanism consistent with this rate law is

(a) \mathbf{2 A B}+\mathbf{B}_{\mathbf{2}} \xrightarrow{\text { slow }} \mathbf{2} \mathbf{A B}_{\mathbf{2}}
(b) \mathbf{A B}+\mathbf{A B} \leftrightharpoons \mathbf{A}_{\mathbf{2}} \mathbf{B}_{\mathbf{2}} (fast)

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Answer: (c)

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Q12 · Dec 2011 · 2 marks

One of the assumptions made in the conventional activated complex theory is:

(a) Equilibrium is maintained between reactants and the activated complex
(b) Equilibrium is maintained between the reactants and the products.
(c) Equilibrium is maintained between the products and the activated complex
(d) Equilibrium is maintained between the reactants, the activated complex and the products.

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Answer: (a)

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Q13 · Dec 2011 · 2 marks

An example of a relaxation method of measuring rates is:

(a) Spectroscopic monitoring of product concentration.
(b) Stopped flow technique
(c) Temperature jump experiments.
(d) Measurement of spectral line widths.

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Answer: (c)

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Q14 · Dec 2011 · 2 marks

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Answer: (d)

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Q15 · Dec 2011 · 2 marks

"Phosphorescence" is represented as

(a) T_1 \rightarrow S_0 + h\nu
(b) T_1 \rightarrow S_0 + \Delta
(c) S_1 \rightarrow S_0 + h\nu
(d) S_1 \rightarrow T_1 + \Delta

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Answer: (a)

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Q16 · Dec 2011 · 2 marks

Using cuvettes of 0.5 cm path length, a 10^{-4} M solution of a chromophore shows 50% transmittance at certain wave length. The molar extinction coefficient of the chromophore at this wave length is (\log 2 = 0.301)

(a) 1500\ \mathrm{M^{-1}\,cm^{-1}}
(b) 3010\ \mathrm{M^{-1}\,cm^{-1}}
(c) 5000\ \mathrm{M^{-1}\,cm^{-1}}
(d) 6020\ \mathrm{M^{-1}\,cm^{-1}}

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Answer: (d)

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