Chemical Kinetics MCQ Quiz in తెలుగు - Objective Question with Answer for Chemical Kinetics - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:-

• According to the Lindemann–Hinshelwood mechanism it is supposed that a reactant molecule A becomes energetically excited by collision with another A molecule.

A + A \(\overset{K_a}{\rightarrow} \) A* + A

and the rate of the reaction will be,

\(\frac {d[A^*]}{dt}= K_a[A]^2\) 

• The energized molecule (A*) might lose its excess energy by collision with another molecule:

A + A* \(\overset{K_a^{{}'}}{\rightarrow} \)  A + A (where, \(K_a^{{}'}\) is the rate constant for  deactivation)

and the rate of the deactivation reaction will be,

\(\frac{d[A^{*}]}{dt}=-K_a^{{}'}[A][A^*] \)

• Alternatively, the excited molecule might shake itself apart and form product P. That is, it might undergo the unimolecular decay

A* \(\overset{K_b}{\rightarrow}\) P (where K_b is the rate constant of unimolecular decay)

and the rate of unimolecular decay will be

\(\frac{d[A^*]}{dt}=-K_{b}[A^*]\)

Explanation:-

• If the unimolecular step is slow enough to be the rate-determining step, the overall reaction will have first-order kinetics, as observed. This conclusion can be demonstrated explicitly by applying the steady-state approximation to the net rate of formation of A*:

\(\frac{d[A^*]}{dt}\)= k_a[A]² − k′_a[A][A*] − k_b[A*] ≈ 0

• From the above equation, we got

\(\left [ A^* \right ]=\frac{K_a[A]^2}{K_b+K_a^{{}'}[A]}\)

• Thus, the rate law for the formation of P is

\(\frac{d[P]}{dt}=K_b[A^*]\)

\(=\frac{K_aK_b[A]^2}{K_b+K_a^{{}'}[A]}\)..............(i)

• Iif the rate of deactivation by collisions are much greater than the rate of unimolecular decay, in the sense that

k′_a[A*][A] >> k_b[A*]

or, k′_a[A] >> k_b

• Hence, under this condition from equation (i) we got,

\(\frac{d[P]}{dt}=\frac{K_aK_b[A]^2}{K_a^{{}'}[A]}\)

\(\frac{d[P]}{dt}=\frac{K_aK_b[A]}{K_a^{{}'}}\)

• Thus, the order of the reaction under this condition is one.

Conclusion:-

• Hence, the reaction will be first order if the rate of deactivation by collisions is much greater than the rate of unimolecular decay.

Explanation:

→ In a diffusion-controlled bimolecular reaction, the rate constant is determined by the diffusion of the two reactants towards each other in the solution. The rate at which the reactants diffuse is affected by the viscosity of the solution, with higher viscosity leading to slower diffusion.

→ The Stokes-Einstein equation relates the diffusion coefficient (D) of a molecule in a solution to the viscosity (η) and temperature (T) of the solution, such that:

D = k_BT/6πηr

where kB is the Boltzmann constant, r is the radius of the molecule, and π is the mathematical constant pi.

For a bimolecular reaction, the rate constant (k_D) is proportional to the diffusion coefficient (D), such that:

kD = 4πDσ

where σ is the collision cross-section of the reactants.

Substituting the Stokes-Einstein equation into the expression for kD yields:

kD = 4π(k_BT/6πηr)σ = (4πk_BT/6πηr)σ = (2k_BT/3πηr)σ

Conclusion: Thus, we see that the rate constant for a diffusion-controlled bimolecular reaction is inversely proportional to the viscosity of the solution, and directly proportional to the temperature and the collision cross-section of the reactants. Specifically, we can see that k_D is proportional to t/η, where t represents the temperature (in Kelvin) and η represents the coefficient of viscosity of the medium.

Concept:

Zero-Order Reactions

• A zero-order reaction is characterized by a rate that is independent of the concentration of reactants.
• The rate law is expressed as:

  Rate = k

  where:
  • k = rate constant
• In zero-order reactions, the reactant concentration does not affect the reaction rate, but the rate constant depends on temperature (Arrhenius equation).
• Examples include reactions catalyzed by enzymes or those occurring on solid surfaces, such as the thermal decomposition of HI on a gold surface.

Explanation:

• (A): Correct. The rate of a zero-order reaction is independent of reactant concentration.
• (B): Incorrect. The rate constant is dependent on temperature as per the Arrhenius equation.
• (C): Correct. The rate constant of the reaction is independent of reactant concentration, as concentration does not influence the rate of a zero-order reaction.
• (D): Correct. The thermal decomposition of HI on a gold surface is a classic example of a zero-order reaction.

Correct features: (A), (C), and (D).

Concept:

In a photochemical reaction, the number of photons absorbed by a reactant can be determined using the quantum efficiency (or quantum yield). The quantum efficiency (\(\Phi \)) is defined as the number of molecules reacted per photon absorbed. It is related to the total number of photons absorbed by the system.

The equation for quantum efficiency is given by:

\(\Phi = \frac{\text{molecules reacted}}{\text{photons absorbed}}\)

Rearranging the equation to solve for photons absorbed:

\(\text{Photons absorbed} = \frac{\text{molecules reacted}}{\Phi} \)

Explanation: 

• Given data:

  • Quantum efficiency (\( \Phi \)) = ( \(1.2 \times 10^2 \, \text{mol eistein}^{-1}\) )
  • Moles of B formed = 1.77 mmol = ( \(1.77 \times 10^{-3} \, \text{mol}\) )

• Since each molecule of B formed corresponds to one photon absorbed, the number of photons absorbed is calculated as:

  • \(\text{Photons absorbed} = \frac{1.77 \times 10^{-3} \, \text{mol}}{1.2 \times 10^2 \, \text{mol eistein}^{-1}} \)

  • \(\text{Photons absorbed} = 1.475 \times 10^{-5} \, \text{einstein}\approx 1.5\times 10^{-5} einstein\)

• Therefore, the total number of photons absorbed is approximately:
  • \(1.5 \times 10^{-5} \, \text{einstein}\)

​Conclusion:

The number of photons absorbed by A is \(1.5 \times 10^{-5}\)

Concept:

Catalytic efficiency is a measure of how efficiently an enzyme converts a substrate into a product. It is defined as the ratio of the turnover number (k_cat) to the Michaelis constant (K_m). The equation for catalytic efficiency is given by:

\(\text{Catalytic Efficiency} = \frac{k_{cat}}{K_m}\)

and

\(k_{cat}=\frac{v_{max}}{[E]_o}\)

The Michaelis-Menten equation is used to describe the rate of enzymatic reactions and is expressed as:

\(V_0 = \frac{V_{max} [S]}{K_m + [S]}\)

On simplifying the equation:

\(\frac{1}{V_o}=\frac{1}{V_{max}}+\frac{k_m}{V_{max}[S]}\)

Where:

• ( V₀ ) is the initial reaction velocity.
• ( V_max ) is the maximum reaction velocity.
• ( S ) is the substrate concentration.
• ( K_m ) is the Michaelis constant.

Explanation: 

• Given the intercept is ( \(4 \times 10^3 \, \text{mol}^{-1} \text{dm}^3 \text{s}^{-1}\) )

  • \(\frac{1}{v_{max}}=4 \times 10^3\)

  • \(v_{max}=2.5 \times 10^{-4}\)

• The turnover number ( k_cat ) is:

  • \(k_{cat} = \frac{2.5 \times 10^{-4}}{10^{-4}}= 2.5 \)

• Thus, the catalytic efficiency is:

  • \(\text{Catalytic Efficiency} = \frac{2.5}{0.04} = 62.5 \)

Conclusion:

The calculated catalytic efficiency is 62.5.

Concept:

In chemical reactions involving molecules, the degree of freedom (DOF) is crucial to understanding the energy distribution within the system. These degrees of freedom represent the different ways in which a molecule can store energy (translational, rotational, and vibrational). The total DOF depends on the molecular structure and complexity.

• Linear Complex: For a linear molecule or complex, the total degrees of freedom are given by:

  • f = 3N - 5

  • Where (N) is the number of atoms in the molecule. The DOF includes 3 translational and 2 rotational degrees of freedom.

• Non-Linear Complex: For a non-linear molecule or complex, the total degrees of freedom are given by:

  • f = 3N - 6

• This includes 3 translational and 3 rotational degrees of freedom. The additional DOF for non-linear molecules is due to the complexity of their structure.

• Activated Complex Theory: According to the activated complex theory (or transition state theory), a temporary high-energy intermediate complex is formed during the reaction. This activated complex has a significant number of degrees of freedom, influencing the temperature dependence of the reaction rate. The energy distribution in the activated complex controls the reaction rate and its response to temperature changes.

Explanation: 

•  

  \(\text{A (diatomic)} + \text{B (triatomic)} \xrightarrow{\text{Activated Complex}} \text{(X)}^* \rightarrow \text{C (product)} \)

  •  

    DOF	A(Diatomic, linear)	B(Triatomic, non linear)	C(Pentatomic, non linear)
    Translational	3	3	3
    Rotational	2	3	3
    Vibrational	3	3	9

     

• Rate constant(k) according to activated complex theory:

  • k = \(\frac{k_BT}{h}.\frac{partition\ function\ of\ product}{partition\ function\ of\ reactant}.e^{\frac{-E_a}{RT}}\)

  • \(\frac{k_BT}{h}.\frac{q_v^9.q_t^3.q_r3}{(q_v^3.q_t^3.q_r^2)_A(q_v^3.q_t^3.q_r^3)_B}.e^{\frac{-E_a}{RT}}\)

• \(A = \frac{k_BT}{h}.\frac{q_v^9.q_t^3.q_r3}{(q_v^3.q_t^3.q_r^2)_A(q_v^3.q_t^3.q_r^3)_B}\)

  • on simplifying the equation:

    • \(A=\frac{k_BT}{h}.\frac{q_v^3}{q_t^3.q_r^2}\)

• q_v is temperature independent

• \(q_t \propto T^{1/2}\)

• \(q_r \propto T^{1/2}\)

• Putting the value in the equation:

  • \(A\propto T \times\frac{1}{T^{3/2}.T^{2/2}}\)

  • \(A \propto T^{-3/2}\)

Conclusion:

The temperature dependence of the reaction is directly proportional to (T^-3/2), making the correct answer option 3.

The correct answer is 2, 1, 1

Explanation:-

. \(r=k[P]^{\mathrm{a}}[\mathrm{Q}]^{\mathrm{b}}[\mathrm{R}]^{\mathrm{c}}\)

\(8.0 \times 10^{-5}=k[0.2]^a[0.5]^b[0.4]^c \quad-(I)\)

\(3.2 \times 10^{-4}=k[0.4]^{\mathrm{a}}[0.5]^{\mathrm{b}}[0.4]^{\mathrm{c}}-(\mathrm{II})\)

\(1.28 \times 10^{-3}=\mathrm{k}[0.4]^{\mathrm{a}}[2.0]^{\mathrm{b}}[0.4]^{\mathrm{c}}-(\mathrm{III})\)

\(4.0 \times 10^{-5}=k[0.1]^{\mathrm{a}}[0.25]^{\mathrm{b}}[1.6]^{\mathrm{c}}-(\mathrm{IV})\)

Firstly, compare date (1) & (2) from the table given in the ques, Conc. of Q & R is constant

but doubling the conc. of P results the 4 times increase in rate constant.

This shows [P] follows second order kinetics.

So, a = 2 divide eq. (II) & (III)

\(\frac{I I}{I I I}=\frac{3.20 \times 10^{-4}}{1.28 \times 10^{-3}}=\frac{K[0.4]^a[0.5]^b[0.4]^c}{K[0.4]^a[2.0]^b[0.4]^c}\)

\(\frac{32}{128}=\left[\frac{5}{20}\right]^b\)

\(\left(\frac{1}{4}\right)^1=\left(\frac{1}{4}\right)^b\)

b = 1

again, divide eq. (II) & (IV)

\(\frac{I I}{I V}=\frac{3.2^8 \times 10^{-4}}{4.0 \times 10^{-5}}=\frac{K[0.4]^2[0.5]^1[0.4]^c}{K[0.1]^2\left[0.25[0.6]^c\right.}\)

\(\frac{8}{1}=\frac{0.16 \times 0.50}{0.01 \times 0.25}\left(\frac{0.4}{1.6}\right)^C\)

\(\frac{8}{1}=16 \times 2 \times\left(\frac{0.4}{1.6}\right)^C\)

\(\frac{8}{16 \times 2}=\left(\frac{1}{4}\right)^C\)

\(\left(\frac{1}{4}\right)=\left(\frac{1}{4}\right)^C\)

c = 1 Thus,

a = 2; b =1; c = 1

Conclusion:-

The order of the reaction with respect to P, Q and R respectively is 2, 1, 1

The correct answer is Negative

Explanation:-

According to conventional bimolecular transition state theory

A + B → [AB]≠ → product

• Decrease in Freedom of Movement: When two separate molecules (A and B) come together to form a transition state ([AB]≠), they lose some of their translational freedom as individual entities, leading to a more ordered system.
• Volume Constriction: The space or volume available to the transition state complex is less than that available to the individual molecules when they were separate. This reduction in volume also suggests a decrease in the number of possible microstates the system can access, leading to lower entropy.
• Energy Barrier: The formation of the transition state requires overcoming an energy barrier which implies a certain level of organization and alignment of the reacting molecules, further decreasing the entropy.

Two reactant molecules combine to form a single complex, thus entropy of the system is negative.

Conclusion:-

So,  the molar entropy of activation ΔS≠ is Negative.

The correct answer is Ea+ 2RT

Concept:-Arrhenius Equation: The Arrhenius equation describes the temperature dependence of reaction rates and is widely used in chemical kinetics. It relates the rate constant (k), temperature (T), activation energy (Ea), and the pre-exponential factor (A).

Natural Logarithm: Taking the natural logarithm of both sides of the Arrhenius equation simplifies the equation and allows us to work with the linearized form, making it easier to solve for Ea. The natural logarithm is commonly used in chemistry and mathematics to handle exponential relationships.

Isolation of Activation Energy: By isolating the activation energy term on one side of the equation, we make it the subject of the formula. This rearrangement is crucial for solving for Ea, which is the goal of the problem.

Units: When using the Arrhenius equation, it's essential to ensure that all units are consistent. In this equation, Ea is typically expressed in joules per mole, T in Kelvin, k in reciprocal seconds (s^-1), and A in reciprocal seconds (s^-1). The gas constant R has consistent units of J/(mol·K).

Explanation:-

\(k= AT^2e^{\frac{-E_0}{RT}}\)

\(lnk =ln A +2lnT - \frac{E_0}{RT}\)

\(\frac{d}{dT}lnk=0+ \frac{2}{T}+\frac{E_0}{RT^2}\)

But,  \(\frac{d}{dT}lnk=\frac{E_a}{RT^2} \implies \frac{2}{T}+\frac{E_0}{RT^2} =\frac{E_a}{RT^2} \)

\(E_a=RT^2[\frac{2}{T}+ \frac{E_0}{RT^2}] \implies E_a =2RT+E_0\)

Conclusion:-

So, The Temperature Dependence of a reaction is given by k= AT2 exp(-Eo)/RT. The activation energy Ea of the reaction is given by Ea+ 2RT

The correct answer is \(\frac{8^{n-1}-1}{2^{n-1}-1}\)

Concept:-

Reaction Order and Rates: The order of a reaction refers to the exponent of the concentration term in the rate equation that describes the relationship between the rate of a chemical reaction and the concentrations of its reactants. For an nth-order reaction with only one reactant, it has a general rate law of the form rate = k[A]^n, where [A] is the reactant's concentration, k is the rate constant, and n is the reaction order.

Half-Life of a Reaction: The half-life (t_0.5) of a reaction is the time it takes for the concentration of a reactant to decrease by half. For different orders of reaction, the mathematical expression for half-life varies.

Fractional-Life of a Reaction: The fractional-life (t_x) of a reaction is the time taken for the concentration of the reactant to decrease to x fraction of its initial concentration. In this case, the relevant fractional life is t_0.875.

Explanation:-

\(kt = \frac{1}{n-1}[\frac{1}{A_t^{n-1}}-\frac{1}{A_o^{n-1}}]\\ kt_{0.5} = \frac{1}{n-1} [\frac{2^{n-1}-1}{A_o ^{n-1}}] \\ kt_{0.875} =\frac{1}{n-1}[\frac{8^{n-1}}{A_o^{n-1}}]\\ So,\\ \frac{t_0.875}{t_0.5} = \frac{8^{n-1}-1}{2^{n-1}-1}\)

Conclusion:-