AIIMS Chemistry Chemical Kinetics Class 12 Questions

A(g) → 2B(g) + C(g) Initial: P₀ 0 0 P₀ - P 2P P Pₜ = P₀ - P + 2P + P = P₀ + 2P Therefore, P = (Pₜ + P₀) / 2 Apply the expression shown below to calculate rate constant. k = (2.303 / t) log [P₀ / (P₀ - P)] Substitute the value of P in above equation.

The rate constant can be calculated by using Arrhenius equation as shown below: k = Ae⁻ᴱᵃ/ᴿᵀ = 4 × 10¹⁰ × e⁻⁽⁻⁵⁴.⁷×¹⁰⁰⁰⁾/⁽⁸.³¹⁴×⁸⁰⁰⁾ = 4 × 10¹⁰ × e⁻⁸.²²⁴ = 4 × 10¹⁰ × 2.68 × 10⁻⁴ = 10.7 × 10⁶

The decomposition of NH₃ on Pt surface is a zero order reaction. If the value of rate constant is 2 x 10⁻⁴ mole liter⁻¹ sec⁻¹. The rate of appearance of N₂ and H₂ are respectively:

Find out time period of 1^{st} order reaction. When reaction complete 2/3^{rd}. If the value of rate constant is 4.3×10⁻⁴

lnk = 10 - 69(KJ)/RT ……(i) k = Ae^(-Ea/RT) ln k = ln A - Ea/RT ……(ii) Comparison of equation (i) and(ii) Ea is 69. The rate constants in Arrhenius equation at two different temperatures are expressed as stated below.

What is the activation energy (kJ/mol) for a reaction if its rate constant doubles when the temperature is raised from 300 K to 400 K? (R = 8.314 J mol⁻¹ K⁻¹)

For the reaction A + 2B → C + D The rate of the reaction is^{-1}/1 d[A]/dt = -1/2 d[B]/dt = +d[C]/dt = +d[D]/dt

For the first order reaction, k = \( \frac{2.303}{t} \log \left( \frac{a}{a-x} \right) \) \( k = \frac{2.303}{t} \log a - \frac{2.303}{t} \log (a-x) \) \( \log (a-x) = \log a \left( \frac{-kt}{2.303} \right) \)

Assertion: Some salts are sparingly soluble at room temperature. Reason: The entropy increases on dissolving the salts.

When $0.05 \text{ M}$ dimethyl amine is dissolved in $0.1 \text{ M NaOH}$ solution then the percentage dissociation of dimethyl amine is : $(\text{K}_\text{b})_{(\text{CH}_3)_2\text{NH}} = 5 \times 10^{-4}$

Assertion: If in zero order reaction, the concentration of the reactant is doubled, the half-life period is also doubled. Reason: For a zero order reaction, the rate of the reaction is independent of initial concentration.

Assertion: The rate of reaction is never negative.

For first order reaction as time duration goes from 10 min to 30 min rate of reaction decreases from 0.4 Ms⁻¹ to 0.04 Ms⁻¹. What is the half life of the reaction.

The rate constant for a first order reaction can be calculated as: \( k = rac{2.303}{t} \log rac{a}{a-x} \) Since \( x \) is 80%, therefore,

Which option is valid for zero order reaction?

If the rate of decomposition of N₂O₅, during a certain time interval is 2.4 × 10⁻⁴ mol L⁻¹ min⁻¹ N₂O₅ → 2NO₂ + 1/2 Ω₂ What is the rate of formation of NO₂ and O₂?

At 300 K, activation energy of A is higher than B by 5.75 kJ/mol in presence of catalyst. Calculate \( \frac{K_B}{K_A} \).

log( k₂/k₁ ) = log( A₂/A₁ ) + (Eₐ₂ - Eₐ₁) / (2.3RT) Since Eₐ₂ - Eₐ₁ = 2RT Therefore, log 1 = log( A₂/A₁ ) + 2RT / 2.3RT 0 = log( A₂/A₁ ) + 2RT / 2.3RT ln( A₂/A₁ ) = 2

The difference in activation energy Eₐ – E_b is given below. rₐ/r_b = (A₁e^(-Eₐ/RT))/(A₂e^(-E_b/RT)) 2/1 = e^(-Eₐ/RT)/e^(-E_b/RT) ln 2 = E_b – Eₐ/RT E_b – Eₐ = RT ln 2 The correct reversed answer is Eₐ – E_b = -RT ln 2.

Each of these questions contains two statements, Assertion (A) and Reason (R). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a), (b), (c) and (d) given below. Assertion (A) Chlorine reacts more rapidly with H₂, in comparison to O₂. Reason (R) D⁻Cl bond is stronger in comparison to H⁻Cl bond.

For a zero order reaction, the expression of half life time is, t₁/₂ = [A₀]/2K

For the reaction: $$2\text{NO} + \text{Br}_2 \rightarrow 2\text{NOBr}$$ the following mechanism has been given: $$\text{NO} + \text{Br}_2 \xrightarrow{\text{Fast}} \text{NOBr}_2$$ $$\text{NOBr}_2 + \text{NO} \xrightarrow{\text{Slow}} 2\text{NOBr}$$ Hence, the rate law is:

Consider the below given figure. The correct option for the above presentation is

A first order reaction, which is 30% complete in 30 minutes has a half-life period of

During the decomposition of H₂O₂ to give oxygen, 48 g O₂ is formed per minute at a certain point of time. The rate of formation of water at this point is

If the rate of decomposition of N₂O₅ during a certain time interval is 2.4 × 10⁻⁴ mol L⁻¹ min⁻¹. N₂O₅ → 2NO₂ + 1/2 Ω₂ What is the rate of formation of NO₂ and O₂ mol L⁻¹ min⁻¹?

Assertion (A) Chlorine reacts more rapidly with H₂ in comparison to D₂. Reason (R) D—Cl bond is stronger in comparison to H—Cl bond.

The rate constant for a first order reaction becomes six times when the temperature is raised from 350 K to 400 K. Calculate the activation energy for the reaction. [R = 8.314 J K⁻¹ mol⁻¹]

The plot of a concentration of the reactant versus time for a reaction is a straight line with a negative slope. The reaction follows a

Energy of activation of forward reaction for an endothermic process is 50 kJ. If enthalpy change for forward reaction is 20 kJ then enthalpy change for backward reaction will be

Assertion: Catalyst changes Gibbs free energy of system. Reason: Catalyst changes pre-exponential factor of a chemical reaction.

For a reaction, \( r = K[CH₃COCH₃]^{3/2} \) then unit of rate of reaction and rate constant respectively is

If t₁/₂ ∝ 1/a² is a straight line graph then determine the order of reaction.

Assertion : Rate of reaction doubles when concentration of reactant is doubled if it is a first order reaction. Reason : Rate constant also doubles.

For a first order gas phase reaction- A(g) → 2B(g) + C(g) P₀ be initial pressure of A and Pₜ the total pressure at time 't'. Integrated rate equation is

Assertion : Two different reactions can never have same rate of reaction. Reason : Rate of reaction always depends only on frequency of collision and Arrhenius factor.

Assertion: Two different reactions can never have same rate of reaction.

For a 1^{st} order reaction if concentration is doubled then rate of reaction becomes

For a reaction X → Y, the graph of the product concentration (x) versus time (t) came out to be a straight line passing through the origin. Hence the graph of -d[X]/dt and time would be

Δr ≥ 1×10⁻⁸ m

Assertion: The overall order of the reaction is the sum of the exponents of all the reactants in the rate expression. Reason: There are many higher order reactions.

In a homogenous reaction A → B + C + D the initial pressure was P₀ and after time t it was P. Expression for rate constant k in terms of P₀, P and t will be

Which curve corresponds to the temperature dependence of the rate R of a simple one step reaction?

Assertion : In rate law, unlike in the expression for equilibrium constants, the exponents for concentrations do not necessarily match the stoichiometric coefficients. Reason : It is the mechanism and not the balanced chemical equation for the overall change that governs the reaction rate.

For the reaction, 2NO₂ + F₂ → 2NO₂F, following mechanism has been provided. NO₂ + F₂ → Slow → NO₂F + F NO₂ + F → Fast → NO₂F Thus, rate expression of the above reaction can be written as

A catalyst

Assertion : The kinetics of the reaction — mA + nB + pC → xY + yY + zZ obeys the rate expression as — dx/dt = k[A]ⁿ[B]ᵐ. Reason : The rate of reaction does not depend upon the concentration of C.

For the reaction, 2N₂O₅ → 4NO₂ + O₂

Assertion : According to transition state theory, for the formation of an activated complex, one of the vibrational degrees of freedom is converted into a translational degree of freedom. Reason : Energy of the activated complex is higher than the energy of reactant molecules.

An endothermic reaction with high activation energy for the forward reaction is given by the diagram