Exams › NEET › Chemistry › Chemical Kinetics
102 questions with worked solutions.
Q1. Why is catalyst added while preparing oxygen from decomposition of \( K C l O_{3} ? \)
Answer: to increase the rate of reaction
In the decomposition of potassium chlorate, a catalyst is added to speed up the reaction. It helps the oxygen be produced more quickly, but it is not the reason the substance changes chemically or the main point of the process.
Q2. In the Arrhenius equation \( k=A e^{E / R T} \) rate will be constant at:
Answer: infinite \( T \) or zero \( E_{a} \)
In the Arrhenius form, the temperature dependence comes from the exponential term. If temperature becomes extremely large, or if the activation energy is zero, the exponential factor becomes constant, so the rate constant no longer changes with temperature.
Q3. As the concentration of reactants increases:
Answer: rate of the reaction increases
When reactant concentration increases, there are more reactant particles in the same volume, so collisions happen more often. More frequent effective collisions increase the reaction rate.
Answer: Both Assertion and Reason are true but Reason is not the correct explanation to Assertion
The assertion is true because a product can act as a catalyst and increase the reaction rate. The reason is also true, but it only states the general effect of a catalyst; it does not specifically explain why a product acting as a catalyst makes the rate increase with respect to the product.
Answer: Assertion is correct but Reason is incorrect
Chlorine reacts faster with H2 than with O2 because the H–H bond is relatively easier to break than the very strong O=O bond, so the H2 reaction has a lower activation barrier. The reason is incorrect because the Cl–Cl bond is not stronger than the H–Cl bond; in fact, H–Cl is stronger.
Answer: rate of the catalytic reaction is fast enough while the yield is also appreciable for this exothermic reaction at this temperature
The Haber process uses a temperature that balances two competing effects: lower temperatures favor ammonia formation at equilibrium, but they make the reaction too slow. Around 500°C, the catalyst gives a fast enough rate while the equilibrium yield remains acceptable.
Answer: \( K_{2}=0.5 K_{1} \)
For most reactions, the rate constant increases with temperature according to the Arrhenius equation. Since 280 K is lower than 300 K, the rate constant at 280 K must be smaller than at 300 K, matching the given relation.
Answer: Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
If the rate is proportional to [RCl], then reducing [RCl] by half reduces the rate by half. That makes both statements true, and the reason directly explains the assertion.
Answer: Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement
Zinc reacts with hydrochloric acid to form zinc chloride and hydrogen gas. In an open container, the hydrogen escapes, so the reaction is driven forward and effectively goes to completion. Thus both statements are true, and the gas evolution explains why the reaction proceeds to completion.
Answer: the rate of disappearance of \( H I \)
In a reaction rate expression, a negative sign is used for reactants because their concentration decreases with time. The factor of 1/2 comes from the stoichiometric coefficient in 2 HI → H2 + I2, so this term gives the rate at which HI disappears.
Answer: t = 6.909/k
For a first-order reaction, the time for 99% completion is derived using the integrated rate law and the natural logarithm of the remaining 1%.
Answer: 500 s
For a first-order reaction, the time can be calculated using the formula t = (2.303/k) * log10([R]₀/[R]). Substituting k = 4.606 × 10⁻³ s⁻¹, [R]₀ = 2.0 g, and [R] = 0.2 g, we get t = (2.303 / 4.606 × 10⁻³) * log10(2.0 / 0.2) = 500 s.
Answer: k' = 2k ; k'' = k
The stoichiometric coefficients in the reaction dictate the relationship between the rate constants. For NO₂, the coefficient is 4, so k' = 2k. For O₂, the coefficient is 1, so k'' = k.
Answer: Rate = k[A]²[B]
When [B] is doubled, the rate doubles, indicating the reaction is first order with respect to B. When both [A] and [B] are doubled, the rate increases by a factor of 8, implying the reaction is second order with respect to A. Thus, the rate law is Rate = k[A]²[B].
Answer: 0.72 M
The reaction follows first-order kinetics. Using the integrated rate law for first-order reactions, the concentration of B can be calculated as: [B] = [A]₀(1 - e^(-kt)). Substituting k = 0.6 × 10⁻³ s⁻¹, [A]₀ = 5 M, and t = 20 × 60 s, we get [B] ≈ 0.72 M.
Answer: 1000 K
Equating the two rate constants, k₁ = k₂, and solving for T using the Arrhenius equation, we find T = 1000 K. This involves equating the exponential terms and simplifying.
Answer: log k vs 1/T
The Arrhenius equation can be linearized as log k = -Ea/(2.303R)(1/T) + log A. A plot of log k vs 1/T gives a straight line with slope -Ea/(2.303R), allowing the calculation of Ea.
Answer: Can be less than or more than Ea
The activation energy for the reverse reaction depends on the enthalpy change (ΔH) of the reaction. It can be less than or more than Ea depending on whether the reaction is exothermic or endothermic.
Answer: 1.25 × 10⁻² mol L⁻¹ s⁻¹ and 3.125 × 10⁻³ mol L⁻¹ s⁻¹
The stoichiometry of the reaction shows that 1 mole of N₂O₅ produces 2 moles of NO₂ and 0.5 moles of O₂. Thus, the rate of formation of NO₂ is twice the rate of disappearance of N₂O₅, and the rate of formation of O₂ is half the rate of disappearance of N₂O₅. Therefore, the rates are 1.25 × 10⁻² mol L⁻¹ s⁻¹ for NO₂ and 3.125 × 10⁻³ mol L⁻¹ s⁻¹ for O₂.
Q20. Activation energy of a chemical reaction can be determined by:
Answer: evaluating rate constants at two different temperatures
Activation energy can be determined using the Arrhenius equation by evaluating the rate constants at two different temperatures and plotting or calculating accordingly.