JEE Chemistry - MCQ Practice Questions

The color of the reaction vessel does not affect the reaction rate. Temperature, nature of reactants, concentration, pressure, and catalyst are actual factors affecting reaction rate.

For a zero-order reaction, t₁/₂ = [A]₀/(2k), which is directly proportional to initial concentration.

The units s⁻¹ or time⁻¹ indicate a first-order reaction. For first-order reactions, the rate constant has units of time⁻¹.

Activation energy is the minimum energy required for reactants to overcome the energy barrier and form products. It is always positive and independent of temperature.

Using ln([A]₀/[A]ₜ) = kt, ln(0.$\frac{5}{0}$.25) = k × 30, ln(2) = k × 30, k = 0.$\frac{693}{30}$ = 0.0231 s⁻¹

The overall reaction is obtained by adding all steps and canceling intermediates: A + B + D → E + F. Rate law is determined by the slow step: rate = k[A][B]

For a first-order reaction, ln[A] = ln[A]₀ - kt, so a plot of ln[A] vs t gives a straight line with slope -k.

A catalyst provides an alternative reaction pathway with lower activation energy, thus increasing the rate constant k without affecting A or T.

For zero-order reaction: d[A]/dt = -k, integrating gives [A] = [A]₀ - kt, which is a linear equation.

For a first-order reaction, if half-life decreases from 10 to 5 minutes (becomes half) with a 10 K increase, this indicates the reaction rate doubles per 10 K, giving a temperature coefficient of 2.

After n half-lives, fraction remaining = ($\frac{1}{2}$)ⁿ. After 5 half-lives: ($\frac{1}{2}$)⁵ = $\frac{1}{32}$.

From fast equilibrium: K = [B]/[A], so [B] = K[A]. The slow step rate law is rate = k'[B][C] = k'K[A][C] = k[A]^($\frac{1}{2}$)[C] where k combines constants.

Using ln(k₂/k₁) = (Eₐ/R)(1/T₁ - 1/T₂): ln(4) = (Eₐ/8.314)($\frac{1}{300}$ - $\frac{1}{320}$), solving gives Eₐ ≈ 51.2 kJ/mol.

When [A] doubles, rate increases by 8 = 2³, so order w.r.t. A = 3. When [B] doubles, rate increases by 2 = 2¹, so order w.r.t. B = 1. Overall order = 3 + 1 = 4.

Since k₁ > k₂, A converts to B faster than B converts to C, so B accumulates initially and then decreases as it slowly converts to C.

According to collision theory, collisions must have both proper spatial orientation and energy ≥ activation energy to result in a reaction.

In the Lindemann mechanism: Step 1 (fast equilibrium): A + A ⇌ A* + A, Step 2 (slow): A* → products. The slow step is rate-determining.

When [B]₀ >> [A]₀, the concentration of B doesn't change significantly during the reaction, so it can be incorporated into the rate constant, making the reaction appear first-order in A only.

A homogeneous catalyst is in the same phase as reactants. H₂SO₄ (liquid) catalyzes esterification of reactants (liquid), making it homogeneous. Others are heterogeneous catalysts.

Using Arrhenius equation: log(k₂/k₁) = (Ea/2.303R)(T₂-T₁)/(T₁T₂). With Ea = 50,000 J/mol, ΔT = 10 K, this gives log(k₂/k₁) ≈ 0.30, so k₂/k₁ ≈ 2.0