JEE Chemistry - MCQ Practice Questions

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

The exponent of [NO] in the rate law is 2, making the reaction second order with respect to NO

Using ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂): ln(5.$\frac{0}{1}$.7) = (Ea/8.314)($\frac{1}{320}$ - $\frac{1}{330}$), solving gives Ea ≈ 50 kJ/mol

For zero-order reaction: [A]₀ - [A]ₜ = kt. So 0.5 - 0.1 = 0.02 × t, giving t = 20 s

For first-order reaction, [A]ₜ = [A]₀($\frac{1}{2}$)^(t/t₁/₂). For [A]ₜ = $\frac{1}{4}$[A]₀, we need ($\frac{1}{2}$)^(t/30) = $\frac{1}{4}$, so t/30 = 2, giving t = 60 minutes

The rate-determining step (slowest step) controls the overall reaction rate, regardless of how many fast equilibrium steps precede it

Overall order = sum of exponents in rate law = 1 + 2 = 3 (third order reaction)

A = k/e^(-Ea/RT) = (2 × 10⁻⁵)/e^(-$\frac{80000}{2}$.478) ≈ 5.6 × 10¹³ s⁻¹