Govt Exams
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.
For zero-order reaction: d[A]/dt = -k, integrating gives [A] = [A]₀ - kt, which is a linear equation.
A catalyst provides an alternative reaction pathway with lower activation energy, thus increasing the rate constant k without affecting A or T.
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.
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]
Using ln([A]₀/[A]ₜ) = kt, ln(0.5/0.25) = k × 30, ln(2) = k × 30, k = 0.693/30 = 0.0231 s⁻¹
Using Nernst equation: Ecell = E°cell - (0.059/n)log(Q). Increasing [Zn²⁺] increases Q, making the log term positive, which decreases Ecell. ΔE = -(0.059/2)log(10) = -0.0295 ≈ -0.0296 V.
At cathode: Na⁺ + e⁻ → Na; moles of Na = 2.3/23 = 0.1 mol. At anode: 2Cl⁻ → Cl₂ + 2e⁻; for 0.1 mol Na, electrons = 0.1 mol, so Cl₂ moles = 0.1/2 = 0.05 mol. Volume at STP = 0.05 × 22.4 = 1.12 L.
The species with highest reduction potential (+1.36 V) is Cl₂, making it the strongest oxidizing agent. Higher E° values indicate greater tendency to accept electrons.
As the reaction proceeds, product concentrations increase while reactant concentrations decrease, reducing the driving force according to Nernst equation: E = E° - (0.059/n)log(Q).