BSEB Intermediate Chemistry study materials
BSEB Intermediate Chemistry study resources

BSEB Intermediate Chemistry: Chemical Kinetics

BSEB Intermediate Chemistry · Work through your exam, one topic at a time.

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Chemical Kinetics

Chemical Kinetics tests whether you can turn experimental rate data into a rate law: order, integrated rate equations, half-life, and Arrhenius behaviour.

Core rules

  • Rate = k[A]^x[B]^y. Order is experimental; molecularity is theoretical and applies only to an elementary step. They coincide only for a single-step reaction.
  • Zero order: [A] = [A]₀ − kt, t½ = [A]₀/2k, k in mol L⁻¹ s⁻¹.
  • First order: k = (2.303/t) log([A]₀/[A]), t½ = 0.693/k, independent of starting concentration, k in s⁻¹.
  • Arrhenius: k = A·e^(−Ea/RT), so log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂). A catalyst lowers Ea but changes neither ΔH nor the equilibrium constant.
  • The slowest step controls the overall rate; a fast pre-equilibrium followed by a slow step can give fractional or even negative-looking orders.

Worked example

A first-order reaction is 75% complete in 60 minutes. Remaining fraction = 0.25. k = (2.303/60) × log(1/0.25) = (2.303/60) × 0.602 = 0.0231 min⁻¹. t½ = 0.693/0.0231 = 30.0 min. Shortcut check: 75% completion is exactly two half-lives, so t½ = 60/2 = 30 min - the two methods agree, confirming the arithmetic.

Traps that cost marks

  • Reading order off the stoichiometric coefficients of the overall balanced equation.
  • Writing mol L⁻¹ s⁻¹ for a first-order k; it must be s⁻¹.
  • Using log where the Arrhenius equation requires ln, or forgetting the 2.303 conversion.
  • Assuming t½ is always concentration-independent - true only for first order.
  • Losing the stoichiometric factor when relating the rate of disappearance of one species to the formation of another.