For second-order reaction (single reactant), integrated rate law:
A1/[A] - 1/[A]₀ = kt
Bln[A] vs t straight
C[A]₀ - [A] = kt
DSame as first
Answer & Solution
Correct answer: A. 1/[A] - 1/[A]₀ = kt
d[A]/dt = -k[A]² integrates to 1/[A] = 1/[A]₀ + kt. Plot 1/[A] vs t is straight line. Half-life t₁/₂ = 1/(k[A]₀) depends on initial concentration.
Related questions
Chemical kinetics studies the rate of a reaction and also its:Manganese dioxide is given as a catalyst for the decomposition of:A catalyst increases the rate of a reaction without itself being:Lowering the activation energy of a reaction makes the rate:Raising the temperature of a reaction makes the rate:The exponential factor gives the fraction of molecules with kinetic energy:Arrhenius was a chemist from:The physical justification of that equation was provided by: