Zero, First and Second Order Reactions

Check the given clue: rate law, graph, half-life, or concentration-time data. Select the matching integrated equation. Keep time units consistent with k. Substitute values carefully, then state what the result means chemically, such as reactant left or fraction decomposed.

Graph comparison is important: [A] vs t is linear for zero order, ln[A] vs t is linear for first order, and 1/[A] vs t is linear for second order.

Using log formula without the factor 2.303, or mixing seconds and minutes when calculating k.

Zero, first, and second order reactions are classified by how rate depends on reactant concentration. Their integrated rate equations relate concentration and time, allowing calculation of rate constant, concentration remaining, or half-life.

Zero, first, and second order reactions are classified by how rate depends on reactant concentration. Their integrated rate equations relate concentration and time, allowing calculation of rate constant, concentration remaining, or half-life. Zero order: [A] = [A0] - kt, t1/2 = [A0]/2k, k unit = mol L^-1 s^-1. First order: ln[A] = ln[A0] - kt; k = (2.303/t)log([A0]/[A]); t1/2 = 0.693/k; k unit = s^-1. Second order: 1/[A] - 1/[A0] = kt; t1/2 = 1/(k[A0]); k unit = L mol^-1 s^-1. For a first order decomposition with k = 2.31 x 10^-3 s^-1, t1/2 = 0.693/k = 0.693/(2.31 x 10^-3) = 300 s. This means half the reactant remains after every 300 s interval.

In a zero order reaction, rate is independent of reactant concentration and [A] decreases linearly with time. In a first order reaction, rate is proportional to [A], and ln[A] or log[A] gives a straight-line relation with time. In a second order reaction involving one reactant, 1/[A] increases linearly with time. Half-life behaviour is a strong identifier: zero order half-life depends directly on initial concentration, first order half-life is independent of initial concentration, and second order half-life is inversely proportional to initial concentration. Zero order: [A] = [A0] - kt, t1/2 = [A0]/2k, k unit = mol L^-1 s^-1. First order: ln[A] = ln[A0] - kt; k = (2.303/t)log([A0]/[A]); t1/2 = 0.693/k; k unit = s^-1. Second order: 1/[A] - 1/[A0] = kt; t1/2 = 1/(k[A0]); k unit = L mol^-1 s^-1. A first order reaction has [A0] = 0.80 mol L^-1 and [A] = 0.20 mol L^-1 after 40 min. k = (2.303/40)log(0.80/0.20) = (2.303/40)log4 = (2.303/40)(0.6021) = 0.0347 min^-1. Interpretation: concentration becomes one-fourth in 40 min. This concept appears as numerical problems on k, t1/2, concentration left, and graph-based identification of order. Long-answer questions may ask for derivation or comparison of equations and half-life relations. Using log formula without the factor 2.303, or mixing seconds and minutes when calculating k.