Zero Order
Reactions & Half-Life

Zero order reaction

Integrated rate law

Integrating a constant rate gives a linear fall in concentration.

−d[R]/dt = k ⟹ [R] = [R]₀ − kt

1. Zero order means −d[R]/dt = k[R]⁰ = k, a constant.
2. Separate and integrate: d[R] = −k dt gives [R] = −kt + I.
3. At t = 0, [R] = [R]₀, so I = [R]₀.
4. [R] = [R]₀ − kt.

Zero order reaction

Expression for t½

Substitute [R] = [R]₀/2 into the integrated law.

[R] = [R]₀ − kt

1. Set [R] = [R]₀/2: [R]₀/2 = [R]₀ − k·t½.
2. Rearranging: k·t½ = [R]₀ − [R]₀/2 = [R]₀/2.
3. t½ = [R]₀/2k.
4. Note [R]₀ does not cancel — this is the key difference from first order.

k = 0.02 mol L⁻¹ s⁻¹, [R]₀ = 0.5 mol L⁻¹

t½

Direct substitution into the zero order half-life formula.

t½ = [R]₀/2k

1. t½ = 0.5/(2 × 0.02).
2. Denominator = 0.04.
3. t½ = 0.5/0.04 = 12.5 s.

k = 0.02 mol L⁻¹ s⁻¹, [R]₀ = 0.5 mol L⁻¹

Time for complete reaction

Unlike first order, a zero order reaction reaches zero concentration in finite time.

Set [R] = 0 in [R] = [R]₀ − kt ⟹ t = [R]₀/k

1. Setting [R] = 0 gives 0 = [R]₀ − kt.
2. t = [R]₀/k = 0.5/0.02.
3. = 25 s, which is exactly twice the half-life of 12.5 s.
4. Option (c) would be correct for a first order reaction, never for zero order.

[R]₀ doubled, zero order

Effect on t½

For zero order the half-life is directly proportional to initial concentration.

t½ = [R]₀/2k

1. t½ = [R]₀/2k, so t½ ∝ [R]₀ with k fixed.
2. Doubling [R]₀ therefore doubles t½.
3. Contrast with first order, where t½ would be unaffected.

2NH₃ →(1130 K, Pt) N₂ + 3H₂

The condition for zero order behaviour

Zero order arises only once the metal surface is saturated.

Rate = k[NH₃]⁰ = k

1. At high pressure the platinum surface becomes fully covered with gas molecules.
2. Once saturated, further changes in ammonia concentration cannot alter the amount adsorbed and reacting.
3. The rate therefore becomes independent of concentration — zero order.
4. At low pressure the surface is not saturated, and the reaction is not zero order. The qualifier matters.

Four reactions

The zero order one

Recall the named surface-catalysed examples.

—

1. NCERT cites HI decomposition on a gold surface as a zero order reaction, so (a) is correct.
2. Hydrogenation of ethene has Rate = k[C₂H₄], first order.
3. N₂O₅ decomposition is first order.
4. All radioactive decay is first order.

Zero order reaction in progress

Behaviour of the rate

Zero order means the rate equals k, which does not change at fixed temperature.

Rate = k[R]⁰ = k

1. The rate law reduces to Rate = k, a constant.
2. k does not vary with time at constant temperature.
3. So the rate stays fixed until reactant runs out, at which point the reaction simply stops.
4. Option (c) confuses 'zero order' with 'zero rate' — a very common misreading.

[R]₀ = 0.8 M, [R] = 0.4 M, t = 20 min, zero order

k

Rearrange the integrated law for k.

k = ([R]₀ − [R])/t

1. [R]₀ − [R] = 0.8 − 0.4 = 0.4 mol L⁻¹.
2. k = 0.4/20 = 0.02 mol L⁻¹ min⁻¹.
3. Units check: concentration divided by time, correct for zero order.
4. Option (b) has first order units and is the trap.

Zero order, 50% consumed in 10 min

Time for 75% consumption

Because the rate is constant, consumption is directly proportional to time.

[R]₀ − [R] = kt, so amount consumed ∝ t

1. The amount consumed is kt, which grows linearly with time.
2. 50% consumed corresponds to 10 minutes.
3. 75% is 1.5 times as much, so it takes 1.5 × 10 = 15 minutes.
4. Note this is quite unlike first order, where 75% would take exactly two half-lives, i.e. 20 minutes.

k = 0.1 mol L⁻¹ min⁻¹, [R]₀ = 1.0 M, t = 4 min

[R] at 4 min

Substitute into the linear integrated law.

[R] = [R]₀ − kt

1. kt = 0.1 × 4 = 0.4 mol L⁻¹.
2. [R] = 1.0 − 0.4 = 0.6 mol L⁻¹.
3. Sanity check: the reaction would finish at t = 1.0/0.1 = 10 min, so at 4 min there should still be plenty left.

Zero order reaction

The linear plot

Each order has its own characteristic linearising plot.

[R] = [R]₀ − kt

1. The zero order integrated law is already linear in [R] and t.
2. So plotting [R] directly against t gives a straight line.
3. ln[R] vs t is linear for first order, and 1/[R] vs t for second order.
4. log[R] vs t is just the first order plot rescaled, so it is also not correct here.

k doubled, [R]₀ fixed, zero order

Effect on t½

t½ is inversely proportional to k for every order.

t½ = [R]₀/2k

1. t½ = [R]₀/2k, so t½ ∝ 1/k.
2. Doubling k halves the half-life.
3. Note that k appears in the denominator for every order, so this inverse relation is universal — only the [R]₀ dependence differs between orders.

Enzyme-catalysed reaction at high substrate concentration

Reason for zero order behaviour

Saturation of a limited resource is the universal cause of zero order kinetics.

Rate = k when all sites are occupied

1. Enzymes have a fixed number of active sites.
2. At high substrate concentration every site is continuously occupied.
3. Adding more substrate cannot increase the number of sites, so the rate stops rising.
4. The enzyme concentration then sets the pace, making the reaction zero order in substrate — the same saturation logic as ammonia on platinum.

Zero order, 20% complete in 10 min

Time for complete reaction

Consumption is linear in time, so simple proportion applies.

amount consumed ∝ t

1. 20% consumed corresponds to 10 minutes.
2. 100% is five times as much, so it takes 5 × 10 = 50 minutes.
3. Option (c) would be correct for first order but is wrong for zero order — a zero order reaction genuinely finishes.

Zero order reaction

What the rate depends on

Rate equals k, and k depends on temperature and catalyst but not concentration.

Rate = k = A e^(−Ea/RT)

1. For zero order, Rate = k exactly.
2. By the Arrhenius equation, k depends on temperature and on Ea, which a catalyst can lower.
3. No concentration term appears anywhere.
4. So the rate is controlled by temperature and catalyst only.

[R]₀ = 0.6 M, t½ = 15 min, zero order

k

Rearrange the zero order half-life formula.

t½ = [R]₀/2k ⟹ k = [R]₀/2t½

1. k = 0.6/(2 × 15).
2. Denominator = 30.
3. k = 0.6/30 = 0.02 mol L⁻¹ min⁻¹.

Comparison of zero and first order

The correct distinction

The rate law itself determines whether the rate responds to falling concentration.

Zero order: Rate = k ; First order: Rate = k[R]

1. For zero order the rate law is Rate = k, with no concentration term, so the rate cannot change.
2. For first order the rate law is Rate = k[R], so as [R] falls the rate falls with it.
3. This is exactly what the two integrated laws show: a straight line versus a flattening curve.
4. Note that in both cases the rate constant k itself stays fixed.

0.10 M → 0.06 M in 8 min, zero order

Rate of reaction

For zero order the rate equals k, which is the concentration drop over time.

Rate = k = ([R]₀ − [R])/t

1. Concentration drop = 0.10 − 0.06 = 0.04 mol L⁻¹.
2. Rate = 0.04/8 = 5.0 × 10⁻³ mol L⁻¹ min⁻¹.
3. Because the reaction is zero order, this rate stays the same throughout — it is not an average that changes.

Photochemical reaction at fixed light intensity

Reason for zero order behaviour

The same saturation logic — a fixed external supply is the bottleneck.

Rate = k, set by photon flux

1. Each reacting molecule requires a photon.
2. At fixed light intensity the number of photons delivered per second is fixed.
3. Provided enough reactant is present to absorb every photon, adding more reactant cannot speed anything up.
4. So the rate becomes independent of concentration — zero order, by the same saturation reasoning as a catalyst surface.

Zero order reaction

Shape of the rate versus time plot

The rate equals k throughout, so it does not change with time either.

Rate = k, independent of both [R] and t

1. Since Rate = k and k is fixed at constant temperature, the rate does not vary with time.
2. So the plot is horizontal for as long as reactant remains.
3. Once the reactant is exhausted, the rate drops abruptly to zero.
4. Note the distinction from the [R] vs t plot, which is a falling straight line.

k = 2 × 10⁻³ mol L⁻¹ s⁻¹, [R]₀ = 0.1 M, t = 30 s

[R]

Substitute into the linear law.

[R] = [R]₀ − kt

1. kt = 2 × 10⁻³ × 30 = 0.06 mol L⁻¹.
2. [R] = 0.1 − 0.06 = 0.04 mol L⁻¹.
3. Sanity check: complete consumption would take 0.1/2 × 10⁻³ = 50 s, so at 30 s some reactant should remain. It does.

General half-life relation and the zero order case

The consistent statement

Substituting n = 0 into the general relation must reproduce the derived formula.

t½ ∝ 1/[R]₀^(n−1)

1. Put n = 0 into the general relation: the exponent becomes (0 − 1) = −1.
2. t½ ∝ 1/[R]₀^(−1) = [R]₀.
3. This matches the derived formula t½ = [R]₀/2k exactly.
4. So option (a) is correct, and the general relation is confirmed for this case.