9 min readPathway sheet3-mark derivation of [R] = −kt + [R]₀ with its graph; 2-mark numerical

Integrated rate equation: zero order reactions

In a zero order reaction the rate equals k and does not depend on concentration. So [R] falls in a straight line: [R] = −kt + [R]₀.

Integrated equation[R] = −kt + [R]₀, or k = ([R]₀ − [R])/t; [R]₀ is the initial and [R] the concentration at time t

Graph[R] against t is a straight line with slope −k and intercept [R]₀

Unit of kmol L⁻¹ s⁻¹

Examples2NH₃ → N₂ + 3H₂ on hot platinum (1130 K, high pressure); 2HI → H₂ + I₂ on a gold surface

Derivation: integrated rate equation for a zero order reaction

Consider R → P. Let [R]₀ be the concentration at t = 0 and [R] the concentration at time t.

1
Rate = −d[R]/dt = k[R]⁰ = k (any number to the power zero is 1).
2
Rearrange: d[R] = −k dt.
3
Integrate both sides: [R] = −kt + I, where I is the constant of integration.
4
At t = 0, [R] = [R]₀. So [R]₀ = −k × 0 + I, which gives I = [R]₀.
5★
Substitute: [R] = −kt + [R]₀.
6
So, k = ([R]₀ − [R])/t. This is the integrated rate equation for a zero order reaction.

★ key step

How to draw the zero order graph

Label the x-axis Time (t) and the y-axis Concentration [R].
Draw a straight line that starts on the y-axis at [R]₀ and slopes downwards to the x-axis.
Label the intercept "[R]₀" and the line "slope = −k".
The line meets the time axis at t = [R]₀/k: a zero order reaction is complete in a finite time.

When is a reaction zero order?

Zero order reactions are uncommon and happen under special conditions, often on a metal surface or with enzymes. Ammonia decomposes on hot platinum at 1130 K: 2NH₃(g) → N₂(g) + 3H₂(g). At high pressure the platinum surface is fully covered with NH₃ molecules. Adding more gas cannot put more molecules on the surface, so the rate stays the same: rate = k[NH₃]⁰ = k. The thermal decomposition of HI on a gold surface is another example.

Worked example: time taken in a zero order reaction

Problem

A zero order reaction has k = 2.0 × 10⁻³ mol L⁻¹ s⁻¹ and [R]₀ = 0.50 mol L⁻¹. (a) How long does [R] take to fall to 0.10 mol L⁻¹? (b) When is the reaction complete?

Working

Given: k = 2.0 × 10⁻³ mol L⁻¹ s⁻¹, [R]₀ = 0.50 mol L⁻¹, [R] = 0.10 mol L⁻¹.
Formula: t = ([R]₀ − [R])/k.
(a) t = (0.50 − 0.10) mol L⁻¹ ÷ 2.0 × 10⁻³ mol L⁻¹ s⁻¹ = 0.40 ÷ 0.0020 = 200 s.
(b) Complete means [R] = 0: t = [R]₀/k = 0.50 ÷ 0.0020 = 250 s.

Answer

(a) 200 s; (b) 250 s

Worked example: finding k

Problem

In a zero order reaction, [R] falls from 0.80 mol L⁻¹ to 0.50 mol L⁻¹ in 60 s. Find k.

Working

Formula: k = ([R]₀ − [R])/t.
k = (0.80 − 0.50) mol L⁻¹ ÷ 60 s = 0.30 ÷ 60.
k = 0.0050 = 5.0 × 10⁻³ mol L⁻¹ s⁻¹.

Answer

k = 5.0 × 10⁻³ mol L⁻¹ s⁻¹

In real life

A slow-release tablet is designed to give a steady amount of medicine per hour, much like a zero order reaction, so the dose in the blood stays even through the day.

Exam angle

3 marks: “Derive the integrated rate equation for a zero order reaction. Draw the graph of [R] against t.” −d[R]/dt = k; d[R] = −k dt; integrate: [R] = −kt + I; at t = 0, I = [R]₀; so [R] = −kt + [R]₀. Graph: straight line, intercept [R]₀, slope −k.

2 marks: “For a zero order reaction k = 3.0 × 10⁻³ mol L⁻¹ s⁻¹. How long will [R] take to fall from 0.10 to 0.075 mol L⁻¹?” t = ([R]₀ − [R])/k = 0.025 ÷ 0.0030 = 8.3 s (rounded).

1 marks: “Give one example of a zero order reaction and the unit of its rate constant.” 2NH₃ → N₂ + 3H₂ on hot platinum at high pressure; mol L⁻¹ s⁻¹.

Remember it

Zero order: zero effect of concentration, so a straight line all the way to zero.