Integrated rate equation: first order reactions
For a first order reaction, rate = k[R], which integrates to k = (2.303/t) log([R]₀/[R]). A graph of ln[R] against t is a straight line.
Derivation: integrated rate equation for a first order reaction
Consider R → P. [R]₀ is the concentration at t = 0 and [R] the concentration at time t.
- 1Rate = −d[R]/dt = k[R].
- 2Separate the variables: d[R]/[R] = −k dt.
- 3Integrate: ln[R] = −kt + I, where I is the constant of integration.
- 4At t = 0, [R] = [R]₀, so I = ln[R]₀.
- 5Substitute: ln[R] = −kt + ln[R]₀, so ln([R]₀/[R]) = kt.
- 6★So, k = (1/t) ln([R]₀/[R]) = (2.303/t) log([R]₀/[R]). Taking antilogs, [R] = [R]₀ e^(−kt).
★ key step
How to draw the two first order graphs
- Graph 1: y-axis ln[R], x-axis t. Draw a straight line starting at ln[R]₀ on the y-axis and sloping down. Label "slope = −k" and "intercept = ln[R]₀".
- Graph 2: y-axis log([R]₀/[R]), x-axis t. Draw a straight line through the origin sloping up. Label "slope = k/2.303".
- A plain graph of [R] against t is a curve that falls ever more slowly and never quite reaches zero.
Examples of first order reactions
Hydrogenation of ethene: C₂H₄(g) + H₂(g) → C₂H₆(g), rate = k[C₂H₄]. All natural and artificial radioactive decay, such as ²²⁶₈₈Ra → ⁴₂He + ²²²₈₆Rn, rate = k[Ra]. The decompositions of N₂O₅ and N₂O are also first order.
First order reactions in the gas phase
For A(g) → B(g) + C(g), we follow the total pressure, because partial pressure is proportional to concentration.
- 1At t = 0: p(A) = pᵢ, p(B) = 0, p(C) = 0. Total = pᵢ.
- 2At time t, x atm of A has reacted: p(A) = pᵢ − x, p(B) = x, p(C) = x.
- 3Total pressure pₜ = (pᵢ − x) + x + x = pᵢ + x, so x = pₜ − pᵢ.
- 4p(A) = pᵢ − (pₜ − pᵢ) = 2pᵢ − pₜ.
- 5★So, k = (2.303/t) log[pᵢ/(2pᵢ − pₜ)].
★ key step
Worked example: k from concentrations
In a first order reaction, the concentration of the reactant falls from 0.80 mol L⁻¹ to 0.20 mol L⁻¹ in 30 minutes. Find k. (log 4 = 0.6020)
- Given: [R]₀ = 0.80 mol L⁻¹, [R] = 0.20 mol L⁻¹, t = 30 min.
- Formula: k = (2.303/t) log([R]₀/[R]).
- [R]₀/[R] = 0.80 ÷ 0.20 = 4.
- k = (2.303 ÷ 30 min) × log 4 = (2.303 × 0.6020) ÷ 30.
- 2.303 × 0.6020 = 1.386; 1.386 ÷ 30 = 0.0462.
- k = 4.62 × 10⁻² min⁻¹ (three significant figures).
k = 4.62 × 10⁻² min⁻¹
Worked example: a gas phase first order reaction
A(g) → B(g) + C(g) is first order. The initial pressure of A is 0.40 atm. After 100 s the total pressure is 0.70 atm. Find k. (log 4 = 0.6020)
- Given: pᵢ = 0.40 atm, pₜ = 0.70 atm, t = 100 s.
- p(A) = 2pᵢ − pₜ = 0.80 − 0.70 = 0.10 atm.
- Formula: k = (2.303/t) log[pᵢ/p(A)].
- k = (2.303 ÷ 100 s) × log(0.40/0.10) = 0.02303 × log 4.
- k = 0.02303 × 0.6020 = 0.01386 s⁻¹.
k = 1.39 × 10⁻² s⁻¹ (rounded to three significant figures)
Carbon-14 dating, used to find the age of old wood, bones and cloth from archaeological sites in India, works because radioactive decay is a first order process.
3 marks: “Derive the integrated rate equation for a first order reaction. Sketch the plot of ln[R] against t.” −d[R]/dt = k[R]; d[R]/[R] = −k dt; ln[R] = −kt + I; I = ln[R]₀; so k = (2.303/t) log([R]₀/[R]). Plot: straight line, slope −k, intercept ln[R]₀.
3 marks: “A first order reaction is 75% complete in 60 minutes. Find k. (log 4 = 0.6020)” 75% complete means [R] = 0.25[R]₀, so [R]₀/[R] = 4. k = (2.303/60) × 0.6020 = 1.386 ÷ 60 = 2.31 × 10⁻² min⁻¹.
2 marks: “For A(g) → B(g) + C(g), show that p(A) = 2pᵢ − pₜ.” At time t: p(A) = pᵢ − x, p(B) = p(C) = x. pₜ = pᵢ + x, so x = pₜ − pᵢ and p(A) = 2pᵢ − pₜ.
First order: 2.303 over t, log of start over now.