Chemical Kinetics · Lesson 5 of 10
First-Order Reactions
“Derive exponential concentration decay, interpret logarithmic graphs and recover reactant pressure from total-pressure measurements.”
• Integrate a first-order disappearance law and apply its initial condition. • Calculate k, remaining concentration and elapsed time with either natural or decimal logarithms. • Test first-order behaviour using concentration–time data. • Derive reactant partial pressures from stoichiometry before using a pressure-based kinetic equation.
A constant fractional loss produces an exponential curve
If a reaction removes the same fraction of its remaining reactant per unit time, the absolute amount removed becomes smaller as less remains. This gives a declining instantaneous rate and a curved concentration–time graph. In a first-order disappearance model, rate is proportional to the current concentration, so the proportionality coefficient has units of inverse time.
Separate the variables by dividing by [R] and multiplying by dt: d[R]/[R]=−k₁dt. Integrate from the known initial state to the later state. The integral of 1/[R] with respect to [R] is a natural logarithm. Using a ratio makes the logarithm’s argument dimensionless, because the two concentrations use the same unit.
With indefinite integrals, ln[R]=−k₁t+I and the initial condition gives I=ln[R]₀. These expressions are shorthand for logarithms of concentration numerical values relative to a common reference unit. The ratio form is safer for calculations because it avoids taking a logarithm of an unqualified dimensional quantity.
Problem
A first-order N₂O₅ disappearance process at 318 K has [N₂O₅]₀=1.24×10⁻² M and [N₂O₅]₆₀=0.20×10⁻² M after 60 min. Find k₁.
- 1.The initial-to-final ratio is 1.24/0.20=6.2. Use k₁=ln(6.2)/60=0.0304092 min⁻¹.
- 2.In seconds, divide by 60: k₁≈5.06819×10⁻⁴ s⁻¹. The temperature belongs to the kinetic conditions, not to this unit conversion.
- 3.Substitution gives [R]₀exp(−k₁×60)=0.0020 M, reproducing the measured final concentration.
- 4.The law describes disappearance; if the overall equation is written 2N₂O₅→4NO₂+O₂, normalised r is one half of that species disappearance rate.
Problem
A reactant starts at 1.00 M and has first-order k₁=2.0×10⁻² s⁻¹. What remains after 100 s?
- 1.Use the exponential concentration equation: [R]t=1.00×exp(−0.020×100)=0.135335 M.
- 2.The remaining fraction is about 0.1353, so the fraction consumed is about 0.8647.
- 3.Using 1−k₁t as an exact fraction would give a negative number here. That linear form is not the first-order integrated law.
Straight-line plots test the proposed order
| Plot for a first-order model | Expected shape or slope | Interpretation |
|---|---|---|
| [R] versus t | Decreasing exponential | Equal time intervals remove equal fractions, not equal amounts. |
| ln([R]/Cref) versus t | Straight line, slope −k₁ | Cref is a fixed concentration reference with the same unit. |
| log₁₀([R]₀/[R]) versus t | Straight line, slope k₁/ln10 | Multiplying the fitted slope by ln10 gives k₁. |
A single pair of points always defines a line, so it cannot on its own verify order. Use multiple measurements and examine scatter. A straight logarithmic trend supports a first-order model over that experimental range; substantial curvature, changing slopes or incompatible constants indicate that assumptions need checking.
| Time / s | Measured [N₂O₅] / M |
|---|---|
| 0 | 0.0163 |
| 400 | 0.0136 |
| 800 | 0.0114 |
| 1200 | 0.0093 |
| 1600 | 0.0078 |
| 2000 | 0.0064 |
| 2400 | 0.0053 |
| 2800 | 0.0043 |
| 3200 | 0.0035 |
Problem
Assess the table using a logarithmic straight-line fit and estimate a half-life.
- 1.Plot concentration against time to see the curved decline. Then take natural logarithms of concentration values in a common reference unit and fit a line using all nine points.
- 2.The fitted slope is approximately −0.000479729 s⁻¹, giving k₁=0.000479729 s⁻¹. A two-endpoint check gives ln(0.0163/0.0035)/3200=4.80751×10⁻⁴ s⁻¹, close to the all-point estimate.
- 3.The first-order law implies a half-life ln2/k₁=1444.87 s. Direct graphical interpolation at half the measured initial concentration, 0.00815 M, between 1200 and 1600 s gives about 1506.67 s.
- 4.The estimates are around 1.4–1.5×10³ s rather than exactly equal. Rounded experimental data do not lie on a perfect exponential, and a fitted intercept need not equal the first measured point exactly.
- 5.State the fitting method and uncertainty rather than presenting a graph-read value as an exact independent constant.
Total gas pressure is not generally reactant pressure
For an ideal gas at fixed temperature and volume, each species partial pressure is proportional to its amount. Suppose pure A initially has pressure p₀ and decomposes as A(g)→B(g)+C(g), with no initial products or changing inert-gas contribution. If A loses partial pressure x, each product gains x. The total pressure is (p₀−x)+x+x=p₀+x. Thus x=p_t−p₀ and p_A=p₀−x=2p₀−p_t.
Problem
A stipulated decomposition is 2N₂O₅(g)→2N₂O₄(g)+O₂(g). Initial pressure is 0.500 atm and total pressure after 100 s is 0.512 atm. Find the first-order disappearance constant.
- 1.Let N₂O₅ lose 2x atm, producing 2x atm N₂O₄ and x atm O₂. Total pressure is 0.500+x, so x=0.012 atm.
- 2.Remaining N₂O₅ pressure=0.500−2×0.012=0.476 atm. Equivalently, p_A=3p₀−2p_t for this 2→3 stoichiometry.
- 3.Use the actual reactant-pressure ratio: k₁=ln(0.500/0.476)/100=0.000491902 s⁻¹, about 4.92×10⁻⁴ s⁻¹.
- 4.A formula derived for A→B+C would give the wrong partial pressure here. This calculation treats the product stoichiometry as stipulated; real NO₂/N₂O₄ equilibration would require additional treatment if it changed the gas-mole balance.
Problem
SO₂Cl₂(g)→SO₂(g)+Cl₂(g) starts at 0.500 atm. At 100 s total pressure is 0.600 atm. Find the pressure-based SO₂Cl₂ disappearance rate when total pressure later reaches 0.650 atm.
- 1.This is a 1→2 conversion, so at 100 s p(SO₂Cl₂)=2×0.500−0.600=0.400 atm.
- 2.k₁=ln(0.500/0.400)/100=0.00223144 s⁻¹.
- 3.At total pressure 0.650 atm, remaining reactant pressure=1.000−0.650=0.350 atm.
- 4.Pressure-based disappearance is v_p=k₁p_A=0.000781002 atm s⁻¹. A molar concentration rate additionally requires division by RT; a temperature must be supplied for a unique numerical M s⁻¹ answer.
First-order behaviour can arise from a genuinely unimolecular process or from a more complicated pathway under particular conditions. Even hydrogenation of ethene, C₂H₄+H₂→C₂H₆, can show a first-order dependence on ethene in a suitable regime; its overall equation does not give a universal one-variable law.
Quick Quiz
Quiz
Which expression is the first-order disappearance law?
Explanation: Integrating d[R]/[R]=−kdt gives exponential rather than linear decay.
Which factor belongs with log₁₀([R]₀/[R]t) when calculating k?
Explanation: Natural and decimal logarithms differ by ln10. Use k=(ln10/t)log₁₀(ratio).
On a plot of ln([R]/Cref) versus time, the first-order slope is:
Explanation: The integrated natural-log law is linear with slope −k.
For pure A initially at p₀ undergoing A→B+C at fixed T,V, its later pressure is:
Explanation: The two product pressures each equal the A pressure lost, so total pressure rises by that loss.
Why cannot a 2→3 gas pressure formula be replaced by a 1→2 formula?
Explanation: Stoichiometry determines how total pressure relates to the unreacted species, independently of the integrated kinetic order.
Practice Problems
- A first-order reactant falls from 5.00 g to 3.00 g. Find the time for k₁=1.15×10⁻³ s⁻¹ when mass is proportional to remaining amount.
- Derive the first-order law using the integration constant and explain why the ratio form is dimensionless.
- A first-order law has k₁=0.00200 s⁻¹ and initial concentration 0.100 M. Calculate [R] after 300 s and its instantaneous disappearance rate then.
- Azoisopropane decomposes to a gaseous hydrocarbon and nitrogen at fixed T,V. Pure initial pressure is 35.0 mm Hg; total pressures are 54.0 at 360 s and 63.0 at 720 s. Estimate k from each time for a 1→2 conversion.
- Why is exactly complete conversion not reached at a finite time in the ideal first-order exponential model?
- How would you use the nine-point N₂O₅ dataset to assess first order and compare graph-read and fitted half-lives?
The concentration or amount ratio equals 5/3 under the stated constant-composition relation. Use t=ln(5/3)/0.00115=444.196 s≈444 s. Use the mass remaining, not the mass consumed, in the denominator.
Key Takeaways
• First-order disappearance has rate proportional to the current reactant concentration. • Integration gives ln([R]₀/[R]t)=k₁t and exponential decay. • Natural-log and base-10 forms require the correct ln10 conversion factor. • A logarithmic concentration plot has slope −k₁. • Use several measured points to test a kinetic model and acknowledge scatter. • For gas measurements, derive reactant partial pressure from the actual stoichiometry. • A pressure-based species rate needs RT conversion before reporting molar concentration units. • First-order behaviour alone does not prove a single elementary pathway.