Decomposition and Soil Carbon Turnover
modelling Level 3

Decomposition and Soil Carbon Turnover

Dead plant material doesn't disappear instantly—it decomposes over months to millennia, releasing CO₂ back to the atmosphere. This model models decomposition as a multi-pool system with different decay rates, introduces temperature and moisture controls, and calculates soil carbon storage at steady state.

Prerequisites: exponential decay, multi pool dynamics, turnover rates, temperature dependence

Updated 12 min read

1. The Question

Where does the carbon go after plants die?

Leaves fall, roots die, wood decays. This dead organic matter (litter) accumulates on and in the soil, where microbes and fungi break it down, releasing CO₂ through heterotrophic respiration.

The balance between litter inputs and decomposition outputs determines:

  • Soil carbon storage — soils hold 2–3× more carbon than atmosphere
  • Net Ecosystem Productivity (NEP = NPP - $R_h$)
  • Carbon sequestration potential — can soils store more carbon?
  • Soil fertility — decomposition releases nutrients

The mathematical question: How do we model the flow of carbon from litter to soil organic matter to CO₂, accounting for different decay rates and environmental controls?

2. The Conceptual Model

Litter and Soil Carbon Pools

Three pools with different turnover times:

  1. Litter (fast pool):
    • Fresh leaves, fine roots, woody debris
    • Turnover time: 1–5 years
    • Rapidly decomposed
  2. Soil Organic Matter - Active (medium pool):
    • Partially decomposed material
    • Microbial biomass
    • Turnover time: 10–50 years
  3. Soil Organic Matter - Slow/Passive (slow pool):
    • Humus, recalcitrant compounds
    • Stabilized on clay minerals
    • Turnover time: 100–1000+ years

Decomposition as Exponential Decay

Each pool loses carbon through first-order decay:

\[\frac{dC}{dt} = I - k C\]

Where:

  • $C$ = carbon in pool (kg C/m²)
  • $I$ = input rate (kg C/m²/year)
  • $k$ = decay constant (year⁻¹)

Turnover time ($\tau$):

\[\tau = \frac{1}{k}\]

Example: Litter with $k = 0.5$ year⁻¹ has $\tau = 2$ years (half decomposes each 2 years).

Environmental Controls

Temperature: Decomposition rate increases exponentially with temperature.

Q₁₀ rule: Decomposition rate doubles per 10°C warming.

\[k(T) = k_{20} \times Q_{10}^{(T-20)/10}\]

Where:

  • $k_{20}$ = decay rate at 20°C
  • $Q_{10} \approx 2$ (typically 1.5–3)
  • $T$ = soil temperature (°C)

Moisture: Decomposition maximized at moderate moisture.

\[f(W) = \begin{cases} W/W_{\text{opt}} & \text{if } W < W_{\text{opt}} \\ 1 & \text{if } W = W_{\text{opt}} \\ (1-W)/(1-W_{\text{opt}}) & \text{if } W > W_{\text{opt}} \end{cases}\]

Where $W$ is relative soil moisture (0–1) and $W_{\text{opt}} \approx 0.6$.

Heterotrophic Respiration

Total heterotrophic respiration:

\[R_h = k_L C_L + k_A C_A + k_S C_S\]

Where subscripts L, A, S denote litter, active SOM, slow SOM.

Net Ecosystem Productivity:

\[\text{NEP} = \text{NPP} - R_h\]

Positive NEP: Ecosystem is a carbon sink
Negative NEP: Ecosystem is a carbon source

3. Building the Mathematical Model

Three-Pool Decomposition Model

State variables:

  • $C_L$ = litter carbon (kg C/m²)
  • $C_A$ = active SOM carbon (kg C/m²)
  • $C_S$ = slow SOM carbon (kg C/m²)

Litter dynamics:

\[\frac{dC_L}{dt} = I_L - k_L C_L\]

Where $I_L$ is litterfall from vegetation (from Model 22: turnover of leaves/roots).

Active SOM dynamics:

\[\frac{dC_A}{dt} = f_{LA} k_L C_L - k_A C_A\]

Where $f_{LA}$ is fraction of decomposed litter entering active SOM (typically 0.3–0.5, rest respired).

Slow SOM dynamics:

\[\frac{dC_S}{dt} = f_{AS} k_A C_A - k_S C_S\]

Where $f_{AS}$ is fraction of decomposed active SOM entering slow SOM (typically 0.2–0.3).

Decay Constants

Typical values at 20°C, optimal moisture:

Pool Decay rate ($k$, year⁻¹) Turnover time ($\tau$, years)
Litter 0.5–2.0 0.5–2
Active SOM 0.05–0.20 5–20
Slow SOM 0.001–0.01 100–1000

Variation by climate:

  • Tropical: High $k$ (warm, wet)
  • Temperate: Moderate $k$
  • Boreal: Low $k$ (cold)
  • Arid: Very low $k$ (dry, limited decomposers)

Steady State

At equilibrium:

\[C_L^* = \frac{I_L}{k_L}\] \[C_A^* = \frac{f_{LA} k_L C_L^*}{k_A} = \frac{f_{LA} I_L}{k_A}\] \[C_S^* = \frac{f_{AS} k_A C_A^*}{k_S} = \frac{f_{AS} f_{LA} I_L}{k_S}\]

Total soil carbon:

\[C_{\text{soil}}^* = C_L^* + C_A^* + C_S^*\]

Key insight: Slow pool dominates total carbon despite small input fraction, because $k_S \ll k_L$.

Heterotrophic Respiration at Steady State

\[R_h^* = k_L C_L^* + k_A C_A^* + k_S C_S^*\]

Substituting steady-state values:

\[R_h^* = I_L + f_{LA} I_L + f_{AS} f_{LA} I_L\]

Wait, this doesn’t equal $I_L$. Let me recalculate correctly.

Actually, at steady state, total inputs = total outputs:

\[R_h^* = (1 - f_{LA}) k_L C_L^* + (1 - f_{AS}) k_A C_A^* + k_S C_S^*\]

After all transfers, this sums to $I_L$ (conservation of carbon).

4. Worked Example by Hand

Problem: A temperate forest has:

  • Litterfall: $I_L = 0.4$ kg C/m²/year (from leaf/root turnover)
  • Litter decay rate: $k_L = 1.0$ year⁻¹
  • Active SOM decay rate: $k_A = 0.1$ year⁻¹
  • Slow SOM decay rate: $k_S = 0.005$ year⁻¹
  • Transfer fractions: $f_{LA} = 0.4$, $f_{AS} = 0.3$

Calculate steady-state carbon in each pool.

Solution

Litter pool:

\[C_L^* = \frac{I_L}{k_L} = \frac{0.4}{1.0} = 0.4 \text{ kg C/m}^2\]

Active SOM:

\[C_A^* = \frac{f_{LA} I_L}{k_A} = \frac{0.4 \times 0.4}{0.1} = \frac{0.16}{0.1} = 1.6 \text{ kg C/m}^2\]

Slow SOM:

\[C_S^* = \frac{f_{AS} f_{LA} I_L}{k_S} = \frac{0.3 \times 0.4 \times 0.4}{0.005}\] \[= \frac{0.048}{0.005} = 9.6 \text{ kg C/m}^2\]

Total soil carbon:

\[C_{\text{soil}}^* = 0.4 + 1.6 + 9.6 = 11.6 \text{ kg C/m}^2\]

Interpretation:

  • Slow SOM holds 83% of total soil carbon (9.6 / 11.6)
  • This matches observations: most soil carbon is old, stable material
  • Total = 11.6 kg C/m² = 116 tonnes C/ha (typical for temperate forest soil)

Heterotrophic respiration:

At steady state, $R_h = I_L = 0.4$ kg C/m²/year (all inputs eventually respired).

5. Computational Implementation

Below is an interactive decomposition simulator.

Total soil C: kg C/m²

Slow SOM fraction: %

Rh: kg C/m²/year

Mean residence time: years

Try this:

  • Tropical forest: High temperature → fast decomposition → low soil carbon accumulation
  • Boreal forest: Low temperature → slow decomposition → huge soil carbon stores (peat)
  • Desert: Dry conditions (low moisture) → very slow decomposition
  • Increase temperature: Speeds all decay rates → reduces equilibrium carbon
  • Increase litterfall: More inputs → higher steady-state carbon
  • Notice: Slow SOM takes centuries to equilibrate, dominates total carbon

Key insight: Cold/wet environments accumulate soil carbon because decomposition is slower than production. This is why peatlands and permafrost soils are massive carbon stores.

6. Interpretation

Global Soil Carbon Distribution

Largest soil carbon stores:

  1. Boreal forests & tundra: Cold → slow decomposition
  2. Peatlands: Waterlogged → anaerobic, very slow decay
  3. Grasslands: Deep roots, moderate climate

Smallest stores:

  1. Tropical forests: Fast turnover (hot, wet)
  2. Deserts: Low productivity + dry conditions

Paradox: Tropical forests have low soil C despite high productivity (fast decomposition balances high input).

Climate Change Feedbacks

Warming accelerates decomposition:

If temperature increases 5°C and Q₁₀ = 2:

\[k_{\text{new}} = k_{\text{old}} \times 2^{5/10} = k_{\text{old}} \times 1.41\]

41% faster decomposition → releases soil carbon → positive feedback.

Permafrost thaw:

  • Frozen organic matter begins decomposing
  • Releases CO₂ and CH₄
  • Major climate concern (~1000 Pg C in permafrost)

Soil Carbon Sequestration

Practices to increase soil C:

  1. No-till agriculture: Reduces disturbance → less decomposition
  2. Cover crops: More inputs to soil
  3. Biochar: Very slow decay ($k \approx 0.0001$ year⁻¹)
  4. Organic amendments: Compost, manure

Saturation: Soil can only hold so much carbon (determined by clay content, climate).

7. What Could Go Wrong?

Assuming Single Decay Rate

Real litter has multiple components:

  • Soluble sugars: $k \approx 10$ year⁻¹ (days to decompose)
  • Cellulose: $k \approx 1$ year⁻¹ (months)
  • Lignin: $k \approx 0.1$ year⁻¹ (years)

Better model: Multiple litter pools with different decay rates.

Ignoring Priming Effects

Priming: Fresh litter addition stimulates decomposition of old SOM.

Mechanism: New carbon provides energy for microbes → they also decompose stable SOM.

Effect: Non-linear interactions between pools.

Constant Transfer Fractions

Real $f_{LA}$, $f_{AS}$ vary with:

  • Litter quality: High lignin → less transfer to SOM
  • Soil clay content: More clay → more stabilization
  • Oxygen availability: Anaerobic → more SOM formation

Neglecting Nutrient Limitation

Decomposition requires nitrogen (microbes need C:N ~10:1).

Low N litter (wood, C:N > 100):

  • Slow decomposition
  • Microbes immobilize N from soil
  • Net N immobilization

High N litter (legumes, C:N < 20):

  • Fast decomposition
  • Net N mineralization

Our model is carbon-only. Full model couples C and N cycles.

8. Extension: Century Model

The Century model (Parton et al., 1987) is the classic soil C model:

Five pools:

  1. Structural litter (slow decay)
  2. Metabolic litter (fast decay)
  3. Active SOM (fast)
  4. Slow SOM (medium)
  5. Passive SOM (very slow)

Climate modifiers:

  • Temperature (Q₁₀)
  • Moisture (optimal curve)
  • Soil texture (clay slows decay)

Lignin effects:

  • High lignin → more to structural litter
  • Affects decay rates

Widely used in ecosystem models (CASA, TEM, BIOME-BGC).

9. Math Refresher: First-Order Decay

General Solution

For $\frac{dC}{dt} = I - kC$ with constant input $I$:

Steady state:

\[C^* = \frac{I}{k}\]

Approach to steady state:

\[C(t) = C^* + (C_0 - C^*) e^{-kt}\]

Where $C_0$ is initial value.

Time to 95% of equilibrium:

\[t_{95} = \frac{-\ln(0.05)}{k} = \frac{3}{k}\]

Example: If $k = 0.1$ year⁻¹:

\[t_{95} = \frac{3}{0.1} = 30 \text{ years}\]

Takes 30 years to reach 95% of equilibrium.

For slow pool ($k = 0.005$ year⁻¹):

\[t_{95} = \frac{3}{0.005} = 600 \text{ years}\]

This is why soil carbon responds slowly to land use change.

Summary

  • Decomposition converts dead organic matter to CO₂ via heterotrophic respiration
  • Three pools: Litter (fast, τ ~ 1 year), Active SOM (medium, τ ~ 10 years), Slow SOM (slow, τ ~ 100+ years)
  • First-order decay: $dC/dt = I - kC$, steady state $C^* = I/k$
  • Temperature control: Q₁₀ rule, decomposition doubles per 10°C
  • Moisture optimum: Too dry or too wet slows decomposition
  • Slow pool dominates total soil carbon (long turnover time)
  • Mean residence time: Total C / Input rate
  • NEP = NPP - R_h determines if ecosystem is carbon sink or source
  • Cold/wet environments accumulate large soil carbon stores
  • Climate warming accelerates decomposition → carbon release

References