Chemostat steady state (Monod kinetics): An organism grows on glucose with parameters μm = 0.3 h^-1, Ks = 0.05 g·L^-1, and yield Yxs = 0.3 g·g^-1 (biomass per substrate). It is cultured in a 2 L continuous stirred-tank reactor (chemostat). Feed contains 10 g·L^-1 glucose and is supplied at 1 L·h^-1. What is the steady-state biomass concentration X in the reactor?

Introduction / Context:
Chemostat problems test understanding of dilution rate, Monod kinetics, washout, and how steady-state biomass concentration depends on the balance between growth and removal. Here, we decide whether the reactor operates at a viable steady state or washes out based on μ (specific growth rate) versus D (dilution rate).

Given Data / Assumptions:

• μm = 0.3 h^-1 (maximum specific growth rate).
• Ks = 0.05 g·L^-1 (Monod constant).
• Yxs = 0.3 g·g^-1 (not directly needed if washout occurs).
• V = 2 L (reactor volume).
• Feed flow F = 1 L·h^-1.
• Feed substrate S_in = 10 g·L^-1 (glucose).
• Ideal, well-mixed chemostat; single-limiting substrate; no cell recycle.

Concept / Approach:
In a chemostat, the dilution rate is D = F / V. At non-washout steady state, the net growth rate equals the washout rate: μ(S) = D. If the required μ(S) to match D exceeds μm, the culture cannot sustain itself and the reactor washes out (X → 0).

Step-by-Step Solution:
Compute dilution rate: D = F / V = 1 / 2 = 0.5 h^-1.Compare D with μm: μm = 0.3 h^-1, so D > μm.Growth cannot keep pace with washout because μ(S) ≤ μm for all S.Therefore, steady-state biomass X must go to zero (washout).

Verification / Alternative check:
Even if S rose to very high levels, μ(S) is bounded by μm. Since D = 0.5 h^-1 > 0.3 h^-1, equality μ = D is impossible; the only steady state is washout (X = 0; S approaches S_in).

Why Other Options Are Wrong:
0.4 or 0.5 g·L^-1: imply a viable steady state with μ(S) = D, which cannot happen when D > μm.

10 g·L^-1: that is the feed substrate concentration, not biomass. Biomass cannot exceed limits set by μ and D; here it goes to zero.

Common Pitfalls:

• Forgetting to compute D and compare it with μm.
• Trying to use yield Yxs when washout already determines X = 0.

Final Answer:
0 g·L^-1