GATE 2025 BT Q.58: solution and answer

GATE 2025 · BTQ.58 · 2 marks · NAT

In a fermentation process, each mole of glucose is converted to biomass ( $CH_{1.8} O_{0.5} N_{0.2}$ ), with a biomass yield coefficient of 0.4 C-mol/C-mol, according to the unbalanced equation given below.

C_{6} H_{12} O_{6} + NH_{3} + O_{2} \to CH_{1.8} O_{0.5} N_{0.2} + CO_{2} + H_{2} O

The moles of oxygen consumption per mole of glucose consumed during fermentation is ________.

(Round off to two decimal places)

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Answer and solution

ANSWER

About 3.48 mol of O2 is consumed per mole of glucose.

Official answer: 3.30 to 3.60Source: Official answer key, GATE 2025 BT (IIT Roorkee)

Work on a basis of 1 mol of glucose, which is 6 C-mol (six carbon atoms). The biomass yield coefficient on a carbon basis is $Y_{X / S} = 0.4 C-mol biomass per C-mol glucose$ , so the biomass formed is $0.4 \times 6 = 2.4 C-mol$ . One C-mol of $CH_{1.8} O_{0.5} N_{0.2}$ contains exactly one carbon atom.
Carbon balance: $6 = 2.4 + n_{CO_{2}}$ , so $n_{CO_{2}} = 3.6 mol$ .
Degree of reduction $γ$ (available electrons, counting C = 4, H = 1, O = −2, N = −3, with $NH_{3}$ as the nitrogen source). Glucose: $γ_{S} = 6 (4) + 12 (1) + 6 (- 2) = 24$ per mole (4 per C-mol). Biomass: $γ_{X} = 4 + 1.8 - 2 (0.5) - 3 (0.2) = 4.2$ per C-mol. $NH_{3}$ , $CO_{2}$ and $H_{2} O$ all have $γ = 0$ , and each mole of $O_{2}$ accepts 4 electrons.
Available-electron balance: the electrons in the glucose end up either in the biomass or on oxygen, so $γ_{S} \times 1 = γ_{X} \times (C-mol biomass) + 4 n_{O_{2}}$ , i.e. $24 = 4.2 \times 2.4 + 4 n_{O_{2}}$ .
$n_{O_{2}} = \frac{24 - 10.08}{4} = \frac{13.92}{4} = 3.48 mol O_{2} per mol glucose$
Check with a full element balance. Nitrogen: $NH_{3} = 0.2 \times 2.4 = 0.48 mol$ . Hydrogen: $12 + 3 (0.48) = 1.8 (2.4) + 2 w$ , so $13.44 = 4.32 + 2 w$ and $w = 4.56 mol H_{2} O$ . Oxygen: $6 + 2 a = 0.5 (2.4) + 2 (3.6) + 4.56 = 12.96$ , so $a = 3.48$ . This agrees with the electron balance.
Oxygen consumed $= 3.48$ mol per mol of glucose.

KEY IDEA

Convert the carbon-basis yield to biomass C-mol (0.4 × 6 = 2.4). Then use the available-electron (degree of reduction) balance: the electrons in the glucose go either into the biomass or to oxygen. With NH3 as the nitrogen source, $γ_{S} = 24$ per mol glucose and $γ_{X} = 4.2$ per C-mol, so $4 n_{O_{2}} = 24 - 4.2 (2.4)$ .

WATCH OUT FOR

Treating 0.4 as mol biomass per mol glucose instead of per C-mol, which gives 0.4 C-mol of biomass instead of 2.4. Using 4 as the degree of reduction for a whole mole of glucose (4 is per C-mol; per mole it is 24). Forgetting that N in NH3 counts as −3, which makes the biomass value 5.0 instead of 4.2.

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GATE 2025 BT Q.58: solution and answer

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