GATE 2025 BT – Question 58

Fundamentals of Biological Engineering · Bioreaction Engineering: Kinetics of cell growth, substrate utilization and product formation, structured and unstructured models · 2 marks · Numerical answer

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.

$C_6H_{12}O_6 + NH_3 + O_2 \rightarrow CH_{1.8}O_{0.5}N_{0.2} + CO_2 + H_2O$

The moles of oxygen consumption per mole of glucose consumed during fermentation is _________. (Round off to two decimal places)

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: 3.46 to 3.50

Explanation

One mole of glucose has 6 C-mol, so the biomass is $0.4 \times 6 = 2.4$ C-mol and the $CO_2$ is $6 - 2.4 = 3.6$ mol. Nitrogen: $NH_3 = 0.2 \times 2.4 = 0.48$ mol. Hydrogen: $12 + 3(0.48) = 1.8(2.4) + 2w$ gives $w = 4.56$ mol of water. Oxygen: $6 + 2b = 0.5(2.4) + 2(3.6) + 4.56 = 12.96$, so $b = 3.48$ mol of $O_2$.