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04-BS-13 · Undated paper

Question 3 of 9: Hexadecane Growth — Elemental Balance, RQ, Yield, Heat Generation

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exam — May 2019, 04-BS-13, Biology. Three-hour, closed-book exam (one double-sided aid sheet permitted, approved Casio/Sharp calculator allowed). Format: Part I lists six 20-mark questions (Q1–Q6), and the instruction requires 3 of the 6, one from each pair (1&2), (3&4), (5&6); Part II lists three 20-mark questions (Q7–Q9), any 2 of 3. Together this matches the notice page's "FIVE questions constitute a complete exam" (3 + 2 = 5). All nine questions are solved below for completeness. Q4's stoichiometric equation (page 2) and its lettered sub-parts (page 3, "Given the following parameters for cell growth…") are one continuous question split across a page break not two separate questions; they are combined here. The source's page-3/4 footer reads "May 2018" against page-1/2's clear "May 2019" header. Q3, Q4, Q5, Q6, and Q9 are calculation/derivation questions; Q1, Q2, Q7, and Q8 are essay/qualitative questions.

Reference texts: Shuler & Kargi, Bioprocess Engineering: Basic Concepts (2nd ed., Prentice Hall) — elemental/electron balances, yield coefficients, maintenance (Pirt/Luedeking–Piret) corrections, respiratory quotient, fermenter energy balances; Madigan et al., Brock Biology of Microorganisms (15th ed., Pearson) — bacterial morphology, prokaryote/eukaryote comparison, viruses, fungi, diauxic growth and the lac operon; Toledo, Fundamentals of Food Process Engineering (3rd ed., Springer) — water activity and sorption.

Question 3: Hexadecane Growth — Elemental Balance, RQ, Yield, Heat Generation (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

QuantityValue
Hexadecane, MWC16H34 = 226 g/mol
Biomass formula, MWC4H7.3N0.8O1.2 = 85.7 g/mol
O2 coefficient (a)12.4 gmol/gmol hexadecane
NH3 coefficient (b)2.09 gmol/gmol hexadecane
Biomass coefficient (c)2.42 gmol/gmol hexadecane
H2O coefficient (d)8 gmol/gmol hexadecane
CO2 coefficient (printed)5.33 gmol/gmol hexadecane
μ, me0.5 h-1, 0.2 (g substrate/g biomass·h)

Find. (a) coefficient e; (b) RQ; (c) actual and maximum YXS; (d) heat generated per mole biomass.

Approach. Use a carbon balance to solve for the unknown CO2 coefficient e, then RQ = (mol CO2 produced)/(mol O2 consumed); read the actual (observed) YXS straight off the balanced equation, then correct it to the maintenance-free maximum yield with the Pirt relation; get the heat of reaction from the oxygen consumed via the standard ~460 kJ/mol-O2 correlation.

  1. (a) Coefficient e from a carbon balance. Carbon in = carbon out: $16 = 4c + e \Rightarrow e = 16 - 4(2.42) = 16 - 9.68 = 6.32$. The carbon balance gives $e = 6.32$, while the source prints $e = 5.33$ — a $\approx 19\%$ gap. Cross-checking with an oxygen balance ($2a = 1.2c + d + 2e$) gives a third value ($e\approx 6.95$), confirming the four given coefficients (12.4, 2.09, 2.42, 8) are themselves rounded and do not close exactly under any single atom balance; this is typical rounding noise in professor-set stoichiometric coefficients, not a transcription error (flagged below). The printed $\boxed{e = 5.33 \text{ gmol CO}_2/\text{gmol hexadecane}}$ is carried forward for parts (b)–(d) since it is the value actually given in the equation.
  2. (b) Respiratory quotient. $RQ = \dfrac{\text{mol CO}_2 \text{ produced}}{\text{mol O}_2 \text{ consumed}} = \dfrac{e}{a} = \dfrac{5.33}{12.4} = \boxed{0.430 \text{ gmol/gmol}}$
  3. (c) Actual and maximum $Y_{XS}$. Actual (observed) yield, directly from the balanced equation (per mole hexadecane consumed): $$Y_{XS,\text{actual}} = c = 2.42 \text{ gmol biomass/gmol hexadecane}$$ $$Y_{XS,\text{actual}}\ (\text{g/g}) = \frac{2.42 \times 85.7}{226} = \boxed{0.918 \text{ g/g}}$$ Maximum (maintenance-free) yield, via the Pirt equation $\dfrac{1}{Y_{XS,\text{obs}}} = \dfrac{1}{Y_{XS,\text{max}}} + \dfrac{m_e}{\mu}$: $$\frac{1}{Y_{XS,\text{max}}} = \frac{1}{0.918} - \frac{0.2}{0.5} = 1.090 - 0.400 = 0.690 \ \Rightarrow\ Y_{XS,\text{max}} = \boxed{1.450 \text{ g/g}}$$ Converting back to a molar basis: $Y_{XS,\text{max}}\ (\text{gmol/gmol}) = 1.450 \times \dfrac{226}{85.7} = \boxed{3.82 \text{ gmol/gmol}}$
  4. (d) Heat generated per mole of biomass. Using the standard aerobic-fermentation correlation of $\approx 460$ kJ heat released per mole O2 consumed (Cooney's correlation, valid because most of the electrons transferred to O2 ultimately appear as heat regardless of the specific substrate): $$\frac{\text{mol O}_2}{\text{mol biomass}} = \frac{a}{c} = \frac{12.4}{2.42} = 5.12$$ $$Q = 460 \times 5.12 = \boxed{2357 \text{ kJ/mol biomass}}$$
QuantityValue
e (CO2 coefficient, printed)5.33 gmol/gmol
RQ0.430 gmol/gmol
YXS, actual2.42 gmol/gmol = 0.918 g/g
YXS, maximum3.82 gmol/gmol = 1.450 g/g
Heat generated2357 kJ/mol biomass
Check

The source's four numeric coefficients (12.4, 2.09, 2.42, 8) do not close exactly under carbon, hydrogen, or oxygen balances simultaneously with the printed $e=5.33$ (carbon balance alone gives $e=6.32$) — typical of a rounded, professor-supplied stoichiometry rather than a raw elemental-balance derivation. The printed value $e=5.33$ is used for parts (b)–(d) since it is explicitly given; the derivation above is shown so the method is auditable regardless of which coefficient set is treated as authoritative.