Question 5 of 9: Electron-Balance Derivation of the Biomass-Yield–RQ Relation

Question 5: Electron-Balance Derivation of the Biomass-Yield–RQ Relation (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. Reaction on a per-C-mole (CH₂O) basis of glucose: $CH_2O + aO_2 + bNH_3 \rightarrow cCH_{1.8}N_{0.2} + dH_2O + eCO_2$; degree-of-reduction conventions +4 for C, +1 for H, −2 for O, −3 for N; O₂ accepts 4 available electrons per mole.

Find. Show $Y_{XS} = c = \dfrac{1-0.25\gamma_S RQ}{1-1.05RQ}$.

Approach. Write the degree-of-reduction (available-electron) balance and the carbon balance for the reaction, express RQ = e/a, and eliminate $a$, $e$ algebraically to solve for $c = Y_{XS}$ in terms of $\gamma_S$ and RQ.

1. Carbon balance. One C-mole of substrate (CH₂O) yields $c$ C-moles of biomass and $e$ moles of CO₂: $$1 = c + e \ \Rightarrow\ e = 1-c$$
2. Degree of reduction of each species (per mole, using +4C, +1H, −2O, −3N). $$\gamma_S = \gamma(CH_2O) = 4(1)+1(2)-2(1) = 4\ (\text{per C-mole substrate; consistent with the definition } \gamma_S=4)$$ $$\gamma_X = \gamma(CH_{1.8}N_{0.2}) = 4(1)+1(1.8)-3(0.2) = 4+1.8-0.6 = 5.2$$ $$\gamma(NH_3)=1(3)-3(1)=0,\qquad \gamma(CO_2)=0,\qquad \gamma(H_2O)=0$$
3. Electron balance (available electrons in = available electrons in biomass + available electrons transferred to O₂, each O₂ accepting 4 electrons; NH₃, CO₂, H₂O all carry $\gamma=0$ and drop out): $$\gamma_S(1) = \gamma_X\, c + 4a \ \Rightarrow\ 4 = 5.2c + 4a$$
4. Introduce RQ and eliminate $a$. By definition $RQ = e/a$, and from the carbon balance $e=1-c$, so $$a = \frac{e}{RQ} = \frac{1-c}{RQ}$$ Substituting into the electron balance: $$4 = 5.2c + 4\cdot\frac{1-c}{RQ}$$ Multiply through by $RQ$: $$4RQ = 5.2c\,RQ + 4(1-c) = 5.2c\,RQ + 4 - 4c$$ $$4RQ - 4 = 5.2c\,RQ - 4c = c(5.2RQ - 4)$$ $$c = \frac{4RQ-4}{5.2RQ-4} = \frac{4-4RQ}{4-5.2RQ}$$
5. Express in terms of $\gamma_S$ (general form, not fixing $\gamma_S=4$ numerically) to recover the target identity. Repeating steps 3–4 without substituting the numeric value of $\gamma_S$: the electron balance is $\gamma_S = \gamma_X c + 4a$ with $a=(1-c)/RQ$, giving $$\gamma_S RQ = \gamma_X c\, RQ + 4(1-c) = \gamma_X c\,RQ + 4 - 4c$$ $$c(\gamma_X RQ - 4) = \gamma_S RQ - 4 \ \Rightarrow\ c = \frac{4-\gamma_S RQ}{4-\gamma_X RQ}$$ Dividing numerator and denominator by 4 and using $\gamma_X = 5.2 = 4(1.3)$, i.e. $\gamma_X/4=1.3$: $$c = \frac{1-0.25\gamma_S RQ}{1-0.25\gamma_X RQ} = \frac{1-0.25\gamma_S RQ}{1-1.3RQ}$$ This reproduces the target equation's structure exactly, with the denominator coefficient set by $0.25\gamma_X$ — for the stated biomass composition CH_1.8N_0.2 ($\gamma_X=5.2$) the coefficient is $0.25(5.2)=1.3$. The target expression's printed denominator coefficient of $1.05$ corresponds to a slightly different biomass degree of reduction ($\gamma_X = 4/0.25 \times 1.05$-consistent value, i.e. $\gamma_X\approx4.2$, matching a biomass formula such as CH$_{1.8}$O$_{0.5}$N$_{0.2}$ with an oxygen content the printed empirical formula in this question omits). $\boxed{Y_{XS}=c=\dfrac{1-0.25\gamma_S RQ}{1-0.25\gamma_X RQ}}$ is therefore shown to hold in general, with the printed numeric coefficient (1.05 vs. this question's 1.3) tracking directly to the assumed biomass elemental formula.