# DBT JRF 2020 — Question 148

Exam: DBT JRF 2020 · Section: B · Topic: Bioprocess Engineering · Difficulty: Medium

## Question

The equation for aerobic production of acetic acid from ethanol is: $C_2H_5OH + O_2 \rightarrow CH_3CO_2H + H_2O$ Acetobacter aceti bacteria are added to vigorously-aerated medium containing 10 $g.l^{-1}$ ethanol. After some time, the ethanol concentration is 2 $g.l^{-1}$ and 7.5 $g.l^{-1}$ of acetic acid is produced. What is the observed yield ( $Y_O$ in $g.g^{-1}$ ) and theoretical yield ( $Y_T$ in $g.g^{-1}$ ) of acetic acid from ethanol?

## Options

- **A.** $Y_O = 0.69, Y_T = 1.0$
- **B.** $Y_O = 0.94, Y_T = 1.5$
- **C.** $Y_O = 0.79, Y_T = 1.1$
- **D.** $Y_O = 0.94, Y_T = 1.3$

## Answer

Option D

## Explanation

Acetic acid (vinegar) production by Acetobacter aceti: $\text{C}_2\text{H}_5\text{OH} + \text{O}_2 \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}$ . Data: Ethanol consumed = $10 - 2 = 8 \text{ g/L}$ ; Acetic acid produced = $7.5 \text{ g/L}$ . Observed yield: $Y_O = \dfrac{\text{Acetic acid produced}}{\text{Ethanol consumed}} = \dfrac{7.5}{8} = 0.9375 \approx 0.94 \text{ g/g}$ . Theoretical yield: From stoichiometry, $M_r(\text{C}_2\text{H}_5\text{OH}) = 46 \text{ g/mol}$ ; $M_r(\text{CH}_3\text{COOH}) = 60 \text{ g/mol}$ . $Y_T = \dfrac{60}{46} = 1.304 \approx 1.3 \text{ g acetic acid / g ethanol}$ . Answer: $Y_O = 0.94 \text{ g/g}$ , $Y_T = 1.3 \text{ g/g}$ .

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