GATE 2022 BT Q.62: solution and answer

GATE 2022 · BTQ.62 · 2 marks · NAT

Consider the growth of S. cerevisiae under aerobic condition in a bioreactor and the specific growth rate of yeast is $0.5 h^{- 1}$ . The overall reaction of the process is

2 C_{6} H_{12} O_{6} + 0.2 NH_{3} + 10.35 O_{2} \to CH_{1.8} O_{0.5} N_{0.2} + 0.2 C_{2} H_{6} O + 10.6 CO_{2} + 10.8 H_{2} O

The heat of combustion values for different compounds are tabulated below with the reference to $CO_{2}$ , $H_{2} O$ , $O_{2}$ , and $N_{2}$ at standard conditions.

| Compound | Heat of combustion ( $kJ mol^{- 1}$ ) |
|---|---|
| $C_{6} H_{12} O_{6}$ | 2802 |
| $NH_{3}$ | 383 |
| $CH_{1.8} O_{0.5} N_{0.2}$ | 560 |
| $C_{2} H_{6} O$ | 1366 |

The specific rate of heat production (rounded off to nearest integer) is ________ $kJ mol^{- 1} h^{- 1}$ .

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Answer and solution

ANSWER

The specific rate of heat production is $q_{Q} = μ Y_{Q / X} = 0.5 \times 4847.4 \approx 2424 kJ mol^{- 1} h^{- 1}$ .

Official answer: 2420 to 2430Source: Official answer key, GATE 2022 BT (IIT Kharagpur)

Symbols used: $μ$ is the specific growth rate of the yeast, $0.5 h^{- 1}$ . $Y_{Q / X}$ is the heat released per mol of biomass ( $CH_{1.8} O_{0.5} N_{0.2}$ ) formed, in $kJ mol^{- 1}$ . $q_{Q}$ is the specific rate of heat production, meaning heat released per mol of biomass present per hour, in $kJ mol^{- 1} h^{- 1}$ .
Hess's law with heats of combustion. The heats are given relative to $CO_{2}$ , $H_{2} O$ , $O_{2}$ and $N_{2}$ , so each of these four has a heat of combustion of zero. The heat released by the reaction is therefore $Q = reactants \sum n_{i} (Δ H_{c})_{i} - products \sum n_{j} (Δ H_{c})_{j}$
Reactants: $2 (2802) + 0.2 (383) + 10.35 (0) = 5604 + 76.6 = 5680.6 kJ$
Products: $1 (560) + 0.2 (1366) + 10.6 (0) + 10.8 (0) = 560 + 273.2 = 833.2 kJ$
Heat released: $Q = 5680.6 - 833.2 = 4847.4 kJ$ , so the reaction is exothermic with $Δ H_{r} = - 4847.4 kJ$ . The equation as written forms exactly 1 mol of biomass, so $Y_{Q / X} = 4847.4 kJ$ per mol of $CH_{1.8} O_{0.5} N_{0.2}$ formed.
The specific growth rate $μ = \frac{1}{X} \frac{d X}{d t} = 0.5 h^{- 1}$ means that each mol of biomass present forms $0.5$ mol of new biomass per hour.
Specific rate of heat production: $q_{Q} = \frac{1}{X} \frac{d Q}{d t} = Y_{Q / X} μ = 4847.4 \times 0.5 = 2423.7 kJ mol^{- 1} h^{- 1}$
Rounded to the nearest integer: $q_{Q} \approx 2424 kJ mol^{- 1} h^{- 1}$ .

KEY IDEA

When heats of combustion share one reference state ( $CO_{2}$ , $H_{2} O$ , $O_{2}$ , $N_{2}$ all zero), the heat a reaction releases is (sum over reactants of $n Δ H_{c}$ ) minus (sum over products of $n Δ H_{c}$ ). The equation forms 1 mol of biomass, so this difference is the heat yield $Y_{Q / X}$ per mol of biomass. Multiplying it by the specific growth rate $μ$ gives the heat released per mol of biomass per hour: $q_{Q} = μ Y_{Q / X}$ .

WATCH OUT FOR

Three slips give answers outside the key. (1) Leaving out the ammonia term, which gives $0.5 (5604 - 833.2) \approx 2385$ . (2) Leaving out the ethanol by-product, which gives $0.5 (5680.6 - 560) \approx 2560$ . (3) Using glucose's heat of combustion once instead of twice. Two other errors are stopping at the heat yield (4847) without multiplying by $μ$ , and getting the sign backwards by subtracting reactants from products. In that last case the magnitude is right but the reaction looks heat-absorbing, when it actually releases heat.

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GATE 2022 BT Q.62: solution and answer

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