Question

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 \[ 2C_6H_{12}O_6 + 0.2 NH_3 + 10.35 O_2 \to CH_{1.8} O_{0.5} N_{0.2} + 0.2 C_2 H_6 O + 10.6 CO_2 + 10.8 H_2 O \] The heat of combustion values for different compounds are tabulated below with reference to CO$_2$, H$_2$O, O$_2$, and N$_2$ at standard conditions. Heat of Combustion 

Compound	Heat of combustion (kJ mol-1)
C6H12O6	2802
NH3	383
CH1.8O0.5N0.2	560
C2H6O	1366

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

Hint:

When calculating the specific rate of heat production, use the heat of combustion of the substrates and products along with the specific growth rate to determine the heat released per mole of substrate consumed.

Answer 1

Correct Answer: 2420

To calculate the specific rate of heat production, we must use the stoichiometry of the reaction along with the heat of combustion values. The specific rate of heat production is the rate of heat released per unit of substrate converted, which is related to the heat of combustion of the products and reactants.
Step 1: Write the overall stoichiometry of the reaction.
The reaction is: \[ 2C_6H_{12}O_6 + 0.2 NH_3 + 10.35 O_2 \to CH_{1.8}O_{0.5}N_{0.2} + 0.2 C_2 H_6 O + 10.6 CO_2 + 10.8 H_2 O. \] Step 2: Calculate the heat of combustion for reactants and products.
- Heat of combustion of $2 \, \text{mol} \, C_6H_{12}O_6 = 2 \times 2802 = 5604 \, \text{kJ}$. - Heat of combustion of $0.2 \, \text{mol} \, NH_3 = 0.2 \times 383 = 76.6 \, \text{kJ}$. - Heat of combustion of $10.35 \, \text{mol} \, O_2 = 10.35 \times 0 \, \text{kJ} = 0 \, \text{kJ}$ (since oxygen is not combusted). For the products:
- Heat of combustion of $1 \, \text{mol} \, CH_{1.8} O_{0.5} N_{0.2} = 560 \, \text{kJ}$.
- Heat of combustion of $0.2 \, \text{mol} \, C_2H_6O = 0.2 \times 1366 = 273.2 \, \text{kJ}$.
- Heat of combustion of $10.6 \, \text{mol} \, CO_2 = 10.6 \times 0 \, \text{kJ} = 0 \, \text{kJ}$ (CO$_2$ is also not combusted).
- Heat of combustion of $10.8 \, \text{mol} \, H_2O = 0 \, \text{kJ}$ (water is not combusted).
Step 3: Calculate total heat of combustion for the reaction.
\[ \text{Heat of combustion (products)} = 560 + 273.2 = 833.2 \, \text{kJ}. \] \[ \text{Heat of combustion (reactants)} = 5604 + 76.6 = 5680.6 \, \text{kJ}. \] Step 4: Determine the heat released during the reaction.
The heat released per mole of substrate converted is the difference between the heat of combustion of the reactants and the products: \[ \Delta H = 5680.6 - 833.2 = 4847.4 \, \text{kJ}. \] Step 5: Account for the specific growth rate.
The specific growth rate is given as 0.5 h$^{-1}$, which means that per hour, 0.5 moles of substrate (glucose) are consumed. Therefore, the rate of heat production is: \[ \text{Rate of heat production} = \Delta H \times \text{specific growth rate} = 4847.4 \times 0.5 = 2423.7 \, \text{kJ mol}^{-1} \, \text{h}^{-1}. \] Step 6: Round to nearest integer:
\[ \boxed{2424} \, \text{kJ mol}^{-1} \, \text{h}^{-1}. \]