# Bioprocess Engineering Questions and Answers – Enthalpy Change Due to Reaction

This set of Basic Bioprocess Engineering Questions and Answers focuses on “Enthalpy Change Due to Reaction”.

1. At what standard conditions does the heat of combustion calculated?
a) 26°C, 1 atm
b) 25°C, 0.5 atm
c) 25°C, 1 atm
d) 26°C, 0.5 atm

Explanation: The standard heat of combustion Δh°c is the specific enthalpy change associated with this reaction at standard conditions, usually 25°C and 1 atm pressure. The standard heat of reaction is the difference between the heats of combustion of reactants and products.

2. Hydrogen peroxide decomposes according to the following thermochemical reaction:
H2O2 (l) → H2O (l) + 1/2 O2(g); ΔH = -98.2 kJ
Calculate the change in enthalpy, ΔH, when 1.00 g of hydrogen peroxide decomposes.
a) -2.89 kJ
b) -2.80 kJ
c) -2.00 kJ
d) -2.85 kJ

Explanation: Molecular mass of H2O2 is 34.0 (2 x 1 for hydrogen + 2 x 16 for oxygen), which means that 1 mol H2O2 = 34.0 g H2O2.
Using these values:
ΔH = 1.00 g H2O2 x 1 mol H2O2 / 34.0 g H2O2 x -98.2 kJ / 1 mol H2O2
ΔH = -2.89 kJ.

3. Calculate ΔH if a piece of metal with a specific heat of 0.98 kJ-kg-1-K-1 and a mass of 2kg is heated from 22°C to 28°C.
a) 11.66 kg
b) 11.56 kg
c) 11.76 kg
d) 11.26 kg

Explanation: ΔH = q = cpsp × m × (ΔT) = (0.98) × (2) × (+6°) = 11.76 kg.

4. If a calorimeter’s ΔH is +2001 Joules, how much heat did the substance inside the cup lose?
a) +2000 J
b) – 2001 J
c) – 2000J
d) + 2001 J

Explanation: Since the heat gained by the calorimeter is equal to the heat lost by the system, then the substance inside must have lost the negative of +2001 J, which is – 2001 J.

5. Calculate the ΔH of the following reaction: CO2(g) + H2(g)O -> H2CO3(g) if the standard values of ΔHf are as follows: CO2(g) : -393.509 KJ/mol, H2O(g): – 241.83 KJ/mol, and H2CO3(g) : – 275.2 KJ/mol.
a) +360.139 KJ
b) +350.129 KJ
c) – 360.139 KJ
d) -350.129 KJ

Explanation: ΔH° = ƩΔvpΔH°f (products) – ƩΔvrΔH°f (reactants) so this means that you add up the sum of the ΔH’s of the products and subtract away the ΔH of the products:
(- 275.2 KJ) – (-393.509 KJ + -241.83KJ) = (-275.2) – (-635.339) = + 360.139 KJ.

6. Calculate ΔH if a piece of aluminum with a specific heat of 0.9 kJ-kg-1-K-1 and a mass of 1.6 kg is heated from 286°K to 299°K.
a) 16.52 kJ
b) 17.72 kJ
c) 18.72 kJ
d) 15.52 kJ

Explanation: ΔH = q = cpsp × m × (ΔT) = (0.9) × (1.6) × (13) = 18.72 kJ.

7. Calculate the ΔH value of the reaction:
HCl + NH3 → NH4Cl
(ΔH values for HCl is -92.30; NH2 is -80.29; NH4Cl is -314.4).
a) 141.8
b) 121.8
c) 131.8
d) 151.8

Explanation: ΔH = ΔHproducts – ΔHreactants
ΔHproducts = -314.4
ΔHreactants = -92.30 + (-80.29) = -172.59
ΔH = -314.4 – 172.59 = 141.8.

8. Calculate ΔH for the reaction:
N2 + 3H2 → 2NH3
(The bond dissociation energy for N-N is 163 kJ/mol; H-H is 436 kJ/mol; N-H is 391 kJ/mol).
a) 867
b) 895
c) 847
d) 875

Explanation: ΔH = ΔHproducts – ΔHreactants
To use the bond dissociation energies, we must determine how many bonds are in the products and the reactants. In NH3 there are 3 N-H bonds so in 2 NH3 there are 6 N-H bonds. In N2 there is 1 N-N bond and in 3H2 there are 3 H-H bonds.
ΔHproducts = 6(391) = 2346
ΔHreactants = 163 + 3(436) = 1471
ΔH = 2346 – 1471 = 875.

9. Consider the reaction:
N2 + O2 → 2NO; ΔH = +180.6 kJ
what is the enthalpy change for the formation of one mole of nitrogen (II) oxide?
a) 96.80 kJ
b) 95.30 kJ
c) 96.50 kJ
d) 95.50 kJ

Explanation: Here we use the conversion factor 180.6 kJ/2 mol NO.
ΔH = 1 mol NO × (180.6 kJ/2 mol NO) = 95.30 kJ.

10. Refer to Q9 and answer what is the enthalpy change for the reaction of 100.0 g of nitrogen with excess oxygen?
a) 1164 kJ
b) 1190 kJ
c) 1194 kJ
d) 1160 kJ

Explanation: Here, we need to convert grams of N2 to moles of N2 and use the conversion factor 180.6 kJ/1 mol N2.

180.6 kJ × (1 mol N2/28.01 g N2) × (180.6 kJ/1 mol N2) = 1164 kJ.

11. Calculate the work done in the following reaction when 1 mol of SO2 is oxidised at constant pressure at 50 °C. State whether work is on the system or by the system.
2SO2(g) + O2(g) → 2 SO3(g)
Given: Temperature = T = 5 °C = 50 + 273 = 323 K , R = 8.314 J K-1 mol-1 ,
a) 1435 kJ
b) 1343 kJ
c) 1345 kJ
d) 1433 kJ

Explanation: The reaction is 2SO2(g) + O2(g) → 2 SO3(g)
Given 1 mole of SO2 is used, hence dividing equation by 2 to get 1 mol of SO2

SO2(g) + ½O2(g) → SO3(g)

Δn = nproduct (g) – nreactant (g) = (1) – (1 + 1/2) = 1 – 3/2 = – 1/2
Work done in chemical reaction is given by
∴ W = – Δn RT = – (-1/2) mol × 8.314 J K-1 mol-1 × 323 K = 1343 J
∴ W = + 1343 J
Positive sign indicates that work is done by the surroundings on the system

Work done by the surroundings on the system in the reaction is 1343 J.

12. Calculate the work done in the following reaction when 2 mol of NH4NO3 decomposes at constant pressure at 10 °C. State whether work is on the system or by the system.

NH4NO3(s) → N2O(g) + 2 H2O(g)

Given: Temperature = T = 100 °C = 100 + 273 = 373 K, R = 8.314 J K-1 mol-1
a) – 18.61 kJ
b) – 18.86 kJ
c) – 18.65 kJ
d) – 18.85 kJ

Explanation: The reaction is NH4NO3(s) → N2O(g) + 2 H2O(g)

Given 2 mol of NH4NO3 decomposes, hence multiplying equation by 2
2 NH4NO3(s) → 2 N2O(g) + 4 H2O(g)
Δn = nproduct (g) – nreactant (g) = (2 + 4) -(0) = 6
Work done in chemical reaction is given by
∴ W = – Δn RT = – (6) mol × 8.314 J K-1 mol-1 × 373 K = – 18607 J
∴ W = – 18.61 kJ
Negative sign indicates that work is done by the system on the surroundings
Work done by the surroundings on the system in the reaction is – 18.61 kJ.

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