Enthalpy problems KEY
Chem 440
1. Hammes, ch. 1, #1-2a
Answer:
Write balanced eqn:
C6H12O6(s) + 6O2(g) Æ 2H2O(l) + 6CO2(aq)
∆H°298 = 6*∆H°f(CO2g) + 6*∆H°f(H2O(l)) - 6*∆H°f(O2g) - ∆H°f(C6H12O6(s))
∆H°298 = 6*(-98.7) + 6*(-68.3) – 6* (0) – (-304.3)
∆H°298 = -697.7 kcal/mol
2. (Taken from Chang, Physical Chemistry for the Chemical and Biological Sciences).
Alcoholic fermentation is the process in which carbohydrates are broken down into ethanol and
carbon dioxide. The reaction is very complex and involves a number of enzyme-catalyzed steps.
The overall change is
C6H12O6(s) Æ 2C2H5OH(l) + 2CO2(g)
Calculate the standard enthalpy change for this reactions, assuming that the carbohydrate is α-Dglucose.
Answer:
∆H°298 = 2*∆H°f(C2H5OH(l)) + 2*∆H°f(CO2(g)) - ∆H°f(C6H12O6(s))
∆H°298 = 2* (-290.76) + 2* (-393.51) – (-1267.11)
∆H°298 = -101.43 kJ/mol
3. (Taken from Tinoco et al., Physical Chemistry)
Which of the above two reactions produces more energy per mole of glucose metabolized?
What fraction of heat is liberated in the less efficient metabolic pathway?
Answer:
From 1:
∆H°298 = -697.7 kcal/mol
-697.7 kcal/mol * 4.184 kJ/1 kcal = -2919.2 kJ/mol
From 2:
∆H°298 = -101.43 kJ/mol
So, the reaction in 1 produces more energy.
-101.4 = n * -2919.2
n = 0.035
the energy produced in reaction 2 is 3.5% of the energy produced in reaction 1.
Enthalpy problems KEY
Chem 440
4. (Taken from Tinoco et al., Physical Chemistry)
The enzyme catalase efficiently catalyzes the decomposition of hydrogen peroxide to give water
and oxygen. At room temperature, the reaction goes essentially to completion.
a) Using heats of formation, calculate ∆H°298 for the reaction
2H2O2(g) ' 2H2O(g) + O2(g)
∆H°f for gaseous H2O2 is -133.18 kJ.mol-1
∆H°f for gaseous H2O is -241.8 kJ.mol-1
Answer:
∆H°298 = 2*∆H°f(H2O(g)) + ∆H°f(O2(g)) – 2*∆H°f(H2O2(g))
∆H°298 = 2*(−241.8) + (0) − 2∗ (−133.18)
∆H°298 = -217.24 kJ/mol
b) The enzyme normally acts on an aqueous solution of hydrogen peroxide, for which the
equation is
2H2O2(aq) ' 2H2O(l) + O2(g)
What is ∆H°298 for this process?
Answer:
∆H°298 = ∆H°f(O2(g)) + 2*∆H°f(H2O(l)) – 2*∆H°f(H2O2(aq))
∆H°298 = 0 + 2*(-285.83) – 2*(-191.17)
∆H°298 = -189.3 kJ/mol
5. Hammes, Ch. 1, #1-3
Answer:
Balanced equations
2CH3COCOOH(l) + 5O2 Æ 4H2O + 6CO2
2CH3COOH(l) + 4O2 Æ 4H2O + 4CO2
∆H° = -227 kcal/mol
∆H° = -207kcal/mol
reverse the bottom equation and add them to get the desired equation.
Must also take negative of sign of heat of combustion for acetic acid, when adding these.
So, ∆H° for the oxication of pyruvic to acetic acid is:
(-227) + (207) = 20 kcal/mol
6. Hammes, Ch. 1, #1-6
we’ll be looking at this question is class.