Enthalpy of Formation
Energetics Worksheet
Practice Questions
1.
Given the standard enthalpies of formation of calcium hydroxide, calcium oxide and water are –1003, –635 and –286 kJ mol–1 respectively, calculate the standard molar enthalpy change for the formation of calcium hydroxide from calcium oxide and water.
CaO + H2O
→
Ca(OH)2
Calculation
ΔH = ΣΔHf(Products) – ΣΔHf(Reactants)
ΔH = [-1003] – [-635 + (-286)]
ΔH = -1003 – (-921)
ΔH = [-1003] – [-635 + (-286)]
ΔH = -1003 – (-921)
Answer
-82 kJ mol-1
2.
The reaction that takes place during the combustion of the solid rocket booster fuel has been summarized as:
10Al(s) + 6NH4ClO4 (s) → 4Al2O3 (s) + 2AlCl3 (s) + 12H2O (l) + 3N2 (g)
Given the standard enthalpies of formation of NH4ClO4 (s), Al2O3 (s), AlCl3 (s), H2O (l) are -295.3, -1675.7, -704.2, and -285.8 kJ mol–1 respectively, calculate the standard enthalpy change at 298 K for this reaction as given.
10Al(s) + 6NH4ClO4 (s) → 4Al2O3 (s) + 2AlCl3 (s) + 12H2O (l) + 3N2 (g)
Given the standard enthalpies of formation of NH4ClO4 (s), Al2O3 (s), AlCl3 (s), H2O (l) are -295.3, -1675.7, -704.2, and -285.8 kJ mol–1 respectively, calculate the standard enthalpy change at 298 K for this reaction as given.
ΣΔHf (Products)
4(-1675.7) + 2(-704.2) + 12(-285.8) + 3(0)
-6702.8 – 1408.4 – 3429.6 = -11,540.8
-6702.8 – 1408.4 – 3429.6 = -11,540.8
ΣΔHf (Reactants)
10(0) + 6(-295.3) = -1,771.8
Total Enthalpy Change
-11,540.8 – (-1,771.8) = -9,769 kJ mol-1
3.
Calculate the standard enthalpy change of combustion of ethyne given the standard enthalpies of formation of C2H2, CO2 and H2O are +226, -394 and -286 kJmol-1 respectively.
C2H2 + 2.5O2
→
2CO2 + H2O
Calculation
Products: [2(-394) + (-286)] = -1074
Reactants: [+226]
-1074 – 226
Reactants: [+226]
-1074 – 226
Answer
-1300 kJ mol-1
4.
Calculate the enthalpy change of formation of ethanol, given the standard enthalpy change of formation of ethene and water are +52.3 and -285 respectively, and the enthalpy change for the hydration of ethene is -46.0 kJmol-1.
2C + 3H2 + 0.5O2
→
C2H5OH
Using Hess’s Law Cycle
ΔHf(ethanol) = ΔHf(ethene) + ΔHf(water) + ΔH(hydration)
= +52.3 + (-285) + (-46.0)
= +52.3 + (-285) + (-46.0)
Answer
-278.7 kJ mol-1
5.
Calculate the enthalpy change of formation of CO2 given the standard enthalpy change of formation of KOH, H2O and K2CO3 are -424.8, -285.8, and -1151.2 kJmol-1 respectively and the enthalpy change of the reaction between carbon dioxide and potassium hydroxide is -193.9 kJmol-1.
Reaction
CO2 + 2KOH → K2CO3 + H2O (ΔH = -193.9)
Formula
ΔHr = ΣΔHf(Products) – ΣΔHf(Reactants)
-193.9 = [-1151.2 + (-285.8)] – [x + 2(-424.8)]
-193.9 = [-1151.2 + (-285.8)] – [x + 2(-424.8)]
Rearranging
-193.9 = -1437 – (x – 849.6)
x = -1437 + 849.6 + 193.9
x = -1437 + 849.6 + 193.9
Answer
-393.5 kJ mol-1
6.
The decomposition of hydrogen peroxide in the presence of iodide ion occurs in two steps:
H2O2 (aq) + I– (aq) → H2O(l) + OI– (aq)
H2O2 (aq) + OI– (aq) → H2O(l) + O2 (g) + I– (aq)
Given the enthalpy change of formation of H2O2; H2O, IO– and I– are -187.6 and -285.8, -104.4, and −55.9 kJmol-1 respectively, calculate the enthalpy change for the decomposition of H2O2 in the presence of KI.
H2O2 (aq) + I– (aq) → H2O(l) + OI– (aq)
H2O2 (aq) + OI– (aq) → H2O(l) + O2 (g) + I– (aq)
Given the enthalpy change of formation of H2O2; H2O, IO– and I– are -187.6 and -285.8, -104.4, and −55.9 kJmol-1 respectively, calculate the enthalpy change for the decomposition of H2O2 in the presence of KI.
2H2O2
→
2H2O + O2
Calculation
Products: 2(-285.8) + 0 = -571.6
Reactants: 2(-187.6) = -375.2
-571.6 – (-375.2)
Reactants: 2(-187.6) = -375.2
-571.6 – (-375.2)
Answer
-196.4 kJ mol-1