Enthalpy of Combustion Questions
Energetics Worksheet
Using Enthalpy of Combustion Data
| Standard Enthalpy of Combustion (ΔHcθ) / kJ mol-1 | ||||
|---|---|---|---|---|
| C2H4 -1387.4 |
C2H5OH -1370.7 |
C8H18 -5470 |
C12H26 -8083.2 |
H2 -285.8 |
| PH3 -750.0 |
C6H14 -4163 |
S -297.0 |
C2H4O -1167 |
|
1.
Use the information in the table above to calculate the standard enthalpy change for the hydration of ethene.
C2H4 + H2O
→
C2H5OH
Formula
ΔHr = ΣΔHc(Reactants) – ΣΔHc(Products)
Calculation
Reactants: [-1387.4]
Products: [-1370.7]
-1387.4 – (-1370.7)
Products: [-1370.7]
-1387.4 – (-1370.7)
Answer
-16.7 kJ mol-1
2.
Calculate the enthalpy change for the cracking of dodecane to produce one mole of octane and two moles of ethene.
C12H26
→
C8H18 + 2C2H4
Calculation
Reactants: [-8083.2]
Products: [-5470 + 2(-1387.4)] = -8244.8
-8083.2 – (-8244.8)
Products: [-5470 + 2(-1387.4)] = -8244.8
-8083.2 – (-8244.8)
Answer
+161.6 kJ mol-1
3.
Calculate the enthalpy change for the combustion of phosphorous (P4) given the standard enthalpy change of formation of PH3 is 5.4 kJmol-1 data in the table above.
P4 + 6H2
→
4PH3
Step 1: Calculate ΔH for Formation of 4PH3
ΔHf(PH3) = +5.4 kJ mol-1
Formation of 4 moles = 4 × 5.4 = +21.6 kJ
Formation of 4 moles = 4 × 5.4 = +21.6 kJ
Step 2: Apply Hess’s Law (Combustion Data)
ΔHr = ΣΔHc(Reactants) – ΣΔHc(Products)
+21.6 = [ΔHc(P4) + 6(-285.8)] – [4(-750)]
+21.6 = [ΔHc(P4) – 1714.8] – [-3000]
+21.6 = [ΔHc(P4) + 6(-285.8)] – [4(-750)]
+21.6 = [ΔHc(P4) – 1714.8] – [-3000]
Answer
ΔHc(P4) = 21.6 – 3000 + 1714.8
-1263.6 kJ mol-1
-1263.6 kJ mol-1
4.
Calculate the enthalpy change of formation of sulfur trioxide given that the enthalpy change of the oxidation of sulfur dioxide to sulfur trioxide is -196.2 kJmol-1.
S + 1.5O2
→
SO3
Combine Equations
1. S + O2 → SO2 (ΔH = -297.0)
2. SO2 + 0.5O2 → SO3 (ΔH = -196.2)
Sum: S + 1.5O2 → SO3
2. SO2 + 0.5O2 → SO3 (ΔH = -196.2)
Sum: S + 1.5O2 → SO3
Answer
-297.0 + (-196.2) = -493.2 kJ mol-1
5.
What is the enthalpy change for the oxidation of ethanol to ethanal, given the data in the table.
C2H5OH + [O]
→
C2H4O + H2O
Calculation
ΔHr = ΣΔHc(Reactants) – ΣΔHc(Products)
Reactants: [-1370.7]
Products: [-1167] (Water does not burn)
-1370.7 – (-1167)
Reactants: [-1370.7]
Products: [-1167] (Water does not burn)
-1370.7 – (-1167)
Answer
-203.7 kJ mol-1
6.
A series of experiments were conducted on an alkene with six carbons and an unknown number of hydrogens. It was found to have a standard enthalpy change of hydrogenation of -271.2 kJmol-1 and a standard enthalpy change of combustion of -3862.6 kJmol-1. Using these data and the table above, determine the formula of the alkene.
Alkene + nH2
→
Alkane (C6H14)
Step 1: Set up Equation
ΔHhyd = [ΔHc(Alkene) + nΔHc(H2)] – ΔHc(Alkane)
-271.2 = [-3862.6 + n(-285.8)] – (-4163)
-271.2 = [-3862.6 + n(-285.8)] – (-4163)
Step 2: Solve for n (moles of H2)
-271.2 = -3862.6 – 285.8n + 4163
-271.2 = 300.4 – 285.8n
285.8n = 571.6
n = 2
-271.2 = 300.4 – 285.8n
285.8n = 571.6
n = 2
Conclusion
The alkane is C6H14.
Since 2 moles of H2 (4 H atoms) were added, the original alkene had 4 fewer hydrogens.
Formula: C6H10 (Hexadiene or Hexyne)
Since 2 moles of H2 (4 H atoms) were added, the original alkene had 4 fewer hydrogens.
Formula: C6H10 (Hexadiene or Hexyne)