Question

use the following table of bond energies to calculate the molar enthalpy of combustion (in kj/mol) of acetylene (c₂h₂) gas in oxygen, based on the following chemical equation: c₂h₂(g) + 2.5 o₂(g) → 2 co₂(g) + h₂o(g) bond energy table: single bond (h:432, c:411/346, n:386/305/167, o:459/358/201/142); multiple bonds (c=c:602, c≡c:835, c=n:615, c≡n:887, c=o:799, c≡o:1072, o=o:494, n≡n:942; all values in kj/mol)

Explanation:

Step1: Identify bonds in reactants

In \( \ce{C2H2} \): 1 \( \ce{C≡C} \) bond and 2 \( \ce{C-H} \) bonds.
In \( \ce{O2} \): 2.5 \( \ce{O=O} \) bonds (since coefficient is 2.5).

Calculate energy to break reactant bonds:

• \( \ce{C≡C} \): \( 1 \times 835 \, \text{kJ/mol} \)
• \( \ce{C-H} \): \( 2 \times 411 \, \text{kJ/mol} \)
• \( \ce{O=O} \): \( 2.5 \times 494 \, \text{kJ/mol} \)

Sum: \( 835 + (2 \times 411) + (2.5 \times 494) = 835 + 822 + 1235 = 2892 \, \text{kJ/mol} \)

Step2: Identify bonds in products

In \( \ce{CO2} \): 2 \( \ce{C=O} \) bonds per \( \ce{CO2} \), 2 moles of \( \ce{CO2} \): \( 2 \times 2 \times 799 \, \text{kJ/mol} \)
In \( \ce{H2O} \): 2 \( \ce{O-H} \) bonds (from table, \( \ce{O-H} = 459 \, \text{kJ/mol} \)): \( 2 \times 459 \, \text{kJ/mol} \)

Calculate energy released by forming product bonds:

• \( \ce{CO2} \): \( 4 \times 799 = 3196 \, \text{kJ/mol} \)
• \( \ce{H2O} \): \( 2 \times 459 = 918 \, \text{kJ/mol} \)

Sum: \( 3196 + 918 = 4114 \, \text{kJ/mol} \)

Step3: Calculate enthalpy of combustion

Enthalpy change (\( \Delta H \)) = Energy to break bonds - Energy to form bonds
\( \Delta H = 2892 - 4114 = -1222 \, \text{kJ/mol} \)

Answer:

\(-1222\)