Question

h-c≡c-h(g) + 2hi(g) → product structure using the table of average bond energies below the image, the δh for the reaction in kj is..... bond energy table (kj/mol): c≡c 839, c-c 348, h-i 299, c-i 240, c-h 413. options: +160 kj, -63 kj, -160 kj, -217 kj

Explanation:

Step1: Identify bonds broken and formed

• Bonds Broken (Reactants):
• 1 \( \ce{C\equiv C} \) bond: Energy = \( 839 \, \text{kJ/mol} \)
• 2 \( \ce{C-H} \) bonds: Each \( \ce{C-H} \) is \( 413 \, \text{kJ/mol} \), so total \( 2 \times 413 = 826 \, \text{kJ/mol} \)
• 2 \( \ce{H-I} \) bonds: Each \( \ce{H-I} \) is \( 299 \, \text{kJ/mol} \), so total \( 2 \times 299 = 598 \, \text{kJ/mol} \)
• Total energy to break bonds: \( 839 + 826 + 598 = 2263 \, \text{kJ/mol} \)

• Bonds Formed (Products):
• 1 \( \ce{C-C} \) bond: Energy = \( 348 \, \text{kJ/mol} \)
• 4 \( \ce{C-H} \) bonds: Wait, no—wait the product is \( \ce{I - C - C - I} \) with \( \ce{H} \) on each \( \ce{C} \). Wait, original reactant is \( \ce{H-C\equiv C-H} \) (ethyne) and \( 2 \ce{HI} \). The product is \( \ce{I - CH2 - CH2 - I} \)? Wait, no, let's re-examine the structure. Wait, the reactant is \( \ce{H-C\equiv C-H} \) (1 \( \ce{C\equiv C} \), 2 \( \ce{C-H} \)) and 2 \( \ce{HI} \) (2 \( \ce{H-I} \)). The product is \( \ce{I - C - C - I} \) with each \( \ce{C} \) bonded to 2 \( \ce{H} \) (so 4 \( \ce{C-H} \) bonds), 1 \( \ce{C-C} \) bond, and 2 \( \ce{C-I} \) bonds.
• So bonds formed:
• 1 \( \ce{C-C} \): \( 348 \, \text{kJ/mol} \)
• 4 \( \ce{C-H} \): \( 4 \times 413 = 1652 \, \text{kJ/mol} \)
• 2 \( \ce{C-I} \): Each \( \ce{C-I} \) is \( 240 \, \text{kJ/mol} \), so \( 2 \times 240 = 480 \, \text{kJ/mol} \)
• Total energy released (bonds formed): \( 348 + 1652 + 480 = 2480 \, \text{kJ/mol} \)

Step2: Calculate \( \Delta H \)

\( \Delta H = \) Energy to break bonds \( - \) Energy to form bonds
\( \Delta H = 2263 - 2480 = -217 \, \text{kJ/mol} \)

Answer:

\(-217 \, \text{kJ}\)