Question

estimate the enthalpy of reaction (in kj) for the following reaction, given the information below. report your answer to the nearest whole number. if the answer is negative, include the negative sign. if the answer is positive, do not include the sign.
c₂h₂(g) + 2 h₂(g) → c₂h₆(g)

Explanation:

Step1: Identify bonds broken and formed

• Bonds broken: 1 \( \ce{C\equiv C} \), 2 \( \ce{C-H} \), 2 \( \ce{H-H} \)
• Bonds formed: 1 \( \ce{C-C} \), 6 \( \ce{C-H} \) (since \( \ce{C2H6} \) has 6 \( \ce{C-H} \) bonds)

Step2: Calculate energy to break bonds

• Energy for \( \ce{C\equiv C} \): \( 960 \, \text{kJ} \)
• Energy for 2 \( \ce{C-H} \): \( 2 \times 412 = 824 \, \text{kJ} \)
• Energy for 2 \( \ce{H-H} \): \( 2 \times 436 = 872 \, \text{kJ} \)
• Total energy to break bonds: \( 960 + 824 + 872 = 2656 \, \text{kJ} \)

Step3: Calculate energy released from forming bonds

• Energy for \( \ce{C-C} \): \( 348 \, \text{kJ} \)
• Energy for 6 \( \ce{C-H} \): \( 6 \times 412 = 2472 \, \text{kJ} \)
• Total energy released: \( 348 + 2472 = 2820 \, \text{kJ} \)

Step4: Calculate enthalpy of reaction

• Enthalpy change \( \Delta H = \) Energy to break bonds \( - \) Energy released
• \( \Delta H = 2656 - 2820 = -164 \, \text{kJ} \) (the question says to include the sign if negative)

Answer:

\(-164\)