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Given the following bond dissociation energies, calculated the enthalpy of the following reaction.H-C≡C-H + 2H2 → H3C-CH3C-C            350 kJ/molC≡C           807 kJ/molC-H            410 kJ/molH-H            436 kJ/molHint: There are 2 H-H bonds on the left and a total of 6 C-H bonds on the right.Group of answer choices-111 kJ/mol-211 kJ/mol-311 kJ/mol-11 kJ/mol

Question

Given the following bond dissociation energies, calculated the enthalpy of the following reaction.H-C≡C-H + 2H2 → H3C-CH3C-C            350 kJ/molC≡C           807 kJ/molC-H            410 kJ/molH-H            436 kJ/molHint: There are 2 H-H bonds on the left and a total of 6 C-H bonds on the right.Group of answer choices-111 kJ/mol-211 kJ/mol-311 kJ/mol-11 kJ/mol

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Solution

To calculate the enthalpy of the reaction, we need to subtract the total energy of the bonds broken from the total energy of the bonds formed.

On the reactant side, we have:

1 C-C triple bond (807 kJ/mol) 2 C-H bonds (2 * 410 kJ/mol = 820 kJ/mol) 2 H-H bonds (2 * 436 kJ/mol = 872 kJ/mol)

So, the total energy of the bonds broken is 807 + 820 + 872 = 2499 kJ/mol

On the product side, we have:

1 C-C single bond (350 kJ/mol) 6 C-H bonds (6 * 410 kJ/mol = 2460 kJ/mol)

So, the total energy of the bonds formed is 350 + 2460 = 2810 kJ/mol

The enthalpy of the reaction is the energy of the bonds formed minus the energy of the bonds broken, so:

ΔH = 2810 - 2499 = 311 kJ/mol

So, the enthalpy of the reaction is -311 kJ/mol. The negative sign indicates that the reaction is exothermic, meaning it releases heat. Therefore, the correct answer is -311 kJ/mol.

This problem has been solved

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