Based on the bond energies for the reaction below, what is the enthalpy of the reaction?HC≡CH (g) + 5/2 O₂ (g) → 2 CO₂ (g) + H₂O (g)

Based on the bond energies, the reaction

HC≡CH(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(g) has a standard enthalpy of reaction of - 1222 kJ/mol.

The bond energies (E) of the reactions can help to calculate the standard enthalpy of the reaction (ΔH°rxn) using the following expression.

Solution:

bond energies in the formula:

= 1222 kJ/mol  

Thus, Based on the bond energies, the reaction:

HC≡CH(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(g) has a standard enthalpy of reaction of -1222 kJ/mol.  

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1219.5 kj/mol

Explanation:

To reach this result, you must use the formula:

ΔHºrxn = Σn * (BE reactant) - Σn * (BE product)

ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H).

The BE values are:

BE C = C: 839 kj / mol

BE C-H: 413 Kj / mol

BE O = O: 495 kj / mol

BE C = O = 799 Kj / mol

BE O-H = 463 kj / mol

Now you must replace the values in the above equation, the result of which will be:

ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463) = 1219.5 kj/mol

the answer is d. 74.02