Ethyne (C2 H2 (g), mc032-1.jpgHf = 226.77 kJ/mol) undergoes complete combustion in the presence of oxygen to produce carbon dioxide (CO2 (g), mc032-2.jpgHf = –393.5 kJ/mol ) and water (H2 O(g), mc032-3.jpgHf = –241.82 kJ/mol) according to the equation below.

mc032-4.jpg

What is the enthalpy of combustion (per mole) of C2 H2 (g)?

–2511.2 kJ/mol

–1255.6 kJ/mol

–862.1 kJ/mol

–431.0 kJ/mol

2 C2H2 + 5 02 > 4 CO2 + 2 H2O

Products - Reactants ( all units are kJ/mo1):

(4 x -393.5) + (2 x -241.82) - (2 x 226.77) - (5 x 0) = -2511.2 kJ/mo1

-2511.2 kJ/mo1 is for 2 moles of C2H2.The question asked for 1 mole of C2H2, so: -2511.2 / 2 = -1255.6 kJ/mo1

answer: -1255.6 kJ/mo1

Answer Link

kasliwalalvi24 kasliwalalvi24 · Answer 2 · 2022-02-01T04:07:04-05:00

Enthalpy is defined as the sum of internal energy and the product of the pressure and volume in a thermodynamic system.

What is the enthalpy of combustion of ethyne molecule?

Enthalpy from a chemical reaction can be determined as the difference between the enthalpy of products and reactants.

Given:

Enthalpy of ethyne = 226.77 kJ/mol
Enthalpy of Carbon dioxide = -393.5kJ/mol
Enthalpy of Water = -241.82 kJ/mol

The chemical reaction of complete combustion of ethyne is:

\(\rm 2\;C_2H_2 + 5 \;O_2 \rightarrow4 \;CO_2 + 2\; H_2O\)

Now, subtracting the enthalpy of product from the reactant, we get:

\((4 \times -393.5) + (2 \times -241.82) - (2 \times 226.77) - (5 \times 0) = -2511.2\; \rm kJ/mol\)

Thus, the enthalpy of the combustion of the reaction is -2511.2 kJ/mol for 2 moles, then the enthalpy for the 1 mol will be:

= \(\dfrac{-2511.2}{2}\) = -1255.6 kJ/mol.

Therefore, the enthalpy is -1255.6 kJ/mol. The correct answer is Option B.

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