contestada

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

Respuesta :

PHILIPSTAYHYPE23
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

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:

[tex]\rm 2\;C_2H_2 + 5 \;O_2 \rightarrow4 \;CO_2 + 2\; H_2O[/tex]

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

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

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:

= [tex]\dfrac{-2511.2}{2}[/tex] = -1255.6 kJ/mol.

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

Learn more about enthalpy here:

https://brainly.com/question/7827769

Otras preguntas