Respuesta :
Explanation:
The given reaction is as follows.
[tex]C_{4}H_{4}(g) + 2H_{2}(g) \rightarrow C_{4}H_{8}(g)[/tex]
So, enthalpy change for the given reaction is calculated as follows.
[tex]\Delta H_{rxn} = \sum n \Delta H_{products} - \sum n \Delta H_{reactants}[/tex]
where, n = number of moles.
Now, putting the given values into the above equation and calculate the change in enthalpy as follows.
[tex]\Delta H_{rxn} = \sum (1 mol C_{4}H_8(g)) \times (-2755 kJ / mol) - \sum (2 mol H_2(g)) \times (-286 kJ/mol) + (1 mole C_4H_4(g)) \times (-2341 kJ/mole)[/tex]
= [tex]\sum -2755 kJ - \sum -2913 kJ)[/tex]
= 158 kJ
Thus, we can conclude that enthalpy of the combustion is 158 kJ.
Answer:
The ΔH for the reaction is 158 kJ
Explanation:
The complete question is:
"Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for C4H4 (-2341 kJ/mol), C4H8 (-2755 kJ/mol), and H2 (-286 kJ/mol), calculate ΔH for the reaction: C₄H₄ (g) + 2H₂ (g) ⇒ C₄H₈ (g)"
In first place, the combustion reactions are:
C₄H₄ + 5 O₂ ⇒ 4CO₂ + 2H₂O
2H₂ + O₂ ⇒2H₂O
4CO₂ + 4H₂O ⇒C₄H₈ + 6O₂
And you get: C₄H₄ (g) + 2H₂ (g) ⇒ C₄H₈ (g)
You want to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction. For that you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (quantity of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants
You know that the balanced equation is:
C₄H₄ (g) + 2H₂ (g) ⇒ C₄H₈ (g)
So:
ΔH= 1 mol of C₄H₈*ΔH of C₄H₈ -[1 mol of C₄H₄*ΔH of C₄H₄ + 2 mol of H₂*ΔH of H₂]
ΔH= -2755 kJ/mol * 1mol - [ -2341 kJ/mol * 1 mol + (-286 kJ/mol * 2 mol)]
ΔH= 158 kJ
The ΔH for the reaction is 158 kJ
