Respuesta :
Answer:
The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,
Explanation:
[tex]CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(g) ,[/tex]ΔH° = ?
We are given with:
[tex]\Delta H_{O-O}=142 kJ/mol[/tex]
[tex]\Delta H_{O=O}=498 kJ/mol[/tex]
[tex]\Delta H_{H-O}=459 kJ/mol[/tex]
[tex]\Delta H_{C-H}=411 kJ/mol[/tex]
[tex]\Delta H_{C-O}=358 kJ/mol[/tex]
[tex]\Delta H_{C=O}=799 kJ/mol[/tex]
ΔH° =
(Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)
[tex]\Delta H^o=(1 mol\times 4\times \Delta H_{C-H}+2 mol\times 1\times \Delta H_{O=O})-(1 mol\times 2\times \Delta H_{C=O}+2 mol\times 2\times\Delta H_{H-O})[/tex]
[tex]\Delta H^o=(1 mol\times 4\times 411 kJ/mol+2 mol\times 1\times 498 kJ/mol)-(1 mol\times 2\times 799 kJ/mol+2 mol\times 2\times 459 kJ/mol)[/tex]
[tex]\Delta H^o=-794kJ[/tex]
[tex]\Delta H^o>0[/tex] endothermic reaction
[tex]\Delta H^o<0[/tex] exothermic reaction
The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,
The enthalpy of the reaction CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) given the bond dissociation energies of CH₄ (ΔH° C-H 411 kJ/mol), O₂ (ΔH° O=O 498 kJ/mol), CO₂ (ΔH° C=O 799 kJ/mol), and H₂O (ΔH° H-O 459 kJ/mol) is -794 kJ/mol. The reaction is exothermic.
The balanced reaction is:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) (1)
The enthalpy of reaction (1) is given by:
[tex]\Delta H^{\circ} = \Sigma \Delta H^{\circ}_{r} - \Sigma \Delta H^{\circ}_{p}[/tex] (2)
Where r is for reactants and p for products
The bonds of each compound are:
- CH₄: 1 mol of 4 bonds C-H → 4*411 kJ/mol
- 2O₂: 2 moles of 1 bond O=O → 2*498 kJ/mol
- CO₂: 1 mol of 2 bonds C=O → 2*799 kJ/mol
- 2H₂O: 2 moles of 2 bonds H-O → 2*2*459 kJ/mol
With the above information and using equation (2), we can calculate the enthalpy of reaction (1):
[tex]\Delta H^{\circ} = 4*\Delta H^{\circ}_{CH_{4}} + 2*\Delta H^{\circ}_{O_{2}} - (2*\Delta H^{\circ}_{CO_{2}} + 2*2*\Delta H^{\circ}_{H_{2}O})[/tex]
[tex]\Delta H^{\circ} = 4*\Delta H^{\circ}_{C-H} + 2*\Delta H^{\circ}_{O=O} - (2*\Delta H^{\circ}_{C=O} + 2*2*\Delta H^{\circ}_{H-O})[/tex]
[tex]\Delta H^{\circ} = [4*411 + 2*498 - (2*799 + 2*2*459)] kJ/mol = -794 kJ/mol[/tex]
Since the enthalpy of the reaction is negative, we have that reaction (1) is exothermic, which means that it releases energy.
In an endothermic reaction, the enthalpy is positive. In this case, the reaction needs to absorb energy so the reactants can produce the products.
Therefore, the enthalpy of the exothermic reaction is -794 kJ/mol.
Learn more about enthalpy of reaction here:
- https://brainly.com/question/11753370?referrer=searchResults
- https://brainly.com/question/3315742?referrer=searchResults
I hope it helps you!
