Respuesta :
Answer : The enthalpy change for the reaction is 1043 kJ/mol.
Explanation :
The given chemical reaction is:
[tex]2CO+O_2\rightarrow 2CO_2[/tex]
As we know that:
The enthalpy change of reaction = E(bonds broken) - E(bonds formed)
[tex]\Delta H=[(2\times B.E_{C\equiv O})+(1\times B.E_{O\equiv O})]-[2\times B.E_{C=O}][/tex]
Given:
[tex]B.E_{C\equiv O}[/tex] = 1074 kJ/mol
[tex]B.E_{O\equiv O}[/tex] = 499 kJ/mol
[tex]B.E_{C=O}[/tex] = 802 kJ/mol
Now put all the given values in the above expression, we get:
[tex]\Delta H=[(2\times 1074kJ/mol)+(1\times 499kJ/mol)]-[2\times 802kJ/mol][/tex]
[tex]\Delta H=1043kJ/mol[/tex]
Therefore, the enthalpy change for the reaction is 1043 kJ/mol.
The enthalpy of the given reaction is 1309 kJ/mol
From the question,
We are to calculate the enthalpy of the reaction
2CO + O₂ → 2CO₂
The enthalpy of reaction equals sum of the bond energies of bonds being broken - sum of the bond energies of the bonds being formed
The enthalpy of a reaction is given by the formula
[tex]\Delta H_{reaction} = \sum BE_{reactants} - \sum BE_{products}[/tex]
From the question, the given bond energies (with the bonds properly indicated) are
BE(C≡O) = 1074 kJ/mol
BE(O=O) = 499 kJ/mol
BE(C=O) = 802 kJ/mol
The reactants are CO and O₂
2CO contains 2 C≡O bonds
O₂ contains 1 O=O bond
The product is CO₂
1 CO₂ contains 2 C=O bonds
∴ 2CO₂ will contain 4 C=O bonds
Then,
The enthalpy of the reaction is
[tex]\Delta H_{reaction} =[ 2(C\equiv O) + (O=O)] - 4(C=O)[/tex]
[tex]\Delta H_{reaction} = 2(1074) +499 - 4(402)[/tex]
[tex]\Delta H_{reaction} = 2148 +499 - 1608[/tex]
[tex]\Delta H_{reaction} = 1309 \ kJ/mol[/tex]
Hence, the enthalpy of the given reaction is 1309 kJ/mol
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