I need HELP!!
Use bond energies to calculate the enthalpy change for the following reaction.
H2(g) + CO2(g) -----> H2O(g) + CO(g)

Respuesta :

Answer: Hit not Hard it is Hoc2 (Hydgergon Perocide)

Explanation: Water plus co(g)= HoC2

Answer : The enthalpy change for the reaction is, 269 kJ/mol

Explanation :

The given chemical reaction is:

[tex]H_2+CO_2\rightarrow CO+H_2O[/tex]

As we know that:

The enthalpy change of reaction = E(bonds broken) - E(bonds formed)

[tex]\Delta H=[(B.E_{H-H})+(2\times B.E_{C=O})]-[(2\times B.E_{O-H})+(1\times B.E_{C\equiv C})][/tex]

Given:

[tex]\Delta H[/tex] = enthalpy change

[tex]B.E_{H-H}[/tex] = 436 kJ/mol

[tex]B.E_{O-H}[/tex] = 463 kJ/mol

[tex]B.E_{C=O}[/tex] = 799 kJ/mol

[tex]B.E_{C\equiv C}[/tex]  = 839 kJ/mol

Now put all the given values in the above expression, we get:

[tex]\Delta H=[(436kJ/mol)+(2\times 799kJ/mol)]-[(2\times 463kJ/mol)+(1\times 839kJ/mol)][/tex]

[tex]\Delta H=269kJ/mol[/tex]

Therefore, the enthalpy change for the reaction is, 269 kJ/mol