Question

I need HELP!!

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

Answer 1

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

Explanation: Water plus co(g)= HoC2

Answer 2

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

Explanation :

The given chemical reaction is:

`H_2+CO_2\rightarrow CO+H_2O`

As we know that:

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

`\Delta H=[(B.E_(H-H))+(2* B.E_(C=O))]-[(2* B.E_(O-H))+(1* B.E_(C\equiv C))]`

Given:

`\Delta H` = enthalpy change

`B.E_(H-H)` = 436 kJ/mol

`B.E_(O-H)` = 463 kJ/mol

`B.E_(C=O)` = 799 kJ/mol

`B.E_(C\equiv C)` = 839 kJ/mol

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

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

`\Delta H=269kJ/mol`

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