Question

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

Answer 1

Answer: The enthalpy change for this reaction is 392 Joules

Explanation:

The balanced chemical reaction is,

$H_2(g)+CO_2(g)\rightarrow H_2O(g)+CO(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times B.E_{reactant}]-\sum [n\times B.E_{product}]$

$\Delta H=[(n_{H_2}\times B.E_{H_2})+(n_{CO_2}\times B.E_{CO_2})]-[(n_{H_2O}\times B.E_{H_2O})+(n_{CO}\times B.E_{CO})]$

where,

n = number of moles

Now put all the given values in this expression, we get

$\Delta H=[(1\times 1\times H-H)+(1\times 2\times C=O) ]-[(1\times 2\times O-H)+(2\times 1\times C-O)]$

$\Delta H=[(1\times 1\times 436)+(1\times 2\times 799) ]-[(1\times 2\times 463)+(2\times 1\times 358)]$

$\Delta H=392J$

Therefore, the enthalpy change for this reaction is 392 Joules