Question

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

Answer 1

For the given reaction, the enthalpy change calculated with bond energies is 36 kJ/mol.

   

Given the following balanced reaction:

H₂(g) + CO₂(g) → H₂O(g) + CO(g)     (1)

The enthalpy change ($\Delta H_{rxn}$) can be calculated with the next equation:

$\Delta H_{rxn} = \Sigma (n*E_{r}) - \Sigma (m*E_{p})$

Where:

n and m are the stoichiometric number of moles of the reactants and the products, respectively          

$E_{r}$ and $E_{p}$ are the bond energies of the reactants and the products, respectively

For reaction (1), the enthalpy change is given by:

$\Delta H_{rxn} = \Sigma (n_{H_{2}}*E_{H-H} + n_{CO_{2}}*2*E_{C=O}) - \Sigma (m_{H_{2}O}*2*E_{H-O} + m_{CO}*E_{C\equiv O})$

$\Delta H_{rxn} = \Sigma (1*436 kJ/mol + 1*2*799 kJ/mol) - \Sigma (1*2*463 kJ/mol} + 1*1072 kJ/mol) = 36 kJ/mol$                        

Therefore, the enthalpy change for the given reaction is 36 kJ/mol.

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