A scientist measures the standard enthalpy change for the following reaction to be -115.5 kJ: CO(g) + Cl2(g)___COCl2(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of COCl2(g) is kJ/mol.

-226.0kJ = ΔH°f COCl₂(g)

Explanation:

Using Hess' law, it is possible to obtain the enthalpy of formation of a substance from the enthalpy change of a reaction and the other enthalpies of formation involved in the reaction.

For the reaction:

CO(g) + Cl₂(g) → COCl₂(g)

Hess's law is:

ΔHr = -115.5kJ = ΔH°f COCl₂(g) - (ΔH°f CO(g) + ΔH°f Cl₂(g))

ΔH°f CO(g) is -110.5kJ/mol

ΔH°f Cl₂(g) is 0 kJ/mol

Replacing in Hess's law:

-115.5kJ = ΔH°f COCl₂(g) - (-110.5kJ/mol + 0kJ/mol)

-115.5kJ = ΔH°f COCl₂(g) + 110.5kJ

it means there are 3 atoms of oxygen