Question

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.

Answer 1

The standard enthalpy of formation of COCl2(g) is 115.5 kJ/mol.

Step-by-step explanation:

To determine the standard enthalpy of formation of COCl2(g), we can use the given value for the standard enthalpy change of the reaction and the standard enthalpies of formation for the other substances. The reaction is:

CO(g) + Cl2(g) → COCl2(g)

Using the standard enthalpies of formation for CO(g) and Cl2(g), which are known values, we can calculate the standard enthalpy of formation for COCl2(g) by subtracting the sum of the standard enthalpies of formation for the reactants from the sum of the standard enthalpies of formation for the products. The equation for this calculation is:

ΔH°f = ΣΔH°f(products) - ΣΔH°f(reactants)

Plugging in the values:

ΔH°f(COCl2) = ΣΔH°f(products) - ΣΔH°f(reactants)

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

ΔH°f(COCl2) = 115.5 kJ/mol

Answer 2

-226.0kJ = ΔH°f COCl₂(g)

Step-by-step 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

-226.0kJ = ΔH°f COCl₂(g)