# What is the enthalpy of formation of carbon monoxide in KJ/mol ? C(s) + O2(g) --> CO2(g) ΔH° = -393 kJ 2CO(g) + O2(g) --> 2CO(g) ΔH° = -588 kJ

##### 2 Answers

The standard enthalpy of formation of carbon monoxide is -99 kJ/mol.

Our target equation is

C(s) +½O₂(g) → CO(g);

We have the following information:

**1.** C(s) + O₂(g) → CO₂(g);

**2.** 2CO(g) + O₂ → 2CO₂(g);

To solve this problem, we use Hess's Law.

Our target equation has C(s) on the left hand side, so we re-write equation **1**:

**1.** C(s) + O₂(g) → CO₂(g);

Our target equation has CO(g) on the right hand side, so we reverse equation **2** and divide by 2.

**3.** CO₂(g) → CO(g) + ½O₂;

That means that we also change the sign of ΔH and divide by 2.

Then we add equations **1** and **3** and their ΔH values.

This gives

C(s) +½O₂(g) → CO(g);

Using your numbers, the standard enthalpy of formation of carbon monoxide is

-99 kJ/mol.

I think that your value for the heat of combustion of CO is incorrect. It should be

-566 kJ.

This would give the correct value of -110 kJ for the heat of formation of CO.

Many enthalpy changes are difficult to measure directly under standard conditions, enthalpy of formation being such a case.

It is a lot easier to measure the enthalpy of combustion using calorimetry. Since the elements and the compound from which they are made will have the same products of combustion we can set up an energy cycle.

Hess' Law states that the total enthalpy change of a reaction is independent of the route taken.

We want

The energy cycle is:

You can see that the

The

We divided -588 by 2 because we want the enthalpy change for forming 1 mole of

The

This refers to:

Applying Hess' Law: