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

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Michael Share
Jul 31, 2016

$\Delta {H}_{f} = - 99 \text{kJ/mol}$.

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 $\Delta {H}_{f}$ for ${C}_{\left(s\right)} + \frac{1}{2} {O}_{2 \left(g\right)} \rightarrow C {O}_{\left(g\right)}$

The energy cycle is:

You can see that the $\textcolor{red}{\text{RED}}$ route will be equal in energy to the $\textcolor{g r e e n}{\text{GREEN}}$ route since the arrows start and finish in the same place.

The $\textcolor{red}{\text{RED}}$ route is $\Delta {H}_{f}$ + $\Delta {H}_{c 1}$

$\Delta {H}_{c 1}$ refers to: $C {O}_{\left(g\right)} + \frac{1}{2} {O}_{2 \left(g\right)} \rightarrow C {O}_{2 \left(g\right)} = - \frac{588}{2} \text{kJ}$

We divided -588 by 2 because we want the enthalpy change for forming 1 mole of $C {O}_{2}$ and the value given is for forming 2 moles.

The $\textcolor{g r e e n}{\text{GREEN}}$ route is $\Delta {H}_{c 2}$.

This refers to:

${C}_{\left(s\right)} + {O}_{2 \left(g\right)} \rightarrow C {O}_{2 \left(g\right)} = - 393 \text{kJ}$

Applying Hess' Law:

$\Delta {H}_{f} + \frac{- 588}{2} = - 393$

$\Delta {H}_{f} - 294 = - 393$

$\Delta {H}_{f} = - 99 \text{kJ/mol}$

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Mar 26, 2015

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

Our target equation is

C(s) +½O₂(g) → CO(g); ΔH = "?"

We have the following information:

1. C(s) + O₂(g) → CO₂(g); ΔH = "-393 kJ"
2. 2CO(g) + O₂ → 2CO₂(g); ΔH = "-588 kJ"

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); ΔH = "-393 kJ"

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; ΔH = "+294 kJ"

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); ΔH = "-99 kJ"

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.

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