Question

Determine the heat of formation of CO from the following equations:
C +O= CO2; heat of formation= -393kj
CO + 1/2O2 = CO2; heat of formation= -285

Answer 1

To determine the heat of formation of CO (carbon monoxide), we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for individual reactions that lead to the same final state.

We have the following two reactions and their respective heat changes:

1. \( C + O_2 \rightarrow CO_2 \) ; \(\Delta H = -393 \, \text{kJ}\)
2. \( CO + \frac{1}{2} O_2 \rightarrow CO_2 \) ; \(\Delta H = -285 \, \text{kJ}\)

We need to manipulate these equations to extract the reaction for the formation of CO from its elements:

\[
C + \frac{1}{2} O_2 \rightarrow CO
\]

To achieve this, we can reverse reaction 2 so it has the decomposition of CO instead:

\[
CO_2 \rightarrow CO + \frac{1}{2} O_2 ; \Delta H = +285 \, \text{kJ} \quad (\text{reverse sign})
\]

Now using the first reaction without any changes, we can add this reversed reaction to the first one. The equations now look like this:

1. \( C + O_2 \rightarrow CO_2 \) ; \(\Delta H_1 = -393 \, \text{kJ}\)
2. \( CO_2 \rightarrow CO + \frac{1}{2} O_2 \) ; \(\Delta H_2 = +285 \, \text{kJ}\)

Now, if we add these two reactions:

\[
C + O_2 + CO_2 \rightarrow CO_2 + CO + \frac{1}{2} O_2
\]

The \( CO_2 \) cancels out on both sides:

\[
C + \frac{1}{2} O_2 \rightarrow CO
\]

Next, we can find the total ΔH for the overall reaction:

\[
\Delta H_{formation, CO} = \Delta H_1 + \Delta H_2 = -393 \, \text{kJ} + 285 \, \text{kJ} = -108 \, \text{kJ}
\]

Therefore, the heat of formation of CO from carbon and oxygen is:

\[
\boxed{-108 \, \text{kJ/mol}}
\]