\[ \left ( 1 \right ) \; \ce{2C (s) + O2 (g) -> 2 CO (g)} \quad \Delta H=-221.0 \; \text{kJ} \nonumber \]

\[ \left ( 2 \right ) \; \ce{C (s) + O2 (g) -> CO2 (g)} \quad \Delta H=-393.5 \; \text{kJ} \nonumber \]

Use this information to calculate the enthalpy change per mole of CO for the reaction of CO with O_{2} to give CO_{2}.

###### Solution:

**A** We begin by writing the balanced chemical equation for the reaction of interest:

\[ \left ( 3 \right ) \; \ce{CO (g) + 1/2 O2 (g) -> CO2 (g)} \quad \Delta H_{rxn}=? \nonumber \]

There are at least two ways to solve this problem using Hess’s law and the data provided. The simplest is to write two equations that can be added together to give the desired equation and for which the enthalpy changes are known. Observing that CO, a reactant in Equation 3, is a product in Equation 1, we can reverse Equation (1) to give

\[ \ce{2 CO (g) -> 2 C (s) + O2 (g)} \quad \Delta H=+221.0 \; \text{kJ} \nonumber \]

Because we have reversed the direction of the reaction, the sign of ΔH is changed. We can use Equation 2 as written because its product, CO_{2}, is the product we want in Equation 3:

\[ \ce{C (s) + O2 (g) -> CO2 (s)} \quad \Delta H=-393.5 \; \text{kJ} \nonumber \]

**B** Adding these two equations together does not give the desired reaction, however, because the numbers of C(s) on the left and right sides do not cancel. According to our strategy, we can multiply the second equation by 2 to obtain 2 mol of C(s) as the reactant:

\[ \ce{2 C (s) + 2 O2 (g) -> 2 CO2 (s)} \quad \Delta H=-787.0 \; \text{kJ} \nonumber \]

Writing the resulting equations as a sum, along with the enthalpy change for each, gives

\[ \begin{align*} \ce{2 CO (g) &-> \cancel{2 C(s)} + \cancel{O_2 (g)} } & \Delta H & = -\Delta H_1 = +221.0 \; \text{kJ} \\ \ce{\cancel{2 C (s)} + \cancel{2} O2 (g) &-> 2 CO2 (g)} & \Delta H & = -2\Delta H_2 =-787.0 \; \text{kJ} \\[2ex] \hline \ce{2 CO (g) + O2 (g) &-> 2 CO2 (g)} & \Delta H &=-566.0 \; \text{kJ} \end{align*} \nonumber \]

Note that the overall chemical equation and the enthalpy change for the reaction are both for the reaction of 2 mol of CO with O_{2}, and the problem asks for the amount per mole of CO. Consequently, we must divide both sides of the final equation and the magnitude of ΔH by 2:

\[ \ce{ CO (g) + 1/2 O2 (g) -> CO2 (g)} \quad \Delta H = -283.0 \; \text{kJ} \nonumber \]

An alternative and equally valid way to solve this problem is to write the two given equations as occurring in steps. Note that we have multiplied the equations by the appropriate factors to allow us to cancel terms:

\[ \begin{alignat*}{3}

\text{(A)} \quad && \ce{ 2 C (s) + O2 (g) &-> \cancel{2 CO (g)}} \qquad & \Delta H_A &= \Delta H_1 &&= + 221.0 \; \text{kJ} \\

\text{(B)} \quad && \ce{ \cancel{2 CO (g)} + O2 (g) &-> 2 CO2 (g)} \qquad & \Delta H_B && &= ? \\

\text{(C)} \quad && \ce{2 C (s) + 2 O2 (g) &-> 2 CO2 (g)} \qquad & \Delta H &= 2 \Delta H_2 &= 2 \times \left ( -393.5 \; \text{kJ} \right ) &= -787.0 \; \text{kJ}

\end{alignat*} \]

The sum of reactions A and B is reaction C, which corresponds to the combustion of 2 mol of carbon to give CO_{2}. From Hess’s law, ΔH_{A} + ΔH_{B} = ΔH_{C}, and we are given ΔH for reactions A and C. Substituting the appropriate values gives

\[ \begin{matrix}

-221.0 \; kJ + \Delta H_{B} = -787.0 \; kJ \\

\Delta H_{B} = -566.0

\end{matrix} \nonumber \]

This is again the enthalpy change for the conversion of 2 mol of CO to CO_{2}. The enthalpy change for the conversion of 1 mol of CO to CO_{2} is therefore −566.0 ÷ 2 = −283.0 kJ/mol of CO, which is the same result we obtained earlier. As you can see, there may be more than one correct way to solve a problem.