Questions: What is the change in enthalpy of the first reaction below, given the enthalpies of the other two reactions? 2 C(s) + O2(g) → 2 CO(g) C(s) + O2(g) → CO2(g) ΔH° = -394 kJ/mol CO(s) + 1/2 O2(g) → CO2(g) ΔH° = -283 kJ/mol A. -111 kJ B. -639 kJ C. -222 kJ D. -338 kJ

What is the change in enthalpy of the first reaction below, given the enthalpies of the other two reactions?

2 C(s) + O2(g) → 2 CO(g)

C(s) + O2(g) → CO2(g) ΔH° = -394 kJ/mol

CO(s) + 1/2 O2(g) → CO2(g) ΔH° = -283 kJ/mol

A. -111 kJ
B. -639 kJ
C. -222 kJ
D. -338 kJ

Transcript text: What is the change in enthalpy of the first reaction below, given the enthalpies of the other two reactions? \[ \begin{array}{ll} 2 \mathrm{C}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}(\mathrm{~g}) \\ \mathrm{C}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) & \Delta H^{\circ}=-394 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{CO}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) & \Delta H^{\circ}=-283 \mathrm{~kJ} / \mathrm{mol} \end{array} \] A. -111 kJ B. -639 kJ C. -222 kJ D. -338 kJ

Solution

Solution Steps

Step 1: Identify the Target Reaction

The target reaction for which we need to find the change in enthalpy is: \[ 2 \mathrm{C}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}(\mathrm{~g}) \]

Step 2: List the Given Reactions and Their Enthalpies

We are given the following reactions and their enthalpies:

\(\mathrm{C}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\) with \(\Delta H^{\circ} = -394 \, \mathrm{kJ/mol}\)
\(\mathrm{CO}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\) with \(\Delta H^{\circ} = -283 \, \mathrm{kJ/mol}\)

Step 3: Use Hess's Law to Find the Enthalpy Change

Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step of the reaction. We need to manipulate the given reactions to derive the target reaction.

Reverse the second reaction to get: \[ \mathrm{CO}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \] The enthalpy change for this reversed reaction is \(+283 \, \mathrm{kJ/mol}\).
Add the first reaction and the reversed second reaction: \[ \begin{align_} \mathrm{C}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) &\rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \quad (\Delta H^{\circ} = -394 \, \mathrm{kJ/mol}) \\ \mathrm{CO}_{2}(\mathrm{~g}) &\rightarrow \mathrm{CO}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \quad (\Delta H^{\circ} = +283 \, \mathrm{kJ/mol}) \end{align_} \]
The net reaction becomes: \[ \mathrm{C}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{~s}) \] with \(\Delta H^{\circ} = -394 + 283 = -111 \, \mathrm{kJ/mol}\).
Since the target reaction is for 2 moles of \(\mathrm{CO}\), multiply the enthalpy change by 2: \[ \Delta H^{\circ} = 2 \times (-111) = -222 \, \mathrm{kJ/mol} \]