Questions: Question 4 5 pts Use the following information to determine the standard enthalpy change when 1 mol of PbO(s) is formed from lead metal and oxygen gas. (report as a whole number, remember ( - ) sign if needed, no sign is needed if positive) PbO(s)+C(graphite) → Pb(s)+CO(g) ΔHρ=107 kJ 2 C (graphite)+O2(g) → 2 CO(g) ΔH°=-222 kJ

Question 4
5 pts

Use the following information to determine the standard enthalpy change when 1 mol of PbO(s) is formed from lead metal and oxygen gas. (report as a whole number, remember ( - ) sign if needed, no sign is needed if positive)

PbO(s)+C(graphite) → Pb(s)+CO(g) ΔHρ=107 kJ
2 C (graphite)+O2(g) → 2 CO(g) ΔH°=-222 kJ

Transcript text: Question 4 5 pts Use the following information to determine the standard enthalpy change when 1 mol of $\mathrm{PbO}(s)$ is formed from lead metal and oxygen gas. (report as a whole number, remember ( - ) sign if needed, no sign is needed if positive) \[ \begin{array}{ll} \mathrm{PbO}(s)+\mathrm{C}(\text { graphite }) \longrightarrow \mathrm{Pb}(s)+\mathrm{CO}(\mathrm{~g}) \quad \Delta H^{\rho}=107 \mathrm{~kJ} \\ 2 \mathrm{C} \text { (graphite) }+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}(\mathrm{~g}) \quad \Delta H^{\circ}=-222 \mathrm{~kJ} \end{array} \] $\square$

Solution

Solution Steps

Step 1: Write the Target Reaction

We need to find the standard enthalpy change for the formation of 1 mol of $\mathrm{PbO}(s)$ from lead metal and oxygen gas. The target reaction is: \[ \mathrm{Pb}(s) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{PbO}(s) \]

Step 2: Analyze Given Reactions

We have two given reactions:

$\mathrm{PbO}(s) + \mathrm{C}(\text{graphite}) \rightarrow \mathrm{Pb}(s) + \mathrm{CO}(g)$ with $\Delta H^{\circ} = 107 \, \mathrm{kJ}$
$2 \mathrm{C}(\text{graphite}) + \mathrm{O}_2(g) \rightarrow 2 \mathrm{CO}(g)$ with $\Delta H^{\circ} = -222 \, \mathrm{kJ}$

Step 3: Reverse the First Reaction

To align with the target reaction, reverse the first reaction: \[ \mathrm{Pb}(s) + \mathrm{CO}(g) \rightarrow \mathrm{PbO}(s) + \mathrm{C}(\text{graphite}) \] Reversing the reaction changes the sign of $\Delta H$: \[ \Delta H^{\circ} = -107 \, \mathrm{kJ} \]

Step 4: Adjust the Second Reaction

Divide the second reaction by 2 to match the stoichiometry of $\mathrm{CO}(g)$ in the reversed first reaction: \[ \mathrm{C}(\text{graphite}) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}(g) \] The enthalpy change for this adjusted reaction is: \[ \Delta H^{\circ} = \frac{-222}{2} = -111 \, \mathrm{kJ} \]

Step 5: Combine Adjusted Reactions

Add the adjusted reactions to obtain the target reaction:

$\mathrm{Pb}(s) + \mathrm{CO}(g) \rightarrow \mathrm{PbO}(s) + \mathrm{C}(\text{graphite})$ with $\Delta H^{\circ} = -107 \, \mathrm{kJ}$
$\mathrm{C}(\text{graphite}) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}(g)$ with $\Delta H^{\circ} = -111 \, \mathrm{kJ}$

Adding these reactions cancels out $\mathrm{C}(\text{graphite})$ and $\mathrm{CO}(g)$, resulting in: \[ \mathrm{Pb}(s) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{PbO}(s) \] The total enthalpy change is: \[ \Delta H^{\circ} = -107 \, \mathrm{kJ} + (-111 \, \mathrm{kJ}) = -218 \, \mathrm{kJ} \]