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青州's Question
Chemistry
Posted 5 months ago

Calculate ΔH∘r×n\Delta \mathrm{H}^{\circ}{ }_{r \times n} for the reaction below given the other two reactions.
8 A( g)+2C(g)→A8C2( g)
8 \mathrm{~A}(\mathrm{~g})+2 \mathrm{C}(\mathrm{g}) \rightarrow \mathrm{A}_{8} \mathrm{C}_{2}(\mathrm{~g})


Given:
A2 B2( g)→2 A( g)+2 B( g)ΔH∘rxn=−231 kJ A8C2( g)+8 B( g)→4 A2 B2( g)+2C(g)ΔH∘rxn=+702 kJ
\begin{array}{l}
\mathrm{A}_{2} \mathrm{~B}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{~A}(\mathrm{~g})+2 \mathrm{~B}(\mathrm{~g}) \quad \Delta \mathrm{H}^{\circ}{ }_{\mathrm{rxn}}=-231 \mathrm{~kJ} \\
\mathrm{~A}_{8} \mathrm{C}_{2}(\mathrm{~g})+8 \mathrm{~B}(\mathrm{~g}) \rightarrow 4 \mathrm{~A}_{2} \mathrm{~B}_{2}(\mathrm{~g})+2 \mathrm{C}(\mathrm{g}) \quad \Delta \mathrm{H}^{\circ}{ }_{\mathrm{r} x \mathrm{n}}=+702 \mathrm{~kJ}
\end{array}

1,626 kJ1,626 \mathrm{~kJ}
−222 kJ-222 \mathrm{~kJ}
471 kJ471 \mathrm{~kJ}
222 kJ222 \mathrm{~kJ}
−1,626 kJ-1,626 \mathrm{~kJ}
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Answer from Sia
Posted 5 months ago
Solution
1
Identify the target reaction: The target reaction is given as: 8 A( g)+2C(g)→A8C2( g) 8 \mathrm{~A}(\mathrm{~g}) + 2 \mathrm{C}(\mathrm{g}) \rightarrow \mathrm{A}_{8} \mathrm{C}_{2}(\mathrm{~g})
2
List the given reactions and their enthalpies: A2 B2( g)→2 A( g)+2 B( g)ΔHrxn∘=−231 kJ \mathrm{A}_{2} \mathrm{~B}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{~A}(\mathrm{~g}) + 2 \mathrm{~B}(\mathrm{~g}) \quad \Delta \mathrm{H}^{\circ}_{\mathrm{rxn}} = -231 \mathrm{~kJ}  A8C2( g)+8 B( g)→4 A2 B2( g)+2C(g)ΔHrxn∘=+702 kJ \mathrm{~A}_{8} \mathrm{C}_{2}(\mathrm{~g}) + 8 \mathrm{~B}(\mathrm{~g}) \rightarrow 4 \mathrm{~A}_{2} \mathrm{~B}_{2}(\mathrm{~g}) + 2 \mathrm{C}(\mathrm{g}) \quad \Delta \mathrm{H}^{\circ}_{\mathrm{rxn}} = +702 \mathrm{~kJ}
3
Manipulate the given reactions to match the target reaction: - Reverse the first reaction and multiply by 4: 4(2 A( g)+2 B( g)→A2 B2( g)) 4 (2 \mathrm{~A}(\mathrm{~g}) + 2 \mathrm{~B}(\mathrm{~g}) \rightarrow \mathrm{A}_{2} \mathrm{~B}_{2}(\mathrm{~g})) ΔH∘=4×231 kJ=924 kJ \Delta \mathrm{H}^{\circ} = 4 \times 231 \mathrm{~kJ} = 924 \mathrm{~kJ} - Reverse the second reaction: 4 A2 B2( g)+2C(g)→ A8C2( g)+8 B( g) 4 \mathrm{~A}_{2} \mathrm{~B}_{2}(\mathrm{~g}) + 2 \mathrm{C}(\mathrm{g}) \rightarrow \mathrm{~A}_{8} \mathrm{C}_{2}(\mathrm{~g}) + 8 \mathrm{~B}(\mathrm{~g}) ΔH∘=−702 kJ \Delta \mathrm{H}^{\circ} = -702 \mathrm{~kJ}
4
Combine the manipulated reactions: 8 A( g)+8 B( g)→4 A2 B2( g)ΔH∘=924 kJ 8 \mathrm{~A}(\mathrm{~g}) + 8 \mathrm{~B}(\mathrm{~g}) \rightarrow 4 \mathrm{~A}_{2} \mathrm{~B}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}^{\circ} = 924 \mathrm{~kJ} 4 A2 B2( g)+2C(g)→ A8C2( g)+8 B( g)ΔH∘=−702 kJ 4 \mathrm{~A}_{2} \mathrm{~B}_{2}(\mathrm{~g}) + 2 \mathrm{C}(\mathrm{g}) \rightarrow \mathrm{~A}_{8} \mathrm{C}_{2}(\mathrm{~g}) + 8 \mathrm{~B}(\mathrm{~g}) \quad \Delta \mathrm{H}^{\circ} = -702 \mathrm{~kJ} - Add the two reactions: 8 A( g)+2C(g)→ A8C2( g) 8 \mathrm{~A}(\mathrm{~g}) + 2 \mathrm{C}(\mathrm{g}) \rightarrow \mathrm{~A}_{8} \mathrm{C}_{2}(\mathrm{~g}) ΔH∘=924 kJ−702 kJ=222 kJ \Delta \mathrm{H}^{\circ} = 924 \mathrm{~kJ} - 702 \mathrm{~kJ} = 222 \mathrm{~kJ}
Answer
222 kJ222 \mathrm{~kJ}
Key Concept
Hess's Law
Explanation
Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction. By manipulating and combining the given reactions, we can determine the enthalpy change for the target reaction.

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