QUESTION IMAGE
Question
given the following thermochemical equations, determine the enthalpy of reaction (δhᵣₓₙ) for the reaction: cs₂(l) + 3 o₂(g) → co₂(g) + 2 so₂(g) δhᵣₓₙ = ? (1) c(s) + o₂(g) → co₂(g) δh° = -393.5 kj (2) s(s) + o₂(g) → so₂(g) δh° = -296.8 kj (3) c(s) + 2 s(s) → cs₂(l) δh° = +87.9 kj options: -602.4 kj, -778.2 kj, -1075 kj, -2150 kj
Step1: Identify Target Reaction
The target reaction is: \( \text{CS}_2(l) + 3\text{O}_2(g)
ightarrow \text{CO}_2(g) + 2\text{SO}_2(g) \)
Step2: Analyze Given Reactions
- Reaction (1): \( \text{C}(s) + \text{O}_2(g)
ightarrow \text{CO}_2(g) \quad \Delta H^\circ = -393.5 \, \text{kJ} \)
- Reaction (2): \( \text{S}(s) + \text{O}_2(g)
ightarrow \text{SO}_2(g) \quad \Delta H^\circ = -296.8 \, \text{kJ} \) (need 2 moles, so multiply by 2)
- Reaction (3): \( \text{C}(s) + 2\text{S}(s)
ightarrow \text{CS}_2(l) \quad \Delta H^\circ = +87.9 \, \text{kJ} \) (reverse to get \( \text{CS}_2(l)
ightarrow \text{C}(s) + 2\text{S}(s) \), so \( \Delta H^\circ = -87.9 \, \text{kJ} \))
Step3: Manipulate Reactions
- Reaction (1): Keep as is (produces \( \text{CO}_2 \)).
- Reaction (2): Multiply by 2: \( 2\text{S}(s) + 2\text{O}_2(g)
ightarrow 2\text{SO}_2(g) \quad \Delta H^\circ = 2 \times (-296.8) = -593.6 \, \text{kJ} \)
- Reaction (3): Reverse: \( \text{CS}_2(l)
ightarrow \text{C}(s) + 2\text{S}(s) \quad \Delta H^\circ = -87.9 \, \text{kJ} \)
Step4: Sum Reactions
Add the manipulated reactions:
- Reverse (3): \( \text{CS}_2(l)
ightarrow \text{C}(s) + 2\text{S}(s) \quad \Delta H = -87.9 \, \text{kJ} \)
- (1): \( \text{C}(s) + \text{O}_2(g)
ightarrow \text{CO}_2(g) \quad \Delta H = -393.5 \, \text{kJ} \)
- 2×(2): \( 2\text{S}(s) + 2\text{O}_2(g)
ightarrow 2\text{SO}_2(g) \quad \Delta H = -593.6 \, \text{kJ} \)
Summing these:
\( \text{CS}_2(l) + 3\text{O}_2(g)
ightarrow \text{CO}_2(g) + 2\text{SO}_2(g) \)
Step5: Calculate \( \Delta H_{\text{rxn}} \)
Sum the \( \Delta H \) values:
\( \Delta H_{\text{rxn}} = -87.9 + (-393.5) + (-593.6) \)
\( \Delta H_{\text{rxn}} = -87.9 - 393.5 - 593.6 \)
\( \Delta H_{\text{rxn}} = -1075 \, \text{kJ} \)
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-1075 kJ (corresponding to the option -1075 kJ)