Question

1. calculate the enthalpy of the following chemical reaction: (4)

cs₂(l) + 3o₂(g) ---> co₂(g) + 2so₂(g)
given:
c(s) + o₂(g) ---> co₂(g) δh = -393.5 kj/mol
s(s) + o₂(g) ---> so₂(g) δh = -296.8 kj/mol
c(s) + 2s(s) ---> cs₂(l) δh = +87.9 kj/mol

Explanation:

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

1. Reaction 1: \( \text{C}(s) + \text{O}_2(g)

ightarrow \text{CO}_2(g) \quad \Delta H_1 = -393.5 \, \text{kJ/mol} \)

1. Reaction 2: \( \text{S}(s) + \text{O}_2(g)

ightarrow \text{SO}_2(g) \quad \Delta H_2 = -296.8 \, \text{kJ/mol} \) (Multiply by 2 for 2\( \text{SO}_2 \))

1. Reaction 3: \( \text{C}(s) + 2\text{S}(s)

ightarrow \text{CS}_2(l) \quad \Delta H_3 = +87.9 \, \text{kJ/mol} \) (Reverse this reaction: \( \text{CS}_2(l)
ightarrow \text{C}(s) + 2\text{S}(s) \quad \Delta H_3' = -87.9 \, \text{kJ/mol} \))

Step3: Combine Reactions

• Reverse Reaction 3: \( \text{CS}_2(l)

ightarrow \text{C}(s) + 2\text{S}(s) \quad \Delta H_3' = -87.9 \, \text{kJ/mol} \)

• Keep Reaction 1: \( \text{C}(s) + \text{O}_2(g)

ightarrow \text{CO}_2(g) \quad \Delta H_1 = -393.5 \, \text{kJ/mol} \)

• Multiply Reaction 2 by 2: \( 2\text{S}(s) + 2\text{O}_2(g)

ightarrow 2\text{SO}_2(g) \quad \Delta H_2' = 2 \times (-296.8) = -593.6 \, \text{kJ/mol} \)

Add these three modified reactions:
\( \text{CS}_2(l) + \text{C}(s) + 2\text{S}(s) + \text{O}_2(g) + 2\text{O}_2(g)
ightarrow \text{C}(s) + 2\text{S}(s) + \text{CO}_2(g) + 2\text{SO}_2(g) \)

Simplify (cancel \( \text{C}(s) \) and \( 2\text{S}(s) \)):
\( \text{CS}_2(l) + 3\text{O}_2(g)
ightarrow \text{CO}_2(g) + 2\text{SO}_2(g) \)

Step4: Calculate Total \( \Delta H \)

\( \Delta H = \Delta H_3' + \Delta H_1 + \Delta H_2' \)
\( \Delta H = (-87.9) + (-393.5) + (-593.6) \)
\( \Delta H = -87.9 - 393.5 - 593.6 \)
\( \Delta H = -1075.0 \, \text{kJ/mol} \)

Answer:

The enthalpy of the reaction is \(\boxed{-1075.0 \, \text{kJ/mol}}\)