$$ \left. \begin{array} { l } { ( S _ { 2 } ( l ) + 3 O _ { 2 } ( g ) \div CO _ { 2 } ( g ) + { } _ { 2 } SO _ { 2 } ( g ) \Delta H = ? } \ { C _ { ( s ) } + C _ { 2 } ( g ) ightarrow CO _ { 2 } ( g ) \Delta H = - 39,3.5 kJ } \ { S ( s ) + O _ { 2 } ( g ) \geq SO _ { 2 } ( g ) \quad \Delta H = - 296.8 kJ } \ { C ( s ) + 2 S _ { s } ( s ) ightarrow CS _ { 2 } ( l ) \quad CH = + 87.9 kJJ } \end{array} ight. $$

Question

$$ \left. \begin{array} { l } { ( S _ { 2 } ( l ) + 3 O _ { 2 } ( g ) \div CO _ { 2 } ( g ) + { } _ { 2 } SO _ { 2 } ( g ) \Delta H = ? } \ { C _ { ( s ) } + C _ { 2 } ( g ) ightarrow CO _ { 2 } ( g ) \Delta H = - 39,3.5 kJ } \ { S ( s ) + O _ { 2 } ( g ) \geq SO _ { 2 } ( g ) \quad \Delta H = - 296.8 kJ } \ { C ( s ) + 2 S _ { s } ( s ) ightarrow CS _ { 2 } ( l ) \quad CH = + 87.9 kJJ } \end{array} ight. $$

UpStudy AI · Accepted Answer

To solve the problem, we need to determine the enthalpy change ($$ \Delta H $$) for the reaction involving $$ S_2(l) $$, $$ O_2(g) $$, and $$ SO_2(g) $$. We will use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Known Reactions and Their Enthalpy Changes

1. $$ C(s) + C_2(g) \rightarrow CO_2(g) \quad \Delta H = -39.3 \, \text{kJ} $$
2. $$ S(s) + O_2(g) \rightarrow SO_2(g) \quad \Delta H = -296.8 \, \text{kJ} $$
3. $$ C(s) + 2S(s) \rightarrow CS_2(l) \quad \Delta H = +87.9 \, \text{kJ} $$

### Target Reaction

We need to find $$ \Delta H $$ for the reaction:

$$
S_2(l) + 3O_2(g) \rightarrow SO_2(g)
$$

### Steps to Solve

1. **Reverse the third reaction** to express $$ S_2(l) $$ in terms of $$ S(s) $$:
   $$
   CS_2(l) \rightarrow C(s) + 2S(s) \quad \Delta H = -87.9 \, \text{kJ}
   $$

2. **Use the second reaction** as is to produce $$ SO_2(g) $$:
   $$
   S(s) + O_2(g) \rightarrow SO_2(g) \quad \Delta H = -296.8 \, \text{kJ}
   $$

3. **Use the first reaction** to produce $$ CO_2(g) $$:
   $$
   C(s) + C_2(g) \rightarrow CO_2(g) \quad \Delta H = -39.3 \, \text{kJ}
   $$

### Combining the Reactions

Now we will combine these reactions to find the overall enthalpy change.

1. From the reversed third reaction:
   $$
   CS_2(l) \rightarrow C(s) + 2S(s) \quad \Delta H = -87.9 \, \text{kJ}
   $$

2. From the second reaction:
   $$
   S(s) + O_2(g) \rightarrow SO_2(g) \quad \Delta H = -296.8 \, \text{kJ}
   $$

3. We need to adjust the second reaction to account for the stoichiometry of $$ S_2(l) $$ and $$ O_2(g) $$. Since we need 2 moles of $$ S(s) $$ to form $$ S_2(l) $$, we will multiply the second reaction by 2:
   $$
   2S(s) + 2O_2(g) \rightarrow 2SO_2(g) \quad \Delta H = 2 \times (-296.8) = -593.6 \, \text{kJ}
   $$

### Final Calculation

Now we can sum the enthalpy changes:

$$
\Delta H = (-87.9) + (-593.6) + (-39.3)
$$

Calculating this gives:

$$
\Delta H = -87.9 - 593.6 - 39.3 = -720.8 \, \text{kJ}
$$

### Conclusion

The enthalpy change for the reaction $$ S_2(l) + 3O_2(g) \rightarrow SO_2(g) $$ is:

$$
\Delta H = -720.8 \, \text{kJ}
$$
The enthalpy change for the reaction $$ S_2(l) + 3O_2(g) \rightarrow SO_2(g) $$ is $$ \Delta H = -720.8 \, \text{kJ} $$.

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