Given `C_((s)) + O_(2(g)) rarr CO_(2(g)), Delta H = -395 kJ , S_((g)) + O_(2(g)) rarr SO_(2(g)) , Delta H = -295 kJ ,`
`CS_(2(l)) + 3O_(2(g)) rarr CO_(2(g)) + 2SO_(2(g)) , Delta H = -1110 kJ`
The heat of formation of `CS_(2(l))` is

Options

A. `+125 kJ mol^(-1)`
B. `3.25 kJ mol^(-1)`
C. `62.5 kJ mol^(-1)`
D. `250 kJ mol^(-1)`

Answer

To find the heat of formation of \( CS_2(l) \), we can 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. We will manipulate the given reactions to find the desired enthalpy change.

### Given Reactions:
1. \( C(s) + O_2(g) \rightarrow CO_2(g) \), \( \Delta H = -395 \, \text{kJ} \) (Reaction 1)
2. \( S(g) + O_2(g) \rightarrow SO_2(g) \), \( \Delta H = -295 \, \text{kJ} \) (Reaction 2)
3. \( CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g) \), \( \Delta H = -1110 \, \text{kJ} \) (Reaction 3)

### Step-by-Step Solution:

**Step 1: Write the equations for the formation of products.**
We need to find the heat of formation of \( CS_2(l) \). The formation reaction can be represented as:
\[ C(s) + 2S(g) + 2O_2(g) \rightarrow CS_2(l) \]

**Step 2: Manipulate the given reactions.**
We can rearrange the reactions to derive the formation reaction for \( CS_2(l) \).

- From Reaction 1, we can use \( CO_2(g) \):
\[ CO_2(g) \leftarrow C(s) + O_2(g) \quad (\Delta H = +395 \, \text{kJ}) \]

- From Reaction 2, we can use \( SO_2(g) \):
\[ 2SO_2(g) \leftarrow 2S(g) + 2O_2(g) \quad (\Delta H = +590 \, \text{kJ}) \]

- From Reaction 3, we can reverse it to find \( CS_2(l) \):
\[ CS_2(l) \rightarrow CO_2(g) + 2SO_2(g) \quad (\Delta H = +1110 \, \text{kJ}) \]

**Step 3: Combine the manipulated reactions.**
Now we can combine these reactions:
1. \( CO_2(g) \leftarrow C(s) + O_2(g) \) (from Reaction 1)
2. \( 2SO_2(g) \leftarrow 2S(g) + 2O_2(g) \) (from Reaction 2)
3. \( CS_2(l) \rightarrow CO_2(g) + 2SO_2(g) \) (from Reaction 3)

Adding these gives:
\[ C(s) + 2S(g) + 2O_2(g) \rightarrow CS_2(l) \]

**Step 4: Calculate the total enthalpy change.**
Now, we can calculate the heat of formation of \( CS_2(l) \):
\[ \Delta H_{f} = \Delta H_{1} + \Delta H_{2} + \Delta H_{3} \]
\[ \Delta H_{f} = (+395 \, \text{kJ}) + (+590 \, \text{kJ}) + (+1110 \, \text{kJ}) \]
\[ \Delta H_{f} = 395 + 590 - 1110 = -125 \, \text{kJ} \]

### Final Answer:
The heat of formation of \( CS_2(l) \) is \( -125 \, \text{kJ/mol} \).