`{:(C(s)+O_(2)(g)rarrCO_(2)(g), ,,,,DeltaH= -94.3 kcal//mol),(CO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g),,,,,DeltaH= -67.4 kcal//mol),(O_(2)(g)rarr2O(g),,,,,DeltaH=117.4 kcal//mol),(CO(g)rarrC(g)+O(g),,,,,DeltaH=230.6 kcal//mol):}`
Calculate `Delta H` for `C(s)rarrC(g)` in `kcal//mol`.

Options

A. 171
B. 154
C. 117
D. 145

Explanation

To calculate ΔH for the reaction \( C(s) \rightarrow C(g) \), 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. We will manipulate the given reactions to derive the desired reaction.

### Given Reactions:
1. \( C(s) + O_2(g) \rightarrow CO_2(g) \), ΔH = -94.3 kcal/mol
2. \( CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \), ΔH = -67.4 kcal/mol
3. \( O_2(g) \rightarrow 2O(g) \), ΔH = 117.4 kcal/mol
4. \( CO(g) \rightarrow C(g) + O(g) \), ΔH = 230.6 kcal/mol

### Target Reaction:
5. \( C(s) \rightarrow C(g) \)

### Steps to derive the target reaction:

**Step 1: Reverse Reaction 1**
- We need to reverse the first reaction to have \( CO_2(g) \) on the reactant side:
\[ CO_2(g) \rightarrow C(s) + O_2(g) \quad \text{ΔH = +94.3 kcal/mol} \]

**Step 2: Use Reaction 2 as is**
- We will keep the second reaction as is:
\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \quad \text{ΔH = -67.4 kcal/mol} \]

**Step 3: Reverse Reaction 4**
- We need to reverse the fourth reaction to have \( C(g) \) on the product side:
\[ C(g) + O(g) \rightarrow CO(g) \quad \text{ΔH = -230.6 kcal/mol} \]

**Step 4: Use Reaction 3 divided by 2**
- Divide the third reaction by 2 to get half of \( O_2(g) \):
\[ O(g) \rightarrow \frac{1}{2} O_2(g) \quad \text{ΔH = \(\frac{117.4}{2}\) = 58.7 kcal/mol} \]

### Combine the Reactions:
Now we will combine the modified reactions:
1. \( CO_2(g) \rightarrow C(s) + O_2(g) \) (ΔH = +94.3 kcal/mol)
2. \( CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \) (ΔH = -67.4 kcal/mol)
3. \( C(g) + O(g) \rightarrow CO(g) \) (ΔH = -230.6 kcal/mol)
4. \( O(g) \rightarrow \frac{1}{2} O_2(g) \) (ΔH = +58.7 kcal/mol)

### Final Calculation:
Now we add the ΔH values:
\[ ΔH = 94.3 - 67.4 - 230.6 + 58.7 \]
Calculating this gives:
\[ ΔH = 94.3 - 67.4 - 230.6 + 58.7 = -145.0 \text{ kcal/mol} \]

### Conclusion:
Thus, the ΔH for the reaction \( C(s) \rightarrow C(g) \) is \( -145.0 \text{ kcal/mol} \).