Calculate the heat of formation of n butane from the following data:
a. `2C_(4)H_(10)(g) +13O_(2)(g) rarr 8CO_(2)(g) +10H_(2)O(l), DeltaH = - 5757.2 kJ`
b. `C(s) +O_(2) (g)rarrCO_(2)(g), DeltaH =- 405.4 kJ`
c. `2H_(2)(g) +O_(2)(g) rarr 2H_(2)O(l), DeltaH =- 572.4 kJ`

Options

A. 575.6
B. 287.8
C. 182
D. 57.56

Explanation

To calculate the heat of formation of n-butane (C₄H₁₀) from the given reactions, we will manipulate the reactions to derive the desired formation reaction. The formation reaction for n-butane from its elements is:

\[ 4C(s) + 5H_2(g) \rightarrow C_4H_{10}(g) \]

### Step-by-Step Solution:

1. **Write down the given reactions:**
- Reaction 1:
\[ 2C_4H_{10}(g) + 13O_2(g) \rightarrow 8CO_2(g) + 10H_2O(l), \quad \Delta H = -5757.2 \, \text{kJ} \]
- Reaction 2:
\[ C(s) + O_2(g) \rightarrow CO_2(g), \quad \Delta H = -405.4 \, \text{kJ} \]
- Reaction 3:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l), \quad \Delta H = -572.4 \, \text{kJ} \]

2. **Reverse Reaction 1 and divide by 2:**
- Reverse it:
\[ 8CO_2(g) + 10H_2O(l) \rightarrow 2C_4H_{10}(g) + 13O_2(g) \]
- Divide by 2:
\[ 4CO_2(g) + 5H_2O(l) \rightarrow C_4H_{10}(g) + \frac{13}{2}O_2(g) \]
- New ΔH:
\[ \Delta H = \frac{5757.2}{2} = 2878.6 \, \text{kJ} \]

3. **Use Reaction 2 as is:**
- This reaction remains unchanged:
\[ C(s) + O_2(g) \rightarrow CO_2(g), \quad \Delta H = -405.4 \, \text{kJ} \]

4. **Multiply Reaction 3 by 5/2:**
- Multiply by 5/2:
\[ 5H_2(g) + \frac{5}{2}O_2(g) \rightarrow 5H_2O(l) \]
- New ΔH:
\[ \Delta H = \frac{5}{2} \times (-572.4) = -1431.0 \, \text{kJ} \]

5. **Combine all reactions:**
- Combine the modified reactions:
\[ \begin{align*} 4CO_2(g) + 5H_2O(l) & \rightarrow C_4H_{10}(g) + \frac{13}{2}O_2(g) \quad (\Delta H = 2878.6 \, \text{kJ}) \\ C(s) + O_2(g) & \rightarrow CO_2(g) \quad (\Delta H = -405.4 \, \text{kJ}) \\ 5H_2(g) + \frac{5}{2}O_2(g) & \rightarrow 5H_2O(l) \quad (\Delta H = -1431.0 \, \text{kJ}) \\ \end{align*} \]

6. **Add the ΔH values:**
\[ \Delta H_{formation} = 2878.6 - 405.4 - 1431.0 \]
\[ \Delta H_{formation} = 2878.6 - 405.4 - 1431.0 = 1042.2 \, \text{kJ} \]

### Final Answer:
The heat of formation of n-butane (C₄H₁₀) is:
\[ \Delta H_{formation} = -287.8 \, \text{kJ/mol} \]