The combustion of butane `(C_(4)H_(10))` is exothermic by 2878.7 kJ ` "mol"^(-1)`.Calculate the standard enthalpy of formation of butane given that the standard enthalpies of formation of `CO_(2)(g)` and`H_(2)O(l)` are `-393.5 kJ "mol"^(-1)` and `-285.8 kJ "mol"^(-1)` respectively.

Explanation

To calculate the standard enthalpy of formation of butane (C₄H₁₀), 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 of the reaction.

### Step 1: Write the balanced equation for the combustion of butane.
The balanced equation for the combustion of butane in oxygen is:

\[ C_4H_{10} + 13/2 O_2 \rightarrow 4 CO_2 + 5 H_2O \]

### Step 2: Write the enthalpy change for the combustion reaction.
The combustion of butane is exothermic, which means it releases energy. The enthalpy change for this reaction is given as:

\[ \Delta H_{combustion} = -2878.7 \, kJ/mol \]

### Step 3: Write the formation equations for the products.
The standard enthalpy of formation for the products (CO₂ and H₂O) can be written as:

- For CO₂:
\[ \Delta H_f (CO_2) = -393.5 \, kJ/mol \]

- For H₂O:
\[ \Delta H_f (H_2O) = -285.8 \, kJ/mol \]

### Step 4: Calculate the total enthalpy of formation for the products.
To find the total enthalpy of formation for the products, we multiply the enthalpy of formation of each product by the number of moles produced:

\[ \Delta H_{products} = [4 \times (-393.5 \, kJ/mol)] + [5 \times (-285.8 \, kJ/mol)] \]

Calculating this gives:

\[ \Delta H_{products} = [4 \times (-393.5)] + [5 \times (-285.8)] \]
\[ = -1574 + (-1429) \]
\[ = -3003 \, kJ \]

### Step 5: Use Hess's law to find the enthalpy of formation of butane.
According to Hess's law, the enthalpy change for the formation of butane can be calculated as follows:

\[ \Delta H_{formation} (C_4H_{10}) = \Delta H_{products} - \Delta H_{combustion} \]

Substituting the values we have:

\[ \Delta H_{formation} (C_4H_{10}) = -3003 \, kJ - (-2878.7 \, kJ) \]
\[ = -3003 \, kJ + 2878.7 \, kJ \]
\[ = -124.3 \, kJ/mol \]

### Conclusion:
The standard enthalpy of formation of butane (C₄H₁₀) is approximately:

\[ \Delta H_f (C_4H_{10}) = -124.3 \, kJ/mol \]

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