During complete combustion of one mole of butane ,2658Kj of heat is released. The thermochemical reaction for above change is

Options

A. `2C_(4) H_(10) (g) + 13 O_(2) (g) rarr 8 CO_(2)(g) + 10 H_(2) O (l), Delta_(c) H = -2658.0 kJ mol^(-1)`
B. `C_(4)H_(10) (g) + (13)/(2) O_(2) (g) rarr 4CO_(2) (g) + 5H_(2)O (l) , Delta_(c) H = - 1329.0 kJ mol^(-1)`
C. `C_(4)H_(10) (g) + (13)/(2) O_(2)(g) rarr 4CO_(2) (g) + 5H_(2) O (l), Delta_(c) H = - 2658.0 kJ mol^(-1)`
D. `C_(4) H_(10) (g) + (13)/(2) O_(2) (g) rarr 4CO_(2) (g) + 5H_(2) O (l) , Delta_(c) H = + 2658.0 kJ mol^(-1)`

Explanation

To write the thermochemical reaction for the complete combustion of one mole of butane, we can follow these steps:

### Step 1: Write the chemical formula for butane.
Butane is represented by the chemical formula \( C_4H_{10} \).

### Step 2: Identify the products of combustion.
The complete combustion of hydrocarbons like butane produces carbon dioxide (\( CO_2 \)) and water (\( H_2O \)).

### Step 3: Write the unbalanced reaction.
The unbalanced reaction can be written as:
\[ C_4H_{10} + O_2 \rightarrow CO_2 + H_2O \]

### Step 4: Balance the carbon atoms.
There are 4 carbon atoms in butane, so we need 4 \( CO_2 \) molecules:
\[ C_4H_{10} + O_2 \rightarrow 4 CO_2 + H_2O \]

### Step 5: Balance the hydrogen atoms.
There are 10 hydrogen atoms in butane, so we need 5 \( H_2O \) molecules:
\[ C_4H_{10} + O_2 \rightarrow 4 CO_2 + 5 H_2O \]

### Step 6: Balance the oxygen atoms.
Now, we count the oxygen atoms on the right side. There are:
- 4 \( CO_2 \) contributing \( 4 \times 2 = 8 \) oxygen atoms
- 5 \( H_2O \) contributing \( 5 \times 1 = 5 \) oxygen atoms

Total oxygen needed = \( 8 + 5 = 13 \) oxygen atoms. Since \( O_2 \) provides 2 oxygen atoms per molecule, we need:
\[ \frac{13}{2} = 6.5 \text{ molecules of } O_2 \]

### Step 7: Write the balanced equation.
To avoid fractions, we can multiply the entire equation by 2:
\[ 2 C_4H_{10} + 13 O_2 \rightarrow 8 CO_2 + 10 H_2O \]

### Step 8: Include the enthalpy change.
The problem states that during the complete combustion of one mole of butane, \( 2658 \, kJ \) of heat is released. Since heat is released, we denote this as a negative change in enthalpy:
\[ \Delta H = -2658 \, kJ \]

### Final Balanced Thermochemical Equation:
Thus, the complete thermochemical equation for the combustion of butane is:
\[ 2 C_4H_{10} + 13 O_2 \rightarrow 8 CO_2 + 10 H_2O \quad \Delta H = -2658 \, kJ \]

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