The standard enthalpy of formation of `NH_(3)(g)` is `-91.8 kJ "mol"^(-1)`.The amount of heat required to decompose 34 g of `NH_(3)(g)` into its elements is

Options

A. `183.6 kJ`
B. `91.8 kJ`
C. `45.9 kJ`
D. `137.7 kJ`

Explanation

To find the amount of heat required to decompose 34 g of `NH3(g)` into its elements, we can follow these steps:

### Step 1: Determine the molar mass of `NH3`
The molar mass of ammonia (NH3) can be calculated as follows:
- Nitrogen (N) has an atomic mass of approximately 14 g/mol.
- Hydrogen (H) has an atomic mass of approximately 1 g/mol, and there are three hydrogen atoms in ammonia.

Calculating the molar mass:
\[ \text{Molar mass of } NH3 = 14 \, \text{g/mol (N)} + 3 \times 1 \, \text{g/mol (H)} = 14 + 3 = 17 \, \text{g/mol} \]

### Step 2: Calculate the number of moles of `NH3` in 34 g
To find the number of moles of ammonia in 34 g, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles of } NH3 = \frac{34 \, \text{g}}{17 \, \text{g/mol}} = 2 \, \text{mol} \]

### Step 3: Use the standard enthalpy of formation to find the heat required
The standard enthalpy of formation of `NH3(g)` is given as `-91.8 kJ/mol`. This value indicates the heat released when 1 mole of `NH3` is formed from its elements. Therefore, to decompose `NH3`, we need to absorb the same amount of heat.

For 2 moles of `NH3`, the heat required can be calculated as:
\[ \text{Heat required} = \text{Number of moles} \times \text{Enthalpy of formation} \]
Substituting the values:
\[ \text{Heat required} = 2 \, \text{mol} \times 91.8 \, \text{kJ/mol} = 183.6 \, \text{kJ} \]

### Final Answer
The amount of heat required to decompose 34 g of `NH3(g)` into its elements is **183.6 kJ**.

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