x g sample of `NH_(4)NO_(3)` is decomposed in a Bomb calorimeter. The temperature of calorimeter increase by `4^(@)C`. The heat capacity of the system is `1.25 " kJ"//.^(@)C`. Calculate the value of x. Given molar heat of decomposition of `NH_(4)NO_(3)` at constant volume is 400 kJ `" mol"^(-1)`.

Answer

To solve the problem step by step, we will use the information provided in the question and the principles of thermodynamics related to heat transfer.

### Step 1: Understand the heat transfer in the calorimeter
When the sample of `NH₄NO₃` decomposes in the bomb calorimeter, it releases heat, which causes the temperature of the calorimeter to rise. The heat absorbed by the calorimeter can be calculated using the formula:

\[ Q = C \cdot \Delta T \]

Where:
- \( Q \) = heat absorbed by the calorimeter (in kJ)
- \( C \) = heat capacity of the system (in kJ/°C)
- \( \Delta T \) = change in temperature (in °C)

### Step 2: Substitute the known values
From the problem, we know:
- \( C = 1.25 \, \text{kJ/°C} \)
- \( \Delta T = 4 \, \text{°C} \)

Substituting these values into the equation:

\[ Q = 1.25 \, \text{kJ/°C} \cdot 4 \, \text{°C} = 5 \, \text{kJ} \]

### Step 3: Relate the heat released to the amount of substance
The heat released by the decomposition of `NH₄NO₃` can also be expressed in terms of the number of moles of the substance decomposed. The molar heat of decomposition is given as 400 kJ/mol. If \( x \) grams of `NH₄NO₃` is decomposed, we need to convert this mass to moles using its molar mass.

The molar mass of `NH₄NO₃` (Ammonium Nitrate) is approximately:
- N: 14 g/mol (2 Nitrogen atoms)
- H: 1 g/mol (4 Hydrogen atoms)
- O: 16 g/mol (3 Oxygen atoms)

Calculating the molar mass:
\[ \text{Molar mass of } NH₄NO₃ = 14 \times 2 + 1 \times 4 + 16 \times 3 = 28 + 4 + 48 = 80 \, \text{g/mol} \]

### Step 4: Calculate the number of moles of `NH₄NO₃`
Let \( n \) be the number of moles of `NH₄NO₃` decomposed:
\[ n = \frac{x}{80} \]

### Step 5: Calculate the heat released in terms of moles
The heat released by the decomposition of \( n \) moles of `NH₄NO₃` is:
\[ Q = n \cdot \text{(molar heat of decomposition)} = n \cdot 400 \, \text{kJ/mol} \]

Substituting \( n \):
\[ Q = \frac{x}{80} \cdot 400 \]

### Step 6: Set the heat absorbed equal to the heat released
From Step 2, we found that \( Q = 5 \, \text{kJ} \). Therefore, we can set the two expressions for \( Q \) equal to each other:
\[ 5 = \frac{x}{80} \cdot 400 \]

### Step 7: Solve for \( x \)
Rearranging the equation to solve for \( x \):
\[ 5 = \frac{400x}{80} \]
\[ 5 = 5x \]
\[ x = 1 \, \text{g} \]

### Step 8: Final result
Thus, the value of \( x \) is:
\[ x = 80 \, \text{g} \]