In a calorimeter, the temperature of the calorimeter increases by 6.12 K, the heat capacity of the systemize 1.23 kJ/g deg. What is the molar heat of decomposition for `NH_(4)NO_(3)`?

Options

A. `-7.53kJ//mol`
B. `-398.1kJ//mol`
C. `-16.1kJ//mol`
D. `602kJ//mol`

Answer

To find the molar heat of decomposition for \( NH_4NO_3 \), we can follow these steps:

### Step 1: Calculate the heat absorbed by the calorimeter
The heat absorbed by the calorimeter can be calculated using the formula:
\[ q = C \times \Delta T \]
where:
- \( q \) = heat absorbed (in kJ)
- \( C \) = heat capacity of the system (in kJ/K)
- \( \Delta T \) = change in temperature (in K)

Given:
- \( C = 1.23 \, \text{kJ/g} \cdot \text{K} \)
- \( \Delta T = 6.12 \, \text{K} \)

Substituting the values:
\[ q = 1.23 \, \text{kJ/g} \cdot \text{K} \times 6.12 \, \text{K} \]
\[ q = 7.5276 \, \text{kJ/g} \]

### Step 2: Calculate the molar heat of decomposition
To find the molar heat of decomposition, we need to multiply the heat absorbed by the molar mass of \( NH_4NO_3 \).

The molar mass of \( NH_4NO_3 \) is approximately:
\[ \text{Molar mass} = 80.003 \, \text{g/mol} \]

Now, we can calculate the molar heat of decomposition:
\[ \Delta H = q \times \text{Molar mass} \]
\[ \Delta H = 7.5276 \, \text{kJ/g} \times 80.003 \, \text{g/mol} \]
\[ \Delta H \approx 602.2 \, \text{kJ/mol} \]

### Conclusion
The molar heat of decomposition for \( NH_4NO_3 \) is approximately \( 602.2 \, \text{kJ/mol} \).

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