A 0.05 L sample of 0.2 M aqueous hydrochloric acid is added to 0.05 L of 0.2 M aqueous ammonia in a calorimeter. Heat capacity of entire calorimeter system is 480 J/K. The temperature increase is 1.09 K. Calculate `Delta_(r )H^(@)` in kJ/mol for the following reaction :
`HCl(aq.)+NH_(3)(aq.)rarrNH_(4)Cl(aq.)`

Options

A. `-52.32`
B. `-61.1`
C. `-55.8`
D. `-58.2`

Explanation

To calculate the standard enthalpy change (ΔrH°) for the reaction:

\[ \text{HCl(aq)} + \text{NH}_3\text{(aq)} \rightarrow \text{NH}_4\text{Cl(aq)} \]

we will follow these steps:

### Step 1: Calculate the number of moles of HCl and NH3

Given:
- Volume of HCl solution = 0.05 L
- Molarity of HCl = 0.2 M

Using the formula for moles:
\[ \text{moles} = \text{Volume (L)} \times \text{Molarity (mol/L)} \]

Calculating moles of HCl:
\[ \text{moles of HCl} = 0.05 \, \text{L} \times 0.2 \, \text{mol/L} = 0.01 \, \text{mol} \]

Since the volumes and molarities of HCl and NH3 are the same, the moles of NH3 will also be:
\[ \text{moles of NH3} = 0.01 \, \text{mol} \]

### Step 2: Calculate the heat absorbed by the calorimeter

Given:
- Heat capacity of the calorimeter system = 480 J/K
- Temperature increase (ΔT) = 1.09 K

Using the formula:
\[ q = C \times \Delta T \]

Calculating heat (q):
\[ q = 480 \, \text{J/K} \times 1.09 \, \text{K} = 523.2 \, \text{J} \]

### Step 3: Convert heat to kilojoules

Since we need ΔrH° in kJ:
\[ q = 523.2 \, \text{J} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = 0.5232 \, \text{kJ} \]

### Step 4: Calculate ΔrH° per mole of reaction

Since the reaction consumes 0.01 moles of HCl and NH3, we will calculate ΔrH° per mole:
\[ \Delta rH° = -\frac{q}{\text{moles}} \]

Calculating ΔrH°:
\[ \Delta rH° = -\frac{0.5232 \, \text{kJ}}{0.01 \, \text{mol}} = -52.32 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta rH° = -52.32 \, \text{kJ/mol} \]

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