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 of the reaction \( \Delta_r H^\circ \) 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 total 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 Joules)
- \( C \) = heat capacity of the calorimeter (in J/K)
- \( \Delta T \) = temperature change (in K)

Given:
- \( C = 480 \, \text{J/K} \)
- \( \Delta T = 1.09 \, \text{K} \)

Substituting the values:

\[ q = 480 \, \text{J/K} \times 1.09 \, \text{K} = 523.2 \, \text{J} \]

### Step 2: Convert heat absorbed to kilojoules

Since we need \( \Delta_r H^\circ \) in kJ, we convert Joules to kilojoules:

\[ q = 523.2 \, \text{J} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = 0.5232 \, \text{kJ} \]

### Step 3: Calculate the number of moles of reactants

Next, we need to find the number of moles of the reactants involved in the reaction. The volume and molarity of both HCl and NH3 are given:

- Volume of HCl = 0.05 L
- Molarity of HCl = 0.2 M
- Volume of NH3 = 0.05 L
- Molarity of NH3 = 0.2 M

Calculating moles:

\[ \text{Moles of HCl} = \text{Volume} \times \text{Molarity} = 0.05 \, \text{L} \times 0.2 \, \text{mol/L} = 0.01 \, \text{mol} \]

\[ \text{Moles of NH3} = 0.05 \, \text{L} \times 0.2 \, \text{mol/L} = 0.01 \, \text{mol} \]

### Step 4: Calculate \( \Delta_r H^\circ \) per mole of reaction

The enthalpy change for the reaction can be calculated using the total heat absorbed and the number of moles of the limiting reactant (which is the same for both in this case):

\[ \Delta_r H^\circ = -\frac{q}{\text{moles of limiting reactant}} \]

Substituting the values:

\[ \Delta_r H^\circ = -\frac{0.5232 \, \text{kJ}}{0.01 \, \text{mol}} = -52.32 \, \text{kJ/mol} \]

### Final Answer

Thus, the standard enthalpy change of the reaction is:

\[ \Delta_r H^\circ = -52.32 \, \text{kJ/mol} \]

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