The amount of energy released when 20 mL of 0.5 M `NH_4OH` are mixed with 100 mL of 0.1 M HCl is x kJ. The heat of dissociation of `NH_4OH` will be (heat of neutralization of NaOH & HCl is y kJ/mol).

Options

A. `-100 x`
B. `y - 100 x`
C. `-100 x - y`
D. `100 x`

Explanation

To solve the problem, we need to calculate the heat of dissociation of ammonium hydroxide (NH₄OH) when it reacts with hydrochloric acid (HCl). We will use the information provided about the energy released during the reaction and the heat of neutralization of sodium hydroxide (NaOH) with HCl.

### Step-by-Step Solution:

1. **Calculate Moles of NH₄OH:**
- Given the concentration of NH₄OH is 0.5 M and the volume is 20 mL.
- Convert volume from mL to L:
\[ \text{Volume in L} = \frac{20 \text{ mL}}{1000} = 0.020 \text{ L} \]
- Calculate moles of NH₄OH:
\[ \text{Moles of NH₄OH} = \text{Molarity} \times \text{Volume in L} = 0.5 \, \text{mol/L} \times 0.020 \, \text{L} = 0.01 \, \text{mol} \]

2. **Calculate Moles of HCl:**
- Given the concentration of HCl is 0.1 M and the volume is 100 mL.
- Convert volume from mL to L:
\[ \text{Volume in L} = \frac{100 \text{ mL}}{1000} = 0.100 \text{ L} \]
- Calculate moles of HCl:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume in L} = 0.1 \, \text{mol/L} \times 0.100 \, \text{L} = 0.01 \, \text{mol} \]

3. **Determine the Heat Released (x kJ):**
- The reaction between NH₄OH and HCl releases x kJ of energy.
- Since both reactants are present in equal moles (0.01 mol), the total heat released can be considered as x kJ.

4. **Heat of Neutralization of NaOH and HCl (y kJ/mol):**
- The heat of neutralization for the reaction of NaOH with HCl is given as y kJ/mol.

5. **Calculate Heat of Dissociation of NH₄OH:**
- The heat of dissociation of NH₄OH can be derived from the difference between the heat released in the reaction of NH₄OH with HCl and the heat of neutralization of NaOH with HCl.
- The equation can be set up as follows:
\[ \text{Heat of dissociation of NH₄OH} = \text{Heat released in NH₄OH + HCl} - \text{Heat of neutralization of NaOH + HCl} \]
- Substituting the known values:
\[ \text{Heat of dissociation of NH₄OH} = x - y \]

6. **Final Answer:**
- The heat of dissociation of NH₄OH is:
\[ \text{Heat of dissociation of NH₄OH} = -100x - y \text{ kJ} \]