In the reactions
`HCI + NaOH rightarrow NaCI + H_(2)O + x cal.` ` H_(2)SO_(4) + 2NaOh rightarrow Na_(2)SO_(4) + 2 H_(2)O + y cal.`

Options

A. ` x = y `
B. `x = 2y`
C. `x = y/2`
D. `x = sqrt y`

Answer

To solve the problem, we need to analyze the two given reactions and their associated heat changes (enthalpy changes).

### Step-by-Step Solution:

1. **Identify the Reactions**:
- The first reaction is:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} + x \text{ cal} \]
- The second reaction is:
\[ \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} + y \text{ cal} \]

2. **Understand the Nature of the Reactions**:
- Both reactions are neutralization reactions between an acid and a base.
- Neutralization reactions are exothermic, meaning they release heat.

3. **Determine the Stoichiometry**:
- In the first reaction (HCl + NaOH), 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of water.
- In the second reaction (H2SO4 + 2NaOH), 1 mole of H2SO4 reacts with 2 moles of NaOH to produce 1 mole of Na2SO4 and 2 moles of water.

4. **Relate the Heat Changes**:
- The heat released in the first reaction is denoted as \( x \) calories.
- The heat released in the second reaction is denoted as \( y \) calories.
- Since the second reaction involves 2 moles of NaOH reacting with 1 mole of H2SO4, it releases heat equivalent to twice that of the first reaction.

5. **Establish the Relationship**:
- Therefore, we can establish the relationship:
\[ y = 2x \]
- This means that the heat released in the second reaction (y) is twice that of the first reaction (x).

6. **Final Relation**:
- Rearranging the equation gives:
\[ x = \frac{y}{2} \]

### Conclusion:
The relationship between \( x \) and \( y \) is:
\[ x = \frac{y}{2} \]