A solution is 500 ml of 2M KOH is added to 500 ml of 2 M HCl and the mixture is well shaken. The rise in temperature `T_(1)` is noted. The experiment is then repeated using `250 ml` of each solution and rise in temperature `T_(2)` is again noted. Assume all heat is taken by the solution

Options

A. `T_(1)=T_(2)`
B. `T_(1)` is 2 times as large as `T_(2)`
C. `T_(2)` is twice of `T_(1)`
D. `T_(1)` is 4 times as large as `T_(2)`

Answer

To solve the problem, we will analyze the heat exchange during the neutralization reaction between KOH and HCl. The key points to consider are the volumes and concentrations of the solutions used in both experiments.

### Step 1: Calculate the moles of KOH and HCl in both experiments.

1. **For the first experiment (500 ml of 2M KOH and 500 ml of 2M HCl):**
- Moles of KOH = Molarity × Volume = 2 M × 0.5 L = 1 mole
- Moles of HCl = Molarity × Volume = 2 M × 0.5 L = 1 mole

2. **For the second experiment (250 ml of 2M KOH and 250 ml of 2M HCl):**
- Moles of KOH = Molarity × Volume = 2 M × 0.25 L = 0.5 moles
- Moles of HCl = Molarity × Volume = 2 M × 0.25 L = 0.5 moles

### Step 2: Determine the heat released in both experiments.

The heat released (Q) during the neutralization reaction can be expressed as:
\[ Q = n \cdot \Delta T \]
where \( n \) is the number of moles of the limiting reagent and \( \Delta T \) is the rise in temperature.

1. **For the first experiment:**
- Total heat released \( Q_1 = n \cdot T_1 = 1 \cdot T_1 = T_1 \) (since n = 1 mole)

2. **For the second experiment:**
- Total heat released \( Q_2 = n \cdot T_2 = 0.5 \cdot T_2 \) (since n = 0.5 moles)

### Step 3: Relate the heat released in both experiments.

Since the heat released in both experiments is equal (as the reaction is the same), we can set up the equation:
\[ Q_1 = Q_2 \]
Thus,
\[ T_1 = 0.5 \cdot T_2 \]

### Step 4: Analyze the relationship between T1 and T2.

From the equation \( T_1 = 0.5 \cdot T_2 \), we can rearrange it to find:
\[ T_2 = 2 \cdot T_1 \]

### Conclusion:

This implies that the rise in temperature in the first experiment (T1) is half that of the rise in temperature in the second experiment (T2). Therefore, the correct option is:
- **T1 is equal to T2.**

### Final Answer:
T1 is equal to T2.