Which of the following will produce the highest risein temperature?

Options

A. 67 mL of 1M NaOH `+`33 mL of 0.5M `H_(2)SO_(4)`
B. 33 mL of 1 M NaOH `+`67 mL of 0.5 M `H_(2)SO_(4)`
C. 40 mL of 1 M NaOH `+` 60 mL of 0.5 M `H_(2) SO_(4)`
D. 50 mL of 1 M NaOH `+` 50 mLof 0.5M `H_(2)SO_(4)`

Explanation

To determine which reaction will produce the highest rise in temperature, we need to analyze the neutralization reactions of the given options. The rise in temperature during a neutralization reaction is directly related to the amount of acid and base that react, as well as their concentrations.

### Step-by-Step Solution:

1. **Identify the Reactions**:
We are dealing with acid-base neutralization reactions. The general reaction can be represented as:
\[ \text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water} \]
The heat released during this reaction is proportional to the number of moles of acid and base that react.

2. **Calculate Moles of Reactants**:
For each option, we will calculate the number of moles of acid and base involved in the reaction.
- **Option 1**: 67 mL of 1 M NaOH and 33 mL of 0.5 M H2SO4
- Moles of NaOH = \( 1 \, \text{mol/L} \times 0.067 \, \text{L} = 0.067 \, \text{mol} \)
- Moles of H2SO4 = \( 0.5 \, \text{mol/L} \times 0.033 \, \text{L} = 0.0165 \, \text{mol} \)
- Since H2SO4 has a valency of 2, it will react with 0.033 mol of NaOH (0.0165 mol × 2).

- **Option 2**: 67 mL of 1 M NaOH and 40 mL of 0.5 M HCl
- Moles of NaOH = \( 0.067 \, \text{mol} \)
- Moles of HCl = \( 0.5 \, \text{mol/L} \times 0.040 \, \text{L} = 0.020 \, \text{mol} \)
- The limiting reagent is HCl, which will react with 0.020 mol of NaOH.

- **Option 3**: 60 mL of 1 M NaOH and 40 mL of 1 M H2SO4
- Moles of NaOH = \( 0.060 \, \text{mol} \)
- Moles of H2SO4 = \( 1 \, \text{mol/L} \times 0.040 \, \text{L} = 0.040 \, \text{mol} \)
- The limiting reagent is NaOH, which will react with 0.060 mol of H2SO4 (0.060 mol × 0.5).

- **Option 4**: 50 mL of 1 M NaOH and 50 mL of 1 M HCl
- Moles of NaOH = \( 0.050 \, \text{mol} \)
- Moles of HCl = \( 1 \, \text{mol/L} \times 0.050 \, \text{L} = 0.050 \, \text{mol} \)
- Both react completely, so the total moles reacting is 0.050 mol.

3. **Determine the Limiting Reagent**:
For each option, we find the limiting reagent and the amount of heat released:
- **Option 1**: Limiting reagent is H2SO4 (0.0165 mol).
- **Option 2**: Limiting reagent is HCl (0.020 mol).
- **Option 3**: Limiting reagent is NaOH (0.060 mol).
- **Option 4**: Both react completely (0.050 mol).

4. **Compare the Amounts**:
The amount of heat released (and thus the rise in temperature) is proportional to the moles of the limiting reagent:
- Option 1: 0.033 mol
- Option 2: 0.020 mol
- Option 3: 0.060 mol
- Option 4: 0.050 mol

5. **Conclusion**:
The highest rise in temperature will occur in **Option 3**, as it has the highest amount of moles reacting (0.060 mol).

### Final Answer:
**Option 3 will produce the highest rise in temperature.**