The amount of heat released when `20 ml`,`0.5 M` `NaOH`is mixed with `100 ml`,`0.1 M` `H_(2)SO_(4)` is `xKJ`. The heat of neutralization will be `:-`

Options

A. `-100 x kJ//mol`
B. `-50 x kJ//mol`
C. `+100 x kJ//mol`
D. `+50 x kJ//mol`

Explanation

To solve the problem of calculating the heat of neutralization when `20 ml` of `0.5 M` NaOH is mixed with `100 ml` of `0.1 M` H₂SO₄, we can follow these steps:

### Step 1: Calculate the number of millimoles of NaOH
We can use the formula:
\[ \text{Number of millimoles} = \text{Molarity} \times \text{Volume in ml} \]
For NaOH:
\[ \text{Number of millimoles of NaOH} = 0.5 \, \text{M} \times 20 \, \text{ml} = 10 \, \text{mmol} \]

### Step 2: Calculate the number of millimoles of H₂SO₄
Using the same formula for H₂SO₄:
\[ \text{Number of millimoles of H₂SO₄} = 0.1 \, \text{M} \times 100 \, \text{ml} = 10 \, \text{mmol} \]

### Step 3: Determine the dissociation of H₂SO₄
1 mole of H₂SO₄ dissociates to give 2 moles of H⁺. Thus, 10 mmol of H₂SO₄ will give:
\[ 10 \, \text{mmol H₂SO₄} \rightarrow 20 \, \text{mmol H⁺} \]

### Step 4: Determine the dissociation of NaOH
1 mole of NaOH dissociates to give 1 mole of OH⁻. Thus, 10 mmol of NaOH will give:
\[ 10 \, \text{mmol NaOH} \rightarrow 10 \, \text{mmol OH⁻} \]

### Step 5: Identify the limiting reactant
In the neutralization reaction, H⁺ reacts with OH⁻ to form water:
\[ \text{H⁺} + \text{OH⁻} \rightarrow \text{H₂O} \]
We have 20 mmol of H⁺ and 10 mmol of OH⁻. Therefore, OH⁻ is the limiting reactant.

### Step 6: Calculate the moles of water formed
Since 10 mmol of OH⁻ will completely react with 10 mmol of H⁺, the amount of water formed will be:
\[ 10 \, \text{mmol H₂O} = 10 \times 10^{-3} \, \text{mol} = 0.01 \, \text{mol} \]

### Step 7: Relate the heat evolved to the heat of neutralization
Let the heat released for the neutralization of 0.01 moles be `x kJ`. Therefore, for 1 mole of H⁺ reacting with 1 mole of OH⁻, the heat of neutralization can be calculated as:
\[ \text{Heat of neutralization} = \frac{x \, \text{kJ}}{0.01 \, \text{mol}} = 100x \, \text{kJ/mol} \]

### Step 8: Conclusion
The heat of neutralization is thus:
\[ \text{Heat of neutralization} = -100x \, \text{kJ/mol} \]
The negative sign indicates that heat is released during the reaction.