The amount of energy released when 20 ml of 0.5 M NaOH are mixed with 100 ml of 0.1 M HCl is x kJ. The heat of neutralization (in kJ `"mol"^(-1)`) is:

Options

A. `-100` x
B. `-50` x
C. `+100` x
D. `+50` x

Answer

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

### Step 1: Calculate the number of moles of NaOH
The number of moles can be calculated using the formula:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} \]
For NaOH:
- Molarity = 0.5 M
- Volume = 20 ml = 0.020 L

Calculating the moles of NaOH:
\[ \text{Number of moles of NaOH} = 0.5 \, \text{mol/L} \times 0.020 \, \text{L} = 0.01 \, \text{mol} \]

### Step 2: Calculate the number of moles of HCl
Using the same formula for HCl:
- Molarity = 0.1 M
- Volume = 100 ml = 0.100 L

Calculating the moles of HCl:
\[ \text{Number of moles of HCl} = 0.1 \, \text{mol/L} \times 0.100 \, \text{L} = 0.01 \, \text{mol} \]

### Step 3: Determine the limiting reactant
In this case, both NaOH and HCl have the same number of moles (0.01 mol). Therefore, they will completely neutralize each other.

### Step 4: Calculate the heat of neutralization
The heat of neutralization is defined as the heat released when one mole of acid reacts with one mole of base. Since we have 0.01 moles of both NaOH and HCl reacting, we can express the heat released (x kJ) for this reaction.

To find the heat of neutralization per mole, we scale up the heat released for 1 mole:
\[ \text{Heat of neutralization} = \frac{x \, \text{kJ}}{0.01 \, \text{mol}} = 100x \, \text{kJ/mol} \]

### Step 5: Indicate the sign of the heat of neutralization
Since the reaction releases heat, the heat of neutralization will be negative:
\[ \text{Heat of neutralization} = -100x \, \text{kJ/mol} \]

### Final Answer
The heat of neutralization (in kJ/mol) is:
\[ \text{Heat of neutralization} = -100x \, \text{kJ/mol} \]

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