The amount of heat evolved when `500 cm^(3) 0.1 M HCl` is mixed with `200 cm^(3)` of `0.2 M NaOH` is

Options

A. 2.292kJ
B. 1.292kJ
C. 0.292kJ
D. 3.392kJ

Answer

To solve the problem of calculating the amount of heat evolved when 500 cm³ of 0.1 M HCl is mixed with 200 cm³ of 0.2 M NaOH, we will follow these steps:

### Step 1: Calculate the number of millimoles of HCl
We know that:
\[ \text{Millimoles of HCl} = \text{Volume (cm³)} \times \text{Molarity (M)} \]
For HCl:
\[ \text{Millimoles of HCl} = 500 \, \text{cm}^3 \times 0.1 \, \text{M} = 50 \, \text{mmol} \]

### Step 2: Calculate the number of millimoles of NaOH
Similarly, for NaOH:
\[ \text{Millimoles of NaOH} = 200 \, \text{cm}^3 \times 0.2 \, \text{M} = 40 \, \text{mmol} \]

### Step 3: Identify the limiting reagent
In the neutralization reaction between HCl and NaOH, the reaction can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
From the calculations:
- HCl provides 50 mmol of H⁺ ions.
- NaOH provides 40 mmol of OH⁻ ions.

Since NaOH (40 mmol) is less than HCl (50 mmol), NaOH is the limiting reagent.

### Step 4: Calculate the heat evolved from the reaction
The enthalpy change (ΔH) for the neutralization of a strong acid with a strong base is approximately -57.3 kJ/mol. Since we have 40 mmol of NaOH reacting, we need to convert this to moles:
\[ \text{Moles of NaOH} = 40 \, \text{mmol} = 40 \times 10^{-3} \, \text{mol} \]
Now, we can calculate the heat evolved:
\[ \text{Heat evolved} = \Delta H \times \text{moles of NaOH} \]
\[ \text{Heat evolved} = -57.3 \, \text{kJ/mol} \times 40 \times 10^{-3} \, \text{mol} = -2.292 \, \text{kJ} \]

### Final Answer
The amount of heat evolved when 500 cm³ of 0.1 M HCl is mixed with 200 cm³ of 0.2 M NaOH is approximately **-2.29 kJ**.

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