Enthalpy change for the reaction
`H^(+)(aq)+OH^(-)(aq)toH_(2)O(l)` is `-57.3kJ"mol"^(-1)` If 1 L of 0.5 `M HNO_(3)` is mixed with 2 L of 0.2 M KOH, the enthalpy change would be

Options

A. `-28.65kJ`
B. `28.65kJ`
C. `-22.92kJ`
D. `11.46kJ`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of H⁺ from HNO₃
Given:
- Concentration of HNO₃ = 0.5 M
- Volume of HNO₃ = 1 L

Using the formula:
\[ \text{Number of moles} = \text{Concentration} \times \text{Volume} \]
\[ \text{Number of moles of H}^+ = 0.5 \, \text{mol/L} \times 1 \, \text{L} = 0.5 \, \text{mol} \]

### Step 2: Calculate the number of moles of OH⁻ from KOH
Given:
- Concentration of KOH = 0.2 M
- Volume of KOH = 2 L

Using the same formula:
\[ \text{Number of moles of OH}^- = 0.2 \, \text{mol/L} \times 2 \, \text{L} = 0.4 \, \text{mol} \]

### Step 3: Identify the limiting reagent
The reaction is:
\[ \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} \]

From our calculations:
- Moles of H⁺ = 0.5 mol
- Moles of OH⁻ = 0.4 mol

Since OH⁻ is present in fewer moles, it is the limiting reagent.

### Step 4: Calculate the enthalpy change for the reaction
The enthalpy change for the neutralization reaction is given as:
\[ \Delta H = -57.3 \, \text{kJ/mol} \]

Since only 0.4 mol of OH⁻ will react (the limiting reagent), we calculate the enthalpy change for this amount:
\[ \text{Enthalpy change} = \Delta H \times \text{moles of limiting reagent} \]
\[ \text{Enthalpy change} = -57.3 \, \text{kJ/mol} \times 0.4 \, \text{mol} = -22.92 \, \text{kJ} \]

### Final Answer
The enthalpy change for the reaction when mixing 1 L of 0.5 M HNO₃ with 2 L of 0.2 M KOH is approximately:
\[ \boxed{-22.9 \, \text{kJ}} \]