The enthalpy of neutralisation of 0.4 M `H_(2)SO_(4)` will be how many times the enthalpy of neutralisation of 0.1 M HCl ?

Answer

To determine how many times the enthalpy of neutralization of 0.4 M H₂SO₄ is compared to that of 0.1 M HCl, we can follow these steps:

### Step 1: Understand the concept of enthalpy of neutralization
The enthalpy of neutralization is the heat change that occurs when an acid reacts with a base to form water and a salt. For strong acids and bases, this value is generally consistent and can be approximated as -55.84 kJ/mol for the reaction of one mole of H⁺ with one mole of OH⁻.

### Step 2: Calculate the effective concentration of H⁺ ions in H₂SO₄
H₂SO₄ is a diprotic acid, meaning it can donate two protons (H⁺ ions) per molecule. Therefore, a 0.4 M solution of H₂SO₄ will provide:
\[ \text{Concentration of H}^+ = 0.4 \, \text{M} \times 2 = 0.8 \, \text{M} \]

### Step 3: Calculate the enthalpy of neutralization for H₂SO₄
Using the standard enthalpy of neutralization for strong acids, we can calculate the enthalpy change for the neutralization of 0.8 M H⁺ ions:
\[ \Delta H_{\text{neutralization, H₂SO₄}} = -55.84 \, \text{kJ/mol} \times 0.8 \, \text{mol} = -44.672 \, \text{kJ} \]

### Step 4: Calculate the enthalpy of neutralization for HCl
For HCl, which is a strong acid that provides 0.1 M of H⁺ ions, the enthalpy of neutralization can be calculated as:
\[ \Delta H_{\text{neutralization, HCl}} = -55.84 \, \text{kJ/mol} \times 0.1 \, \text{mol} = -5.584 \, \text{kJ} \]

### Step 5: Compare the two enthalpies of neutralization
To find how many times the enthalpy of neutralization of H₂SO₄ is compared to HCl, we divide the two values:
\[ \text{Ratio} = \frac{\Delta H_{\text{neutralization, H₂SO₄}}}{\Delta H_{\text{neutralization, HCl}}} = \frac{-44.672 \, \text{kJ}}{-5.584 \, \text{kJ}} \]
Calculating this gives:
\[ \text{Ratio} \approx 8 \]

### Conclusion
The enthalpy of neutralization of 0.4 M H₂SO₄ is approximately 8 times that of 0.1 M HCl.

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