What would be the heat released when:

Options

A. `0.5` mol of `HCI` is neutralised with `0.5` mol of `NaOH`
B. `0.5` mol of `HNO_(3)` is neutralised with `0.3` mol of `NaOH`
C. `100 ml` of `0.2M HCI` + `200 ml` of `0.2 M KOH`
D. `200 ml` of `0.1 M H_(2)SO_(4)` + `150 ml` of `0.2 M KOH`

Explanation

To determine the heat released in each part of the question regarding the neutralization reactions, we will follow a systematic approach. The heat released during the neutralization of a strong acid with a strong base is given by the formula:

\[ Q = n \times \Delta H^\circ_{neutralization} \]

where:
- \( Q \) is the heat released,
- \( n \) is the number of moles of the limiting reactant,
- \( \Delta H^\circ_{neutralization} \) is the standard enthalpy of neutralization, which is approximately -57.1 kJ/mol for strong acids and bases.
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