The energy released in the neutralisation of `H_(2)SO_(4)` and `KOH` is `59.1 kJ`. Calculated the value `DeltaH` for the reaction
`H_(2)SO_(4)+2KOH rarr K_(2)SO_(4) +2H_(2)O`

Explanation

To calculate the value of ΔH for the reaction:

\[ \text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O} \]

we start with the information given in the question.

### Step 1: Understand the enthalpy of neutralization
The enthalpy of neutralization is defined as the heat change when one equivalent of an acid reacts with one equivalent of a base to form one equivalent of salt and water. In this case, we are given that the energy released in the neutralization of sulfuric acid (H₂SO₄) and potassium hydroxide (KOH) is -59.1 kJ.
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