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Question: The energy released in the neutralization of \({H_2}S{O_4}\) and \(KOH\)is \( - 59.1kJ\). Calculate ...

The energy released in the neutralization of H2SO4{H_2}S{O_4} and KOHKOHis −59.1kJ - 59.1kJ. Calculate the value of ΔH\Delta H for the reaction.
H2SO4+2KOH→K2SO4+2H2O{H_2}S{O_4} + 2KOH \to {K_2}S{O_4} + 2{H_2}O
A.ΔH=−118.2kJ\Delta H = - 118.2kJ
B.ΔH=+118.2kJ\Delta H = + 118.2kJ
C.ΔH=−236.4kJ\Delta H = - 236.4kJ
D.None of these

Explanation

Solution

The given value of −59.1kJ - 59.1kJ is for the neutralization of H2SO4{H_2}S{O_4}and KOHKOH is the energy released in the simplest reaction, that is with 1mol1mol of KOHKOH. Neutralization reaction gives water and salt.
0.5H2SO4+KOH→0.5K2SO4+H2O ΔH = - 59.1kJ0.5{H_2}S{O_4} + KOH \to 0.5{K_2}S{O_4} + {H_2}O{\text{ }}\Delta {\text{H = - 59}}{\text{.1kJ}}

Complete step by step answer:
We have to understand that after neutralization, a solution is said to be neutral as no excess hydrogen ions are left and the value is  7 \,7\,. The calculation of the amount of heat evolved or consumed is concerned with thermochemistry. The heat of neutralisation is the heat that forms when an acid and a base react to form a salt plus water.
The given energy released is for the neutralization of 1mol1mol of KOHKOH;
0.5H2SO4+KOH→0.5K2SO4+H2O ΔH = - 59.1kJ0.5{H_2}S{O_4} + KOH \to 0.5{K_2}S{O_4} + {H_2}O{\text{ }}\Delta {\text{H = - 59}}{\text{.1kJ}}
The required reaction is,
H2SO4+2KOH→K2SO4+2H2O{H_2}S{O_4} + 2KOH \to {K_2}S{O_4} + 2{H_2}O
So by multiplying the reaction of 1mol1mol KOHKOH by 22 we will get the required reaction.
So do the same for its enthalpy too, that is −59.1×2=−118.2kJ - 59.1 \times 2 = - 118.2kJ
So the right answer is option A.

Additional information:
Enthalpy of neutralization is the enthalpy change when an acid reacts with base to form water and its salt.
If enthalpy for each compound is given, the total enthalpy change can be calculated using,
ΔH=ΔH(of all the compound in product) - ΔH ( of all the compound in reactant)\Delta H = \Delta H({\text{of all the compound in product) - }}\Delta {\text{H ( of all the compound in reactant)}}
In a special vessel called a calorimeter, heat tests are carried out by carrying out the reaction. The reaction solution and the calorimeter absorb the heat given off by the neutralisation reaction. Owing to the absorbed heat, both the sample and the calorimeter rise in temperature and this rise can be calculated using a thermometer.

Note:
The energy is released during this neutralization process implies that it is an exothermic reaction, that is products are more stable than reactants. For an endothermic reaction, energy is absorbed and products are less stable than reactants.