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Question: The energy released in the neutralisation of \({{H}_{2}}S{{O}_{4}}\) and \(KOH\) is -59.1kJ. Calcula...

The energy released in the neutralisation of H2SO4{{H}_{2}}S{{O}_{4}} and KOHKOH is -59.1kJ. Calculate the 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=−436.2kJ\Delta H=-436.2kJ
B. ΔH=+118.2kJ\Delta H=+118.2kJ
C. ΔH=−220.2kJ\Delta H=-220.2kJ
D. ΔH=−118.2kJ\Delta H=-118.2kJ

Explanation

Solution

This question requires the concept of manipulating the thermochemical equations in such a manner that the required equation gets obtained. As the equations change, the enthalpy changes accordingly.

Complete Solution :
In order to answer our question, we need to learn about the enthalpy. Enthalpies are of different types. Some of them are:
Standard enthalpy of reaction: As the enthalpy of a reaction varies with temperature, therefore, for the sake of comparison the values change in enthalpies for different reactions need to be expressed at their standard state conditions. The standard enthalpy of the system is "the enthalpy change for a reaction when all the participating substances are in their standard states".

- Phase transformation enthalpy: The three states of matter - Solid, liquid and gas differ from one another in the arrangement of their constituent particles. The magnitudes of intermolecular forces acting between the particles in these states are also different. It is a common observation that when solid is converted into the liquid state, energy is to be supplied. This energy is used in breaking the intermolecular forces in the solid which are of high magnitude. Whenever there is a change in the state of matter (solid liquid or from liquid gas), the process is called phase change or transition. It is also achieved by the change in enthalpy or heat content of the system.

- Enthalpy of neutralisation: It is defined as the heat evolved or decreased in enthalpy when 1 g equivalent of an acid is neutralised by 1g equivalent of a base in dilute solution.
Apart from these 3, there are other enthalpies too, for different conditions accordingly.
When a reaction is multiplied or divided by a constant, then the enthalpy also gets multiplied or divided respectively, by the same constant. Now, balancing the reaction of H2SO4{{H}_{2}}S{{O}_{4}}and KOHKOHwe have:
12H2SO4+KOH→12K2SO4+H2O,   ΔH=−59.1kJ/mol\dfrac{1}{2}{{H}_{2}}S{{O}_{4}}+KOH\to \dfrac{1}{2}{{K}_{2}}S{{O}_{4}}+{{H}_{2}}O,\,\,\,\Delta H=-59.1kJ/mol

- Now, if we multiply the whole equation by 2, to obtain the required equation, we have:
H2SO4+2KOH→K2SO4+2H2O,   ΔH=2×(−59.1)= −118.2kJ/mol{{H}_{2}}S{{O}_{4}}+2KOH\to {{K}_{2}}S{{O}_{4}}+2{{H}_{2}}O,\,\,\,\Delta H=2\times (-59.1)=\,-118.2kJ/mol
So, we obtain ΔH= −118.2kJ/mol\Delta H=\,-118.2kJ/mol,
So, the correct answer is “Option D”.

Note: It is to be noted that when we talk about enthalpy, we always talk about the enthalpy difference. The absolute enthalpy cannot be found out as the origin or zero point cannot be reached.