
Given $S+{{O}_{2}}\to S{{O}_{2}};\Delta H=-296.1kJ\,,\,\,2S{{O}_{3}}\to 2S{{O}_{2}}+{{O}_{2}};\Delta H=198.2kJ$, find the $\Delta H$ for the reaction $2S+3{{O}_{2}}\to 2S{{O}_{3}}$.
Answer
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Hint: The concept of Hess Law is to be used in this question. The first two reactions have to be manipulated in such a way so as to obtain the final reaction, and the enthalpies need to be added up.
Complete answer:
In order to answer our question, we need to learn about Hess Law. This law was presented by Hess in 1840. The theory was based on the fact that since internal energy change or the change in enthalpy are functions of the state of the system. So the heat absorbed or released in a particular reaction should be independent of the path in which the reaction is brought out. It only depends on the final state (products) and the initial (reactants) of the system and not the steps or the manner in which change undergoes. This phenomenon is known as Hess's law and may be stated as "if a chemical change is occuring in two or more than two different ways, whether in one step or two or more steps, the amount of total enthalpy change is the same no matter by which method the change is brought about".
Now, let us come to the question, we have to somehow deduce the final reaction from the above two reactions. If we multiply the first equation by 2 we have:
\[2S+3{{O}_{2}}\to 2S{{O}_{3,}}\Delta H=2\times 296.1kJ=592.2kJ\]
Now, we need to reverse the second equation so the sign of the enthalpy will get changed. So, we get the equation as:
\[2S{{O}_{2}}+{{O}_{2}}\to 2S{{O}_{3}};\Delta H=-198.2kJ\]
Now, the final reaction can be deduced by adding up the two reactions, and if the reactions are added up their enthalpies will also get added up and we get the final enthalpy as:
\[\Delta H=592.2+(-198.2)kJ=394kJ\]
So, we get the $\Delta H$ for the final reaction as 394kJ, which is the required answer.
Note:
It is to be noted that the $\Delta H$ sign shows us whether reaction is endothermic or exothermic. S, in this way, we can predict the nature of the final reaction using Hess Law without even carrying out the reaction.
Complete answer:
In order to answer our question, we need to learn about Hess Law. This law was presented by Hess in 1840. The theory was based on the fact that since internal energy change or the change in enthalpy are functions of the state of the system. So the heat absorbed or released in a particular reaction should be independent of the path in which the reaction is brought out. It only depends on the final state (products) and the initial (reactants) of the system and not the steps or the manner in which change undergoes. This phenomenon is known as Hess's law and may be stated as "if a chemical change is occuring in two or more than two different ways, whether in one step or two or more steps, the amount of total enthalpy change is the same no matter by which method the change is brought about".
Now, let us come to the question, we have to somehow deduce the final reaction from the above two reactions. If we multiply the first equation by 2 we have:
\[2S+3{{O}_{2}}\to 2S{{O}_{3,}}\Delta H=2\times 296.1kJ=592.2kJ\]
Now, we need to reverse the second equation so the sign of the enthalpy will get changed. So, we get the equation as:
\[2S{{O}_{2}}+{{O}_{2}}\to 2S{{O}_{3}};\Delta H=-198.2kJ\]
Now, the final reaction can be deduced by adding up the two reactions, and if the reactions are added up their enthalpies will also get added up and we get the final enthalpy as:
\[\Delta H=592.2+(-198.2)kJ=394kJ\]
So, we get the $\Delta H$ for the final reaction as 394kJ, which is the required answer.
Note:
It is to be noted that the $\Delta H$ sign shows us whether reaction is endothermic or exothermic. S, in this way, we can predict the nature of the final reaction using Hess Law without even carrying out the reaction.
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