Question

How to determine the heat evolved or consumed by the reaction of \[1.0g\] \[S{O_2}\] (g) with excess oxygen, with results from a Hess’ Law Equation?

Answer 1

Hint: We need to study Hess’ Law and determine the heat evolved or consumed by the reaction of \[1.0g\] \[S{O_2}\] with excess oxygen. A Russian Chemist and Doctor G. Hess formulated the Hess’ Law which laid the foundations of thermochemistry. It helped to calculate the enthalpy change in all steps of the reaction.

Complete step by step answer:
It is given that the reaction of \[1.0g\] \[S{O_2}\] occurs in excess oxygen. This implies to the fact that $S{O_2}$is the limiting reagent. A limiting reagent or reactant is one which gets consumed and hence there is no reactant left to continue the reaction and hence the reaction stops. According to Hess’s Law of Constant Heat summation, if the reaction takes place in several steps then its Total heat change is the sum of the heat changes of the individual reactions.
Sulphur dioxide reacts with excess oxygen to produce Sulphur trioxide with the evolution of heat whose reaction is as follows:
$2S{O_2}(g) + {O_2}(g) \to 2S{O_3}(g) + 198kJ$
Mass of $S{O_2} = 1g$
Molar mass of $S{O_2} = 64.07g/mol$
Enthalpy $\Delta H = - 198kJ$ for two moles of $S{O_2}$
We are to calculate the $\Delta H$ for \[\;1.0gS{O_2}\]$ = 1 \times \dfrac{{1molS{O_2}}}{{64.07gS{O_2}}} \times \dfrac{{ - 198kJ}}{{2molS{O_2}}}$=$ - 1.54kJ$

Hence the heat evolved or consumed by the reaction of \[1.0gS{O_2}\] with excess oxygen is $ - 1.54kJ/mol$

Note: We must be noted that the sign of $\Delta H$ is negative which indicates that the reaction is exothermic that is, heat is evolved. Also, note that Heat formation or heat change is also known as standard enthalpy change and is independent of the path between initial state (reactants) and final state (products). The combination of Kirchhoff’s Law and Hess’s Law can be used to calculate the formation of heat at different temperatures.