Question

Give the following thermochemical equations:

& {{S}_{8}}(s)+8{{O}_{2}}(g)\to 8S{{O}_{2}}(g)\,\,\,\,\Delta H=-2374.6kJ \\\ & {{S}_{8}}(s)+12{{O}_{2}}(g)\to 8S{{O}_{3}}(g)\,\,\,\Delta H=-3165.8kJ \\\ \end{aligned}$$ Find the value of $\Delta H$ for the reaction ${{O}_{2}}(g)+2S{{O}_{2}}(g)\to 2S{{O}_{3}}(g)$. A. -5540.4kJ B. -1385.1kJ C. -791.2kJ D. -197.8kJ E. -700kJ

Answer 1

Try to use the Hess Law and manipulate the above equations to obtain the required chemical equation. The enthalpies will get added, subtracted, multiplied or divided, for finding resultant enthalpy.

Complete step by step solution:
In order to answer our question, we need to know about the enthalpy and heat. Let us learn about Hess Law in thermodynamics. This law was presented by Hess in 1840. The theory was based on the fact that since internal energy change ($\Delta U$) or the change in enthalpy ($\Delta H$) are functions of the state of the system. So the heat evolved or absorbed in a given reaction must be independent of the path in which the reaction is brought upon. It only depends on the initial (reactants) and the final state (products) of the system and not the manner or the steps in which change takes place. This generalisation is known as Hess's law and may be stated as "if a chemical change is made to take place in two or more than two different ways, whether in one step or two or more steps the amount of total enthalpy change is same no matter by which method the change is brought about".
Let a substance A be changed to C in two steps involving a change in enthalpy at each step.

& A\to B+{{\Delta }_{r}}{{H}_{1}} \\\ & B\to C+{{\Delta }_{r}}{{H}_{2}} \\\ \end{aligned}$$ So, total enthalpy change from A to C will be ${{\Delta }_{r}}{{H}_{steps}}=\Delta {{H}_{1}}+\Delta {{H}_{2}}$. There are some properties which need to be remembered: i.When a reaction is reversed, the sign of $\Delta H$changes. ii. When a reaction is multiplied or divided with a constant, the $\Delta H$also gets multiplied or divided accordingly. Now, let us try to manipulate the equation. On reversing the first equation in the question, we have: $$8S{{O}_{2}}(g)\to {{S}_{8}}(s)+8{{O}_{2}}(g)\,\,\,\,\,\,\Delta H=2374.6kJ$$…….(1) Adding this with the second equation in the question, the enthalpies also get added up and we have: $$\begin{aligned} & 8S{{O}_{2}}(g)+4{{O}_{2}}(g)\to 8S{{O}_{3}}(g) \\\ & \Delta H=(2374.6-3165.8)kJ=\,-791.26kJ \\\ \end{aligned}$$ Now, in order to obtain the desired equation, we will divide the above equation y 4, the enthalpy i'll also get divided. So, we have: $$\begin{aligned} & 2S{{O}_{2}}(g)+{{O}_{2}}(g)\to 2S{{O}_{3}}(g) \\\ & \Delta H=\dfrac{-791.26}{4}kJ=-197.8kJ \\\ \end{aligned}$$ **So, we obtain the value of $\Delta H$ of the reaction as -197.8 kJ which gives us option D as the correct answer.** **Note:** It is to be noted that enthalpy is a state function. Moreover, when we talk about enthalpy, it always represents the change of enthalpy $\Delta H$. It is so because absolute zero temperatures can be found on the kelvin scale, so there is no origin or zero point to measure absolute enthalpy.