Question 9.7: Sulfur trioxide reacts with water to form sulfuric acid, a m......
Sulfur trioxide reacts with water to form sulfuric acid, a major contributor to acid rain. One origin of SO_3 is the combustion of sulfur, which is present in small quantities in coal, according to the following equation.
\mathrm{S}(s)+{\frac{3}{2}}\,\mathrm{O}_{2}(\mathrm{g})\to\mathrm{SO}_{3}(\mathrm{g})
Given the thermochemical information below, determine the heat of reaction for this reaction.
S(s) + O_2(g) → SO_2(g) ΔH° = –296.8 kJ
2 SO_2(g) + O_2(g) → 2 SO_3(g) ΔH° = –197.0 kJ
Strategy We need to use the reactions whose enthalpy changes are known to construct a pathway that results in the desired reaction. In this case, we can see that SO_2 is formed in the first reaction we are given and then is consumed in the second. It is critical to pay attention to the precise stoichiometry of the given and desired reactions. The reaction of interest produces just one mole of SO_3, whereas the second reaction we are given produces two moles. So we will need to correct for that.
We start with the first of the two given reactions.
{\mathrm{S}(s)+O_{2}(\mathrm{g})\longrightarrow\mathrm{S}\mathrm{O}}_{2}(\mathrm{g})\qquad\Delta H^{\circ}=-2 96.8\mathrm{~kJ}
Next we multiply the second given reaction by one-half. This will account for the fact that the desired reaction produces only one mole of SO_3.
\frac{1}{2}\times[2\;\mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\rightarrow2\;\mathrm{SO}_{3}(\mathrm{g})\qquad\qquad \Delta H^{\circ} = −197.0\text{ kJ}]
This gives an equation in which one mole of SO_2 is consumed, along with its enthalpy change.
\mathrm{SO}_{2}(\mathrm{g})+\frac{1}{2}\:\mathrm{O}_{2}(\mathrm{g})\to\mathrm{SO}_{3}(\mathrm{g})\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Delta H^{\circ}=-98.5\:\mathrm{kJ}
Adding this to the first reaction above gives the desired amount of SO_3 and the heat of reaction:
Analyze Your Answer We may not have any intuition regarding the particular chemical reactions involved, so we’ll have to look at the structure of the problem to see if our answer makes sense. Both of the reactions we added are exothermic, so it makes sense that the combination would be more exothermic than either of the individual reactions.
Check Your Understanding Use the following thermochemical equations as needed to find the heat of formation of diamond:
C(diamond) + O_2(g) → CO_2(g) ΔH° = –395.4 kJ
2 CO_2(g) → 2 CO(g) + O_2(g) ΔH° = 566.0 kJ
C(graphite) + O_2(g) → CO_2(g) ΔH° = –393.5 kJ
2 CO(g) → C(graphite) + CO_2(g) ΔH° = –172.5 kJ