Calculate enthalpy changes for the following: a. 0.041 g of sulfur (rhombic) burns, forming SO2(g) (ΔH° for SO2(g) = -296.84 kJ/mol) Enthalpy change kJ
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041 g. To do this, we can use the molar mass of sulfur, which is 32.06 g/mol. Number of moles of S = mass of S / molar mass of S Number of moles of S = 0.041 g / 32.06 g/mol Show more…
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From these data, S(rhombic) + O2(g) → SO2(g) ΔH°rxn = −296.06 kJ/mol S(monoclinic) + O2(g) → SO2(g) ΔH°rxn = −296.36 kJ/mol calculate the enthalpy change for the transformation S(rhombic) → S(monoclinic) (Monoclinic and rhombic are different allotropic forms of elemental sulfur.)
Sri K.
From the following enthalpy changes: 1) 3/2 O2 (g) + SO2 (g) -> SO3 (g) ΔH = -395.2 kJ 2) 2 SO2 (g) + O2 (g) -> 2 SO3 (g) ΔH = -198.2 kJ Calculate the value of ΔH for the reaction S (s) + O2 (g) -> SO2 (g) using Hess's law of Heat Summation.
Adi S.
Calculate the enthalpy change for the reaction $$\mathrm{P}_{4} \mathrm{O}_{6}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)$$ given the following enthalpies of reaction: $$\begin{array}{ll}{\mathrm{P}_{4}(s)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{6}(s)} & {\Delta H=-1640.1 \mathrm{kJ}} \\ {\mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)} & {\Delta H=-2940.1 \mathrm{kJ}}\end{array}$$
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