Hydrogen sulfide, H2S, is a foul-smelling gas. It burns to form sulfur dioxide. 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g) ΔH = -1036 kJ. Calculate the enthalpy change to burn 30.9 g of hydrogen sulfide.
Added by Matthew R.
Step 1
First, we need to find the molar mass of H2S. The molar mass of hydrogen is approximately 1 g/mol and the molar mass of sulfur is approximately 32 g/mol. Therefore, the molar mass of H2S is (1*2) + 32 = 34 g/mol. Show more…
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Hydrogen sulfide, $\mathrm{H}_{2} \mathrm{~S}$, is a poisonous gas with the odor of rotten eggs. The reaction for the formation of $\mathrm{H}_{2} \mathrm{~S}$ from the elements is $$ \mathrm{H}_{2}(g)+\frac{1}{8} \mathrm{~S}_{8}(\text { rhombic }) \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g) $$ Use Hess's law to obtain the enthalpy change for this reaction from the following enthalpy changes: $$ \begin{array}{c} \mathrm{H}_{2} \mathrm{~S}(g)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{SO}_{2}(g) ; \Delta H=-518 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H=-242 \mathrm{~kJ} \\ \frac{1}{8} \mathrm{~S}_{8}(\text { rhombic })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) ; \Delta H=-297 \mathrm{~kJ} \end{array} $$
Hydrogen sulfide gas is a poisonous gas with the odor of rotten eggs. It occurs in natural gas and is produced during the decay of organic matter, which contains sulfur. The gas burns in oxygen as follows: $$ 2 \mathrm{H}_{2} \mathrm{~S}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{SO}_{2}(g) $$ Calculate the standard enthalpy change for this reaction using standard enthalpies of formation.
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