Question

3. During volcanic eruptions, hydrogen sulfide gas is given off and oxidized by air according to the following chemical equation: 2H2S + 302 (→ 2502 (+2H30 A Hx? Calculate the standard enthalpy change for the above reaction (AH "working equations" (4 pts) (see pages 136-137): ) using the following 35. + 2H: 0 2H:S + SO2001 AH° - 146.9 kJ/mol Sa+02 → SO2 AH° = -296.4 kJ/mol

Answer 1

Consider the equation

35(s) + 2H2O(g) + 2H25(9) + Sozlg) AH = 146.9 kJ/mol

Multiply it with -1 we get

-35(S) – 2H2O(g) +-2H25(9) - S02(g) AH° = –146.9 kJ/mol
i.e.
2H2S(9) + S02(g) +35(s) + 2H2O(9) AH = -146.9 kJ/mol
....(1)

Now consider

Multiply this with 3, we get

S(s) + O2(g) → SO2(g) AH = –296.4 kJ/mol

​​​​​​.....(2)

35(s)+302(g) + 3502(9) AH = 3x(-296.4 kJ/mol) = -889.2 kJ/mol

Now add (1) and (2)

2H25(9)+S02(9)+33(s)+302(g) + 3S(s)+2H2O(g)+3502(9) -146.9 kJ/mol - 889.2 kJ/mol AH° =
Cancel common terms on both sides (3S(s)) and take SO​​​​​2(g) to right side
2H2S(9)+302(g) → 2H2O(g) +3502(9) - S02(9) AH° = –1036.1 kJ/mol

i.e. This is the required equation and standard enthalpy change for the required reaction=-1036.1 kJ/mol

2H2S(g) + 302(g) → 2H2O(g) +2502(9) AH° = -1036.1 kJ/mol