Question

please show me how to get the correct answers in preperation for my upcoming exam

22.) E 23.) D

22) During volcanic eruptions, hydrogen sulfide gas is given off and oxidized by air according to the following chemical equation: 2H2S(g) + 302(g) - 2802(g) + 2H2O(g) Calculate the standard enthalpy change for the above reaction given: 3S(s) + 2H2O(g) → 2H2S(g) + SO2(g) AH° = 146.9 kJ/mol S(s) + O2(g) → SO2(g) AH°= -296.4 kJ/mol A) 443.3 kJ/mol B) 742.3 kJ/mol c) - 742.3 kJ/mol D) -149.5 kJ/mol E) -1036.1 kJ/mol 23) Given the data in the table below, AHºrxn for the reaction 3NO 2 (g) + H20 (1) 2HNO3 (aq) + NO (8) Substance AH® (kJ/mol) H20 (1) -286 NO (8) 90 NO2 (8) 34 HNO3 (aq) -207 NH3 (8) - 46 A) -64 B) 64 C) -508 D) -140 E) 140

Answer 1

22.

Given that

2H2S +3O2 = 2SO2 +2H2O   dH=?

3S + 2H2O = 2H2S +SO2     dH = 146.9 kj/ mole -----1

S+ O2 = SO2   dH -296.4 kj/ mole    -----2

First reverse eq 1 and eq 2 multiply by 3 and add them as follows:

2H2S +SO2     = 3S + 2H2O dH = -146.9 kj/ mole

3S+ 3O2 =3 SO2   dH -889.2 kj/ mole

===============================

2H2S +3O2 = 2SO2 +2H2O   dH=-1036.1 KJ/ mole

23

Given that

3NO2 + H2O = 2HNO3 +NO

dH reaction = dH products - dH reactants

= 2 *(-207)+ 90 –[ 3*34 + -286] KJ/ mole

=(-414+90-102+286 ) KJ/ mole

=-140 KJ/ mole