Question

1.Using the enthalpies of formation given below, calculate ΔH°_rxn in kJ, for the following reaction.

Report your answer to two decimal places in standard notation.



H₂S(g) + 2O₂(g) → SO₃(g) + H₂O(l)



H₂S (g): -20.60 kJ/mol

O₂ (g): 0.00 kJ/mol

SO₃ (g): -395.77 kJ/mol

H₂O (l): -285.83 kJ/mol

2. Calculate the amount of heat absorbed/released (in kJ) when 22.54 grams of SO₃ are produced via the above reaction.

Report your answer to two decimal places, and use appropriate signs to indicate heat flow direction.

Answer 1

1) Given reaction is H2S(g) + 2O2(g) → SO3(g) + H2O(l)

ΔH°rxn = ΔH_f^o(products) - ΔH_f^o( reactants)

= ΔH_f^o [SO3(g)] + ΔH_f^o [H₂O(l)] - { ΔH_f^o [H2S(g)]+ 2ΔH_f^o [O2(g)]}

=  -395.77 kJ/mol + (-285.83 kJ/mol) - {-20.60 kJ/mol + 2 x0.00 kJ/mol }

= -661 kJ/mol

Therefore, ΔH°rxn = -661 kJ/mol

2)

Given reaction is H2S(g) + 2O2(g) → SO3(g) + H2O(l)   ΔH°rxn = -661 kJ/mol   

ΔH°rxn = -ve, it means heat energy is released.

H2S(g) + 2O2(g) → SO3(g) + H2O(l)   ΔH°rxn = -661 kJ/mol     

80 g 661 kJ energy is released.

  22.54 g ?

? = (22.54 g / 80 g ) x 661 kJ energy is released

= 186.24 kJ energy is released

Therefore,

186.24 kJ heat energy is released when 22.54 grams of SO3 are produced .