Question

a) Determine the mass (in grams) of bromine that will result in a change in energy of -2.58 kJ when bromine reacts with excess potassium iodide according to the following balanced thermochemical equation:

2KI(s) + Br₂(l) → 2KBr(s) + I₂(s)  ΔH_r° = -131.80 kJ

b) Using the enthalpies of formation given below, Calculate the amount of heat absorbed/released (in kJ) when 8.39 grams of SO₂ are produced via the above reaction.

2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(l)



H₂S (g): -20.60 kJ/mol

O₂ (g): 0.00 kJ/mol

SO₂ (g): -296.84 kJ/mol

H₂O (l): -285.83 kJ/mol

Answer 1

a) 131.80 kJ energy released when 159.8 g of Bromine reacts

   2.58 kJ energy released when (159.8 X 2.58)/131.8 = 3.13 g of Bromine reacts.

b) ΔH_r° = ΔH products - ΔH reactants

             = [(2 X -296.84) + (2 X -285.83)] - [2 X -20.60]

             = (593.68 + 571.66) - (41.2)

             = 1124.14 KJ

128 g of SO2 releases 1124.14 KJ then 8.39 g of SO2 releases 73.68 KJ