Question

Calculate the total enthalpy change that occurs when 85.00 grams of H₂S (g) reacts with 45.00 grams of O₂ (g) according to the equation below.
8 H₂S (g) + 4 O₂ (g) → S₈ (g) + 8 H₂O (g)

Answer 1

The reaction is

8H2S + 4O2 -------> S8 + 8H2O

We know enthalpy of elements in thier standard state is zero. thus enthalpy of oxygen and sulphur is zero.

delta H^oformation of H2S=-20.63kJ/mol

delta H^oformation of H2O(g)= -241.82kJ/mol

Thus heat of given reaction

delta H reaction = (8x delta H water ) - ( 8x delta H of H2s)

= (- 8 x241.82 ) - (--8x20.63) kJ

= - 1,769.52 kJ

According to the equation 8 moles of H2S reacts with 4 moles of oxygen

Thus 8x 34 g of H2S reacts with 4x32 g of oxygen

then 85g of H2S requires = (85x4x32 ) / (8x34 ) = 40g

Ths H2s is the limiting reagent that is consumed in the reaction completely.

Therefore the heat of reaction is calculated according to H2s

8x 34 g O H2S on reaction gave -1769.52kJ

85 g gives = 85x1769.52 / (8x34) = -552.975 kJ