Question

The value of ΔH° for the reaction below is -790 kJ. The enthalpy change accompanying the reaction of 0.95 g of S is __________ kJ.
23 2S (s) + 3O (g) 2SO (g

Answer 1

The enthalpy change accompanying the reaction of 0.95 g of S is -11.85 kJ

Further explanation

Given

ΔH° for the reaction is -790 kJ

Required

The enthalpy change

Solution

Reaction

2S (s) + 3O₂(g) = 2SO₃ (g)

mol of S(MW=32 g/mol) :

$\tt mol=\dfrac{0.95}{32}=0.03$

For 2 mol S, the entalphy = -790kJ

The enthalpy change for 0.03 mol :

$\tt \dfrac{0.03}{2}\times -790~kJ=-11.85~kJ$