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

The enthalpy change accompanying the reaction will be "-11.85 kJ".

Given:

Value of ΔH°,

According to the question,

→

Mol of S will be:

=

=

hence,

The enthalpy change for 0.03 mol will be:

=

=

Thus the answer above is correct.

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

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) :

For 2 mol S, the entalphy = -790kJ

The enthalpy change for 0.03 mol :

i think that the answer is c