contestada

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

Respuesta :

ardni313

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) :

[tex]\tt mol=\dfrac{0.95}{32}=0.03[/tex]

For 2 mol S, the entalphy = -790kJ

The enthalpy change for 0.03 mol :

[tex]\tt \dfrac{0.03}{2}\times -790~kJ=-11.85~kJ[/tex]

Cricetus

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

Given:

Value of ΔH°,

  • -790 kJ

According to the question,

→ [tex]2S(s) +3O_2(g) =2SO_3(g)[/tex]

Mol of S will be:

= [tex]\frac{0.95}{32}[/tex]

= [tex]0.03[/tex]

hence,

The enthalpy change for 0.03 mol will be:

= [tex]\frac{0.03}{2}\times -790[/tex]

= [tex]-11.85 \ kJ[/tex]

Thus the answer above is correct.

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https://brainly.com/question/13799877

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