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 :

1. they vary with a higher number because boiling is way hotter than melting

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :