Given the enthalpies of reaction S(s)+O2(g)→SO2(g) ΔH = -297 kJ

2S(s)+3O2(g)→2SO3(g) ΔH=-791kJ

Calculate the enthalpy change (ΔH) for the following reaction using the data above and Hess's Law:

2SO2+O2⟶2SO3

2SO₂+O₂⇒2SO₃  ΔH=-197 kJ

Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways

Reaction

2SO₂+O₂⇒2SO₃

Given :

1. S(s)+O₂(g)→SO₂(g) ΔH = -297 kJ

Reverse

SO₂(g) ⇒S(s)+O₂(g ΔH = +297 kJ (sign change to +) x 2

2SO₂(g) ⇒2S(s)+2O₂(g ΔH =  +594 kJ

2.2S(s)+3O2(g)→2SO3(g) ΔH=-791kJ

Add both reactions and remove/subtract the same compound for different sides

1. 2SO₂(g) ⇒2S(s)+2O₂(g) ΔH =  +594 kJ

2.2S(s)+3O₂(g)→2SO₃(g) ΔH=-791kJ

+

2SO₂+O₂⇒2SO₃  ΔH=-197 kJ

c8h10 is your answerr

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.