Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 ( g ) ⟶ 2 SO 3 ( g ) The standard enthalpies of formation for SO 2 ( g ) and SO 3 ( g ) are Δ H ∘ f [ SO 2 ( g ) ] = − 296.8 kJ / mol Δ H ∘ f [ SO 3 ( g ) ] = − 395.7 kJ / mol Calculate the amount of energy in the form of heat that is produced when a volume of 2.67 L of SO 2 ( g ) is converted to 2.67 L of SO 3 ( g ) according to this process at a constant pressure and temperature of 1.00 bar and 25.0 °C . Assume ideal gas behavior.

ΔHorxn = - 11.79 KJ

Explanation:

2 SO 2 ( g ) + O 2 ( g ) ⟶ 2 SO 3 ( g )

The standard enthalpies of formation for SO 2 ( g ) and SO 3 ( g ) are Δ H ∘ f [ SO 2 ( g ) ] = − 296.8 kJ / mol Δ H ∘ f [ SO 3 ( g ) ] = − 395.7 kJ / mol

From the reaction above, 2 mol of SO2 reacts to produce 2 mol of SO3. Assuming ideal gas behaviour,

1 mol = 22.4l

x mol = 2.67l

Upon cross multiplication and solving for x;

x = 2.67 / 22.4 = 0.1192 mol

0.1192 mol of SO2 would react to produce 0.1192 mol of SO3.

Amount of heat is given as;

ΔHorxn = ∑mΔHof(products) − ∑nΔHof(reactants)

Because O2(g) is a pure element in its standard state, ΔHοf [O2(g)] = 0 kJ/mol.

ΔHorxn = 0.1192 mol * (− 395.7 kJ / mol) - 0.1192 mol * ( − 296.8 kJ / mol)

ΔHorxn = - 47.17kj + 35.38kj

ΔHorxn = - 11.79 KJ

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