Question

The following are the thermochemical equations for the

combination of oxygen with sulfur to form sulfur dioxide and for the
combination of oxygen with sulfur to form sulfur trioxide:
S(s) + O₂(g) →→→SO₂(g)
AH = -296.8 kJ
2S(s) + 302(g)2SO3(g)
AH = -791.4 kJ
Using the above thermochemical equations, and applying Hess' Law,
determine AH for the following reaction:
2SO2(g) + O2(g) 2SO3(g) AH = ?

Answer 1

Using Hess's Law, you can determine the enthalpy change (ΔH) for the formation of sulfur trioxide from sulfur dioxide and oxygen by reversing and scaling the equation for the formation of sulfur dioxide, then adding it to a scaled equation for the formation of sulfur trioxide, which gives ΔH for the desired reaction as -198.4 kJ.

Step-by-step explanation:

The calculation of ΔH for the reaction 2SO2(g) + O2(g) ⇒ 2SO3(g) involves using Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for any set of reactions that leads to the final reaction.

From the provided equations:

• S(s) + O2(g) ⇒ SO2(g), ΔH = -296.8 kJ
• 2S(s) + 3O2(g) ⇒ 2SO3(g), ΔH = -791.4 kJ

To find the ΔH for 2SO2(g) + O2(g) ⇒ 2SO3(g), we need to manipulate the original equations so that when added together, they give the desired reaction.

The first reaction represents the formation of SO2 and needs to be reversed and multiplied by 2 in order to cancel SO2 on the reactant side:

• 2SO2(g) ⇒ 2S(s) + 2O2(g), ΔH = +2(× -296.8 kJ)
• = +593.6 kJ

The second equation is already in the correct direction but needs to be divided by 2:

• 2S(s) + 3O2(g) ⇒ 2SO3(g), ΔH = × ½ (× -791.4 kJ) = -395.7 kJ

By adding these modified equations, we cancel out the S(s) and O2(g) to obtain the desired reaction:

2SO2(g) + O2(g) ⇒ 2SO3(g), ΔH

= -395.7 kJ + (+593.6 kJ)

= -198.4 kJ