Question

Given the following standard enthalpy change at 298 K:

S(s)+O₂​(g)→SO₂​(g),δH=−296.8kJ mol⁻¹
2SO₂​(g)+O₂​(g)→2SO₃​(g),δH=−197.8kJ mol⁻¹

Calculate the standard enthalpy change for the reaction: S(s)+23/2O₂​(g)→SO₃(g),δH=?
a. δH=−99.0kJ mol⁻¹
b. δH=−198.0kJ mol⁻¹
c. δH=−297.0kJ mol⁻¹
d. δH=−395.6kJ mol⁻¹

Answer 1

To calculate the standard enthalpy change for the reaction S(s) + 23/2O₂​(g) → SO₃(g), we need to modify and combine the given reactions. The standard enthalpy change is -692.5 kJ mol⁻¹.

Step-by-step explanation:

To calculate the standard enthalpy change for the reaction S(s) + 23/2O₂​(g) → SO₃(g), we need to use the given enthalpy changes for the two reactions:

1. S(s) + O₂​(g) → SO₂​(g), δH = -296.8 kJ mol⁻¹
2. 2SO₂​(g) + O₂​(g) → 2SO₃​(g), δH = -197.8 kJ mol⁻¹

First, we need to modify the given reactions so that they match the desired reaction:

1. Multiply the first reaction by 2 to get the same number of moles of SO₂​(g) as the second reaction.
2. Multiply the second reaction by 1/2 to get the same number of moles of O₂​(g) as the first reaction.

The modified reactions are:

1. 2S(s) + 2O₂​(g) → 2SO₂​(g), δH = 2 × -296.8 kJ mol⁻¹ = -593.6 kJ mol⁻¹
2. 2SO₂​(g) + 1/2 O₂​(g) → 2SO₃​(g), δH = 1/2 × -197.8 kJ mol⁻¹ = -98.9 kJ mol⁻¹

Now we can add the modified reactions to get the desired reaction:

2S(s) + 2O₂​(g) + 2SO₂​(g) + 1/2 O₂​(g) → 2SO₂​(g) + 2SO₃​(g), δH = -593.6 kJ mol⁻¹ + (-98.9 kJ mol⁻¹) = -692.5 kJ mol⁻¹

Therefore, the standard enthalpy change for the reaction S(s) + 23/2O₂​(g) → SO₃(g) is δH = -692.5 kJ mol⁻¹.