Question

Find ΔHrxn for the following reaction:

2CdS(s)+3O₂(g)→2CdO(s)+2SO₂(g)

Use the following reactions with known ΔH values:
2H₂S(g)+O₂ → S(s,rhombic) + 2H₂O (g); ΔH=−442.4 kJ
S(s,rhombic) + O₂(g) → SO₂ (g); ΔH=−296.8 kJΔ
CdO(s) + 2H₂S(g) → CdS(s) + H₂O; ΔH=−124.7 kJ
Express the energy to one decimal place and include the appropriate units.

Answer 1

The ΔHrxn for the reaction 2CdS(s) + 3O₂(g) → 2CdO(s) + 2SO₂(g) is calculated using Hess's law to be -911.3 kJ, taking into account the manipulation and summing of provided reactions with their respective enthalpy changes.

Step-by-step explanation:

To find the ΔHrxn for the reaction 2CdS(s) + 3O₂(g) → 2CdO(s) + 2SO₂(g), we can use Hess's law to add and manipulate the given reactions with known enthalpy changes (ΔH) to match the target reaction. Here is how we can approach this:

• First, we need to reverse the given reaction for the formation of CdS, which will be 2CdS(s) → 2CdO(s) + 2H₂S(g) and the ΔH will become +124.7 kJ because reversing a reaction changes the sign of ΔH.
• Next, we use the reaction 2H₂S(g) + O₂(g) → S(s,rhombic) + 2H₂O(g) directly, noting ΔH is -442.4 kJ.
• Lastly, we multiply the reaction S(s,rhombic) + O₂(g) → SO₂(g) by 2 to match the 2 moles of SO₂ in our target reaction and thus the ΔH becomes 2 x (-296.8 kJ) = -593.6 kJ.

Now we can sum the ΔH values of these manipulated reactions:

(+124.7 kJ) + (-442.4 kJ) + (-593.6 kJ) = -911.3 kJ

Therefore, the enthalpy change, ΔHrxn, for the reaction 2CdS(s) + 3O₂(g) → 2CdO(s) + 2SO₂(g) is -911.3 kJ.