Question

Use the standard reaction enthalpies given below to determine δh°rxn for the following reaction:

8 so₃ (g) → 8 s(s) +12o₂ (g) δh°rxn = ?
given:
so ₂(g) → s(s) o₂(g) δh°rxn = 296.8 kj
2 so ₂(g) +o₂(g) → 2so₃ (g) δh°rxn =-197.8 kj

A. -494.6 kj
B. 3166 kj
C. -3166 kj
D. -692.4 kj
E. -791.4 kj

Answer 1

The standard reaction enthalpy for the given decomposition reaction of SO3 is calculated using Hess's law, by reversing and adjusting the enthalpies of known reactions. The correct enthalpy change for the reaction is approximately 3166 kJ. option b is correct.

Step-by-step explanation:

To determine the standard reaction enthalpy (ΔH°rxn) for the decomposition of sulfur trioxide (SO3) into sulfur (S) and oxygen (O2), we will use Hess's law. This principle states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps into which the reaction can be divided. We are given the following reactions and their enthalpies:

SO2(g) → S(s) + O2(g), ΔH°rxn = 296.8 kJ

2 SO2(g) + O2(g) → 2 SO3(g), ΔH°rxn = -197.8 kJ

To obtain the desired reaction, we can reverse the second reaction and divide its enthalpy by 2 to match the stoichiometry, and then add it to the first reaction multiplied by 8 (the coefficient of SO2 in the desired reaction). Here's the calculation:

1. Reverse the second reaction:
   
   2 SO3(g) → 2 SO2(g) + O2(g), ΔH°rxn becomes +197.8 kJ
2. Divide the reversed reaction's ΔH°rxn by 2 to match stoichiometry:
   
   SO3(g) → SO2(g) + ½ O2(g), ΔH°rxn becomes +98.9 kJ
3. Multiply the first given reaction by 8:
   
   8 SO2(g) → 8 S(s) + 8 O2(g), ΔH°rxn becomes 8 × 296.8 kJ = 2374.4 kJ
4. Add the enthalpies of the adjusted reactions:
   
   8 SO3(g) → 8 SO2(g) + 4 O2(g) (from step 2 × 8), ΔH°rxn = 8 × 98.9 kJ = 791.2 kJ.
   
   Finally, add the enthalpies:
   
   2374.4 kJ + 791.2 kJ = 3165.6 kJ

Therefore, the correct answer is close to 3166 kJ, which corresponds to option B.