Question

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction: 4 SO3(g) → 4 S(s) + 6 O2(g) ΔH°rxn = ? Given: SO2(g) → S(s) + O2(g) ΔH°rxn = +296.8 kJ 2 SO2(g) + O2(g) → 2 SO3(g) ΔH°rxn = -197.8 kJ

Answer 1

Answer:

Explanation:

We are going to manipulate the equations:

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

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

So that after we add them together we will get the desired equation:

4 SO3(g) → 4 S(s) + 6 O2(g) ΔH°rxn = ?

The  ΔH°rxn  will then be added according to Hess´s law to get our answer.

Notice that S has a coefficient of 4, and it only appears in the first as 1 mole. Therefore we will multiply  it by 4. Also, notice in the final equation we want the sulfur as product, so we need to multiply by 4 the first equation:

    4 SO2(g) → 4 S(s) + 4 O2(g)      ΔH°rxn =( +296.8 kJ  ) x 4 = 1,187.20 kJ

Working with the second equation, we can see that if we multiply this equation by 2 and reverse it, when we add it  to the previously manipulated equation, we can cancel the SO2s which are not present in our desired reaction:

4 SO3(g)   → 4  SO2(g) +  4 O2(g)  ΔH°rxn = ( + 197.8 kJ) x 2   = 395.6 kK

adding our worked equations:

    4 SO2(g) →   4 S(s) +          4 O2(g)               ΔH°rxn = 1187.2 kJ

    4 SO3(g)   → 4  SO2(g) +  2 O2(g)                ΔH°rxn =  395.6 kJ

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4 SO3(g)  →   4 S(s)  + 6  O2(g)                        ΔH°rxn =  1582.8 kJ