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professor190 [17]

3 years ago

8

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

1 answer:

siniylev [52]3 years ago

8 0

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

____________________________________________________

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

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<em />

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