4.66 - Thermodynamics study questions - 2

Q466-02 Two reactions occurring in the manufacture of sulfuric acid are shown below:
reaction 1 S(s) + O2(g) SO2(g) ΔH = -297 kJ
reaction 2 SO2(g) + ½O2(g) SO3(g) ΔH = -92 kJ

State the name of the term ΔH. and whether reaction 1 would be accompanied by a decrease or increase in temperature.

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ΔH is the standard enthalpy change for the reaction.

Reaction 1 has negative enthalpy change. Chemical energy is changing to heat energy, the temperature increases.


At room temperature sulfur trioxide SO3 is a solid. Deduce with a reason whether the ΔH value would be more negative, or less negative, if SO3(s) instead of SO3(g) were formed in reaction 1.

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Energy is required to change from a solid state to a gaseous state. If sulfur(VI) oxide were formed in the solid state the energy change would be greater, i.e. more negative.

Deduce the ΔH value of this reaction:

S(s) + 1½O2(g) SO3(g)
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Constructing this equation from the two given above. S(s) appears in reaction 1 on the left hand side, which is what we want.

SO3(g) appears in reaction 2 on the right hand side, again just what we want. So it is a matter of adding reaction 1 to reaction 2:

reaction 1 S(s) + O2(g) SO2(g) ΔH = -297 kJ
reaction 2 SO2(g) + ½O2(g) SO3(g) ΔH = -92 kJ

S(s) + 1½O2(g) + SO2(g) SO2(g) + SO3(g) ΔH = -389 kJ

The enthalpy of reaction, ΔH = -389 kJ


Predict the sign of ΔS for reaction 2 and explain your choice.

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In reaction 2 1½ moles of gas become 1 mole of gas. There is a decrease in the number of moles of gas, so the entropy goes down, i.e. ΔS is negative.

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