Given the thermochemical equation 2SO2(g) + O2(g) → 2SO3(g) ΔH = −198.2 kJ/mol calculate the heat evolved when 87.9 g of SO2 (molar mass = 64.07 g/mol) is converted to SO3. Strategy The thermochemical equation shows that for every 2 moles of SO2 reacted, 198.2 kJ of heat are given off (note the

Question

Given the thermochemical equation

[latex]2SO_{2}(g) + O_{2}(g) → 2SO_{3}(g) ΔH = −198.2 kJ/mol[/latex]

calculate the heat evolved when 87.9 g of [latex]SO_{2}[/latex] (molar mass = 64.07 g/mol) is converted to [latex]SO_{3}[/latex].

Strategy The thermochemical equation shows that for every 2 moles of [latex]SO_{2}[/latex] reacted, 198.2 kJ of heat are given off (note the negative sign). Therefore, the conversion factor is

[latex]\frac{-198.2 KJ}{2 mol SO_{2}}[/latex]

How many moles of [latex]SO_{2}[/latex] are in 87.9 g of [latex]SO_{2}[/latex]? What is the conversion factor between grams and moles?

Accepted Answer

We need to first calculate the number of moles of [latex]SO_{2}[/latex] in 87.9 g of the compound and then find the number of kilojoules produced from the exothermic reaction. The sequence of conversions is as follows:

grams of [latex]SO_{2}[/latex] → moles of [latex]SO_{2}[/latex] → kilojoules of heat generated

Therefore, the enthalpy change for this reaction is given by

[latex]\Delta H=87.9 \cancel{g SO_{2}} imes \frac{1 \cancel{mol SO_{2}}}{64.07 \cancel{g SO_{2}}} imes \frac{-198.2 KJ}{2 \cancel{mol SO_{2}}} =-136 KJ[/latex]

and the heat released to the surroundings is 136 kJ. Keep in mind that the ΔH for a reaction can be positive or negative, but the heat released or absorbed is always a positive quantity. The word “released” or “absorbed” gives the direction of heat transfer, so no sign is needed.

Check Because 87.9 g is less than twice the molar mass of [latex]SO_{2}[/latex] (2 × 64.07 g) as shown in the preceding thermochemical equation, we expect the heat released to be smaller than 198.2 kJ.

Question 6.3: Given the thermochemical equation 2SO2(g) + O2(g) → 2SO3(g) ...

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