CH 202 Problem 5.44

5.44 At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO3:

KClO3(s) ----------> 2KCl(s) + 3O2(g)

DH = -89.4 kJ

For this reaction, calculate DH for the formation of:

(a) 0.632 mol of O2

Enthalpy changes are extensive; that is, they depend on amount of material. The enthalpy change given in the problem (-89.4 kJ) is for the formation of 3 mol O2; so we get -89.4 kJ per 3 mol O2 formed. we set up a conversion:

(b) 8.57 g KCl

the enthalpy change above is given for the formation of 2 mol of KCl. 8.57 g of KCl is

molar mass KCl

moles of KCl

so we make another conversion:

make sure that this makes sense - we liberate 89.4 kJ by forming two mol of KCl; we should get less heat if we form less KCl (here we form 0.115 mol ).

(c) The decomposition of KCl proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of KClO3 from KCl and O2, is likely to be feasible under ordinary conditions? Explain your answer.

We're asked about the formation KClO3 from KCl and O2, which is the reverse reaction from the above. We have to flip the reaction around and change the sign of DH; what was very exothermic is now very endothermic:

2KCl(s) + 3O2(g) ------> 2KClO3(s)

DH = +89.4 kJ

Without having precisely defined a quantity called the entropy yet, it's a little early in the term for a question like this one, but here goes. One of the characteristics of exothermic reactions is that they are usually (but not always!) spontaneous - that is, they take place without outside intervention. Conversely, reactions that are endothermic usually need a little nudge to get them going. Since this decomposition reaction is endothermic, we suspect that it wouldn't be very spontaneous under 'ordinary' conditions (like room temperature and 1 atm pressure.)

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DAC 7/9/08