At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO3:
2 KClO3(s)→2 KCl(s) + 3 O2(g) →H = –89.4 kJ
For this reaction, calculate ΔH for the formation of (a) 1.36 mol of O2 and, (b) 10.4 g of KCl. (c)The decomposition of KClO3 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.
Chapter 7 Relevant Key Terms 1. SN2: 1 step mechanism, x leaving and y inserting at same time, both arrows drawn at once, 1 transition state, 0 intermediates, Back-Side Attack, Inversion Configuration, “RDS” Rate Determining step - bimolecular - 2 molecules involved, 2. SN1: 2 step mechanism, “RDS” Rate Determining Step - Unimolecular - 1 molecule...