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Can someone explain this:
S2O8 + 3I --> 2SO4 + I3
The amount of I3 formed can be determined by adding a known amount of S2O3 and allowing it to react:
2S2O3 + I3 --> S4O6 + 3I
Then starch is added, which reacts with any excess I3 to form a black-blue complex. The rate of the first reaction can then be determined, where t is the time elapsed since the addition of the starch
rate = .5 S2O3 / t
The answer says that the first reaction is the slow step and the second reaction is the fast step. But I thought that since the rate law uses the S2O3 from the second reaction, that the second reaction would be the slow step. Also, I was under the impression that any reaction after the slow step is irrelevant to the rate law, therefore if the first reaction is the slow step, then the second one should be irrelevant to the rate law. Could someone explain this to me?
Thanks!
S2O8 + 3I --> 2SO4 + I3
The amount of I3 formed can be determined by adding a known amount of S2O3 and allowing it to react:
2S2O3 + I3 --> S4O6 + 3I
Then starch is added, which reacts with any excess I3 to form a black-blue complex. The rate of the first reaction can then be determined, where t is the time elapsed since the addition of the starch
rate = .5 S2O3 / t
The answer says that the first reaction is the slow step and the second reaction is the fast step. But I thought that since the rate law uses the S2O3 from the second reaction, that the second reaction would be the slow step. Also, I was under the impression that any reaction after the slow step is irrelevant to the rate law, therefore if the first reaction is the slow step, then the second one should be irrelevant to the rate law. Could someone explain this to me?
Thanks!
