Integrated Rate Law Example 2

by Jules Bruno
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A certain reaction has a rate constant of .289 seconds inverse. How long in seconds would it take for the concentration of reaction reaction a to decrease from 1.43 moler. 2.850 moller. Alright, so they tell us our rate constant is .289 seconds inverse. Remember a dead giveaway for a first order reaction? Is that K is in units of time inverse. So that's unique to first order. The fact that it's seconds in verse tells us that its first order. So since its first order, we know that we're dealing with Ln. 80 equals negative Katie plus Ln A. Up Now here our initial is 1.43. so Ln of 1.43 And then our final is .850. So Elena .850 over here equals Alright, so here our rate constant is .289. And then here we're looking for time T What we're gonna do here is we're gonna subtract Ellen of 1.43 from both sides. When we do that we're going to get .520193 equals negative .289. T The viable size by -289. Okay, so here since K is in seconds. That means time would also be in seconds. Remember that their units must always match if Kay was in minutes inverse. Then time would be in minutes. Okay, they're gonna match in terms of the units. So here when we work all this out we get one point seconds here are answer has three sig figs because 30.289 has three sig figs, as well as 1.43 and 0.850. So again, 1.80 seconds will be the time for this first order process.