The gas phase decomposition of HI has the following rate law: 2 HI1g2¡H21g2 + I21g2 Rate = k3HI42 At 443 °C, k = 30.1 M-1 min-1. If the initial concentration of HI is 0.010 M, what is the concentration after 1.5 hours? (LO 14.8) (a) 6.9 * 10-3 M (b) 1.8 * 10-3 M (c) 3.6 * 10-4 M (d) 8.9 * 10-4 M

Video transcript

Well everyone in this video want to calculate for the concentration of S. 03 after 15 minutes if its initial concentration is 150.5 molar. So for a rate law, even though we're given this already here, we're gonna go ahead and actually recognize that that's equal to K. Times A. To the X. Times concentration of B. To the Y. So what K. Here represents is of course the rate constant, our A. And R. B. Here, our reactant and then the X. And Y. Exponents are reacting orders here since we see A. Two that also represents what the rate order is and that's just a second order. So in that second order with respect to S. 03 of course. So the second order integrated rate law is equal to one over the final concentration of the molecule that were interested in Equalling two K. times team Plus one over the concentration of the molecule as initial states. Alright, so T. Here They said is 15 minutes. We're gonna convert this into seconds. So using a direct conversion factor for every 60 seconds, we have one minute We see here that the minute units will council giving us that T. is equal to 900 seconds. Now using this equation now, one over the concentration of A. At its final state is equal to K. Is 0.1 to two M. To the negative one times X. To the negative one. So M. Is just moller and S. Is just seconds times team and that's what we just saw for in red and that is 900 seconds Plus one over the initial concentration. And that is 0.500 molar. Alright, so just do some simplifications here and that. We get one oh and 9.8 M. To the negative one Plus 2.00 M. to the -1. So here we just took that and cover it into a decimal. And then here we see that for the units The second will cancel leaving us with just M. to the -1. So here, now that we simplified it, we see that these units are the same. So we go ahead and combine these two numerical values. We do. So we get that one over the final concentration of the molecule that we're interested in. 0 to 111.8 M to the negative one. And I'll just isolate this denominator here. We get the final value of this concentration to be 8. times 10 to the negative three molars. And that's the concentration of S. 03. After 15 minutes of this initial concentration is at 150.5 molars. So A. B. C. And D. The one that matches our final answer is going to be answer choice eight