The reaction 2 NO2¡2 NO + O2 has the rate constant
k = 0.63 M- 1s...

Question

The reaction 2 NO2¡2 NO + O2 has the rate constant
k = 0.63 M- 1s - 1. (b) If the initial concentration
of NO2 is 0.100 M, how would you determine
how long it would take for the concentration to decrease
to 0.025 M?

Accepted Answer

Hey everyone today, we're given the rate constant for the decomposition of hydrogen iodide at 25°C. And we're being asked to find how long it would take for the concentration of hydrogen iodide to fall from 0.156 moller to 0. Mohler. So before doing anything, I'd like to take a look at the rate constant that's given and that's because it gives us a very crucial piece of information. Now, the units here says more clarity to the power of negative one time, second, to the power of -1. Which is another way of writing one over polarity times seconds, times time. So because we have these units, we can actually go ahead and say that we're dealing with a second order equation or a second order reaction as such. We can go ahead and use the integrated rate law for a second order reaction to go ahead and find how long it would take for the concentration to change. So let's go ahead and write out that integrated rate law recall. And this can this can be found in your textbook as well. But the rate law goes as such. One over the concentration of the the final concentration of the reactant is equal to the rate constant times the time it takes for the reaction to proceed plus one over the initial concentration. And we already have most of these values. We know that the initial concentration, 0.156 Mohler. The final concentration, a sub t 0.0956, smaller. And that K. Is equal to 1.6 times 10 to the negative three polarity times. To the power of negative one times seconds. The power of negative one. So, setting all of this in excuse me, we get that one over 0. morning is equal to 1.6 times 10 to the Native three. And I'll leave out the units for now, just for the sake of space and simplicity, times T. Which which we're trying to solve for plus one over the initial concentration. So simplifying and rearranging for tea. Now we'll eventually get that T. Is equal to 4.05 More to Power Native one divided by 1.6 times 10 to the negative three moller to the Native one times seconds to the negative one. Which gives us a final value of 2531 seconds. Therefore, the time it would take for the concentration of hydrogen iodide to decrease from 0.156 moller to 0.0956 Mohler is seconds. I hope this helps. And I look forward to seeing you all in the next one

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