Nitrosyl bromide decomposes at 10 °C. NOBr1g2S NO1g2 + 1>2 Br21g2 Use the following kinetic data to determine the order of the reaction and the value of the rate constant for consumption of NOBr.

Video transcript

Hello. Everyone in this video being told that at 127 degrees Celsius H. Two C. 03 decomposes to produce H. 20. And C. 02. So we're being asked was the order of the reaction. And what is the rate constant for the consumption of H. Two C. 03. Based on the following connect data connect data here is just this right here. So let's go ahead and recall That if our order is zero then the integrated rate law is the concentration of the molecule that are interested in. His final concentration is equal to negative K. Times T. Plus the initial concentration of the molecule that are interested in the graph then is going to be the concentration of the molecule that we're interested in versus T. For time. And the slope is equal to negative cape. Now if the order is equal to one. And the integrated rate law is the natural log of the final concentration of the molecule that were interested in going to negative K. Times T. Plus the natural log of the initial concentration of the molecule that were interested in. The graph is going to be the natural log of the concept of the concentration of the molecule that are interested in versus teat. And then our slope is equal to again negative cave. Now if the order is equal to two. The integrated rate law is one over the concentration of the molecule that are interested in its final state Equalling two K times T. Plus one over the concentration or the initial concentration of the molecule that were interested in. The graph is going to be one over the concentration of eight which is the market are interested in versus T. And this time the slope is equal to positive Cape. Alright, so again this here is the concentration at time t our case that we see here is the red law. The teeth that we see here is time. And this right here with a little zero which is just not is the initial concentration. So here we're gonna go ahead and plot the given data using zero first and second integrated rate law to determine the order of the reaction. Again, the given data is just right over here. So if we go ahead and break this all down. So again we have time in seconds we have a concentration of H two C. 03. And that is what we're given. Then we of course need the natural log of H two c. 03. And then we also want the one over the concentration of H two C. 03. Alright, so times we have zero 60 and 80. No this here is just all this. I'll just go ahead and basically rewrite everything. That's 0.278 for 20 seconds. That's 0.214 at 40 seconds. We have 0.174 at 60 seconds. Or one minute we have 0.146. And lastly at 80 seconds we have 0.126. Now we take the natural log of these concentrations at time zero we have negative 3.583 at 20 seconds we have negative 3.845 at 40 seconds. We have negative 4.53 at 60 seconds. We have negative 4.225. And lastly at 80 seconds we have negative 4.372. And lastly we take one over this concentration that we have for time zero. We have 35.971 at seconds. We have 46.771 at 40 seconds. We have 57.571 at 60 seconds. We have 68 371. And lastly at 80 seconds we have 79.171. Again. We wanna plot this given data in the zero first and second integrated law just to determine the order of our reaction. So if I plot everything, we get that the concentration of H two C. 03 versus T. What we get for this graph is that the r squared value is equal to 0.9477. To every plot the natural log of the concentration of H two, C. 03 versus teeth. We get that the r squared value is equal to 0.9866. And lastly every plot one over the concentration of H two C. 03 versus T. We get that the r squared value is equal to one so well determined is that from this plot here it is linear because we have R squared value equal into one, so the reaction is second order, so then our K. Which is just equal to our slope, which we can get from our equation that we have. This equaling to 0.54 per M. Times us and similarity and S. Is just seconds. So the final answer is that the reaction is second order and that our rate constant, which is just equal to K. And in our case is equal to slope is 0.54, her Moeller time seconds. And this right here is going to be my final answer for this problem.