The gas-phase decomposition of NO2, 2 NO21g2¡ 2 NO1g2 + O21g2, is studied at 383 C, giving the following data: Time (s) 3no2 4 (M) 0.0 0.100 5.0 0.017 10.0 0.0090 15.0 0.0062 20.0 0.0047 (c) Predict the reaction rates at the beginning of the reaction for initial concentrations of 0.200 M, 0.100 M, and 0.050 M NO2.

Video transcript

Hey everyone. So today we're being told that phosphorus Penta chloride decomposes to phosphorous, tri chloride and chlorine, gas and the gas phase reaction. And were given the values for the reaction. As time progresses, we're being asked to calculate the rate of the reaction at the following three concentrations $0.15 0.2 molar and 0.23 molar. Now to do this, the first thing we need to do is actually identify what is the rate order of the reaction or sorry, what is the order of the reaction exactly? So to do this, we also need to plot the values for the natural log with respect to the concentration of PCL five versus time. As well as one over the concentration of P C 05 versus time. So in order to do this, we take the values and we plot them and we get a or calculate them and we get a table and these are the values. So with this in mind, we now have to go ahead and plot both the natural log and one over concentration graphs using these values because whichever one is linear. If the natural log data is linear, that will mean the reaction is first order and if the one over is linear then that means it is second order. So plotting the natural log first and let me scroll down. So we have a little more space plotting it. We get a graph that looks like this and as you can see the graph is linear. So we don't have to look any further, we know that this is a first order reaction. Now, since it is a first order reaction, we know that we need to use the integrated rate law for first order reactions. And that'll be the natural log, oops. The natural log of the final concentration will be equal to negative K times T. Where K is the rate constant? Easy time plus the natural log of the initial reacting concentration, which in this case is PCL five. Now looking at this, we can see that this can relate really closely to the linear linear line equation where Y is equal to mx plus B. Thus we can see that the line or the slope of the line is the rate constant. And we can use the values on this graph to find the rate constant. So using em and let's write this in blue. Using point slope we get that M is equal to y two minus Y one over M two minus or sorry, X two minus X one X one. So using these values let us use the values of X and Y. From here as well as over here let us use the ones above. Yeah, there we go. Using these values for X one, X 2, Y one and Y two. We can substitute and get -1. Or sorry plus 1.61949. Because we're subtracting a negative so it adds it divided by 200 minus 100. Therefore M Will be equal to negative 2. Times 10 to the negative 3rd 2nd to the -1. And since m is just negative K. Again relating back to the relation between the slow formula and the integrated rate law, that means K. The rate constant is just too positive. 2.58 times 10 to the negative three seconds to the negative one. So with this in mind, since the reaction is indeed first order, we can also go ahead and say that the total rate law is the rate law Is simply K times the concentration of PCL- five. This is indeed a first action. First order reaction. Oops. So with this in hand we can actually just go ahead and substitute in the concentrations that we're trying to solve for. So if the rate is equal to 2.5, 8 times 10 to the negative third seconds to the native one And we have a concentration of 0.15 moller, Then the rate is 3.87 times 10 to the negative 4th molar per second. We can go ahead and do the same thing for the other two concentrations. 2.58 times 10 to the negative three seconds, two negative one times 0.2 molar equals 5.16 times 10 to the negative to the -4 molar per second. And finally, When the concentration is 0.23, eight times 10 to the native three seconds, the native one Into 0.23 Molar. The final concentration, Or the final rate is 5. Times 10 to the -4 m/s. Therefore the reaction is first order, and these are the rates. When the concentrations are $0.15 0.2 molar and 0. molar, respectively, I hope this helps, and I look forward to seeing you all in the next one.