The reaction 2 NO1g2 + Cl21g2¡2 NOCl1g2 was performed and the following data were obtained under conditions of constant 3Cl24:

(a) Is the following mechanism consistent with the data? NO1g2 + Cl21g2ΔNOCl21g2 1fast2 NOCl21g2 + NO1g2¡2 NOCl1g2 1slow2

Video transcript

Hello everyone today. We have the following problem. The reaction of hydrogen with iodine produces hydrogen iodide, which is the formula here. This is the chemical reaction. The reaction performed was performed in such a way that the concentration of iodine was kept constant. Over the course of the reaction, the natural log of the concentration of hydrogen gas was plotted against time. And the following graph was obtained determined if the mechanism below is consistent with the obtained data. So we know that we have a graph here that's graft the natural log of the concentration of hydrogen gas versus our time here in a straight line with a negative slope. So this slope is negative because a natural log is going to be the inverse of whatever the number was. And so this means that we have a first order reaction here. But to confirm that we have to write the rate law. And so the right law is determinant on the slow step only, not the fast step. So we can ignore that. And it has only made by multiplying the concentration of our reactant. So we have hydrogen gas here in brackets multiplied by iodine. And since there's a two coefficient, there is a two exponent exponent of two over the iodine. And so the rate law written here using this slow step confirms that the reaction that we have is first order with respect. So it's first order with respect to hydrogen gas with respect to hydrogen gas. However, there's something addition that we need to discuss the term this iodine with the two exponents comes from the intermediate I of course, there should never be an intermediate in a rate law. And since this is a reaction with april a tive lee fast irreversible, so it has a fast irreversible first step. We can go ahead and that's to note it with these double arrows here, we can use a steady state approximation to remove the intermediate term from the rate law. However, there's just no need to do that because the intermediate comes from the first step, which doesn't have hydrogen gas in it. So applying to state, the steady state approximation, this study state approximation will not matter in this situation. So, over all, we can conclude that the written mechanism is consistent with the data with data, and that will be our final answer overall, I hope this helped, and until next time.