Consider the diagram that follows, which represents two steps in an overall reaction. The red spheres are oxygen, the blue ones nitrogen, and the green ones fluorine. (d) Write the rate law for the overall reaction if the first step is the slow, rate-determining step. [Section 14.6]

Video transcript

hi everyone for this problem. It reads what is the rate law for the overall reaction represented by the diagram below where the black circles are carbon and red circles are oxygen if the second step is the rate determining step. So here we want to write a rate law. So let's go ahead and write out our reactions based off of the diagram that was given. Okay. And we're told that the black circles are carbon and the red circles are oxygen. So that means our first reaction is going to be C 02 plus C yields C two oh four. And the second reaction is C 204 plus 02 yields to C 03. Okay, so now that we know our two reactions. Okay, The rate law of the reaction depends on the rate determining step. And we're told in the problem that the second step is the rate determining step. So let's go ahead and put a star here that this is our rate determining step. And let's go ahead and define what is a rate law. It is the mathematical equation that correlates how changing the concentrations of each reactant affects the rate of the reaction. So here we see, we have to react ints okay, and what we're going to do to write our rate law is rate is equal to k where K represents the rate constant of the reaction. And we're going to multiply this by the concentrations of each of the reactant. Okay, So it's going to be K times the concentration of C 204 times the concentration of 02. Okay. And we need to make sure we really understand that. We need to use the rate determining step when we write a rate law. Okay? So here our rate determining step is the second step, and we're able to write our rate law using the reactant in the second step. Okay? So let's go ahead and highlight this, and this is our final answer, and this is the end of this problem. I hope this was helpful.