For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: (c) N21g2 + 3 H21g2¡2 NH31g2

Video transcript

Hello everyone today, we are being asked to give the relationship between the rate of disappearance of the reactant and the rate of appearance of the products in the gas phase reaction below. And we have the reaction below here recall that for reaction A to be the real reaction is given by the following formula. So we have the rate is equal to negative one over a times the change in our compound over the change in time. And that's also going to equal one over B. Which is also multiplied by the change in our second product over our change in time. And so we know that A delta is the change in concentration of reactant of products and then T. Is a delta change in time. So for our first reaction here, we have an 02 which is our reactant. And that's going to have a rate of the disappearance. And that rate is going to be equal to negative 1/4 because there's four moles of that N. 02 times the change in the concentration of that N. 02 over delta. T. For our oxygen gas that is also going to be reacted. And that is also going to have a rate of disappearance. So that is going to have a rate of disappearance. And that rate is going to be equal to simply negative the concentration of that oxygen gas over the change in time. Since there's only one mold that 0. 2. And lastly, for our product, our di nitrogen Penta oxide Or in 205. That is going to be our product. And therefore we'll have a rate of appearance. And that rate will be equal to one half since there's two moles of that times the change in the concentration of our di nitrogen Penta oxide here, over the change in time. And so putting this all together, we can conclude that the answer is going to be equal to negative 1/4. The change in the concentration of our nitric oxide over the delta T. Or change in time, which is equal to negative change in our concentration of oxygen gas over delta T. And lastly, that's equal to one half times the change in the concentration of our diet, nitrogen Penta oxide over that change in T. And with that we have our final answer. I hope this helped until next time.