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Ch.14 - Chemical Kinetics

Chapter 14, Problem 32a

Consider the reaction: NO2(g) → NO(g) + 1/2 O2(g) The tabulated data were collected for the concentration of NO2 as a function of time: a. What is the average rate of the reaction between 10 and 20 s? Between 50 and 60 s?

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Hey everyone in this example, we need to calculate the rate of formation of oxygen gas between 10.20 seconds. Based on the gathered data for the given reaction. So based on this given reaction below, we want to write out a rate expression So that we can say our rate is equal to the change in concentration of our reactant. CO2 divided by the change in time. And we're going to set this equal to the change in concentration of our product Divided by the change in time. Now we need specific coefficients here and according to our given reaction, we have a coefficient of one in front of our CO2. So we would just have a negative one in front of our rate expression here in terms of its concentration because CO2 is a reactant and then as far as our oxygen, we have a coefficient of one half which we want in the denominator. So we would go ahead and make this positive or we would actually make this no longer a fraction. So it would just be positive too since we have this oxygen gas as a product. So plugging in what we know from our given equation here or the given data, we would go ahead and say that our rate is equal to The change in concentration for 02. We should have in our numerator, we can pick at any time and so let's do between Or from 20 seconds to 10 seconds. So we can have the concentration at 20 seconds because we want final, minus initial. Which is given in the chart as .819 Molar minus the concentration for C. 02 at 10 seconds which is given as 100.932 moller according to the chart. So this would be our numerator and then in our denominator we're gonna have our change in time. So because we picked 20 seconds we're going to have final time 20 seconds minus initial time 10 seconds. And so what we're going to do is set this equal to The right hand side which is two times are changing concentration of co two divided by our change in time. And so simplifying the left hand side we would say and sorry this should still have that negative one there. So in our next line we would have that our rate is equal to Positive 0.0113. And this is in units of polarity times inverse seconds. And this is still set equal to two Times are changin concentration of 02 divided by the change in time. And so now we want to go ahead and isolate for our change in concentration of 02. So we would have Both sides divided by two So that it can be canceled out here on the right hand side. And what we would get is that our change sorry our change in concentration of 02 divided by the change in time is equal to A value of 5.7 times 10 to the negative third power, and we still have units of polarity times inverse seconds. And so this would be our final answer for our rate of formation of according to the prompt. So I hope that everything I explained was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.