Ozone is unstable with respect to decomposition to ordi-nary oxygen:
2 O31g2 ∆ 3 O21g2 Kp = 1.3 * 1057
How many O3 molecules are present at equilibrium in 10 mil-lion cubic meters of air at 25 °C and 720 mm Hg pressure?

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Accepted Answer

Hello. Everyone in this video trying to calculate for the numbers of ammonia present at equilibrium. So this K. P. Here stands for the Librium constant calculated from the partial pressures of a reaction equation. And this is the reaction that we're talking about. So let's first hear kind of dissect this word problem and read all the information that we're given. So first piece of information that we're given is the volume of air And this is equal to 11 million cubic meters. Were also given the pressure Which is 730 of Mercury. They were given the were given the percent of nitrogen and hydrogen will consider this as the mole fractions. So the mole fractions of end too. So we'll convert the percentage into decimal form and that is going to be 0.781 here and the mole fraction for H two is equal to 0.5. Again, these are the percentages in the air. Alright, so again we're being told that the air consists of 78.1% of N two. We're here trying to calculate for the partial pressure of N. To this equation then is the mole fraction of N two, multiplied by the given pressure. So now plugging in the numerical values again, the mole fraction is 0.781. The pressures that were given is 730 millimeters of mercury. Want to quickly convert this into a. T. M. So for every 1 80 M we have 760 millimeters of mercury. Then we can see that these two units will cancel. Leaving us with the final answer to be 0.75017 units being an A. T. M. Now the same exact thing for the pressure or pressure pressure of H two. So we're given that we have 0.00005% of H2. So the equation here is basically the same. The pressure pressure of H two is equal to the mole fraction of H two, multiplied by the pressure that is given to us. So, plugging in the numerical value again, our mole fraction here is 0.5. And then the given pressure is 730 millimeters of mercury. Again, we want this in A. T. M. So for everyone A. T. M. We have 760 millimeters of mercury. You can see here that these two municipal council leaving us with the partial pressure of H 22 equal to 4.8026 times 10 to the negative 80 M. Alright, so again, we will recognize that our chemical equation which is already up there. So we're gonna go ahead and just rewrite it down below. So our chemical equation for this problem is our end to end this caches form reacts with three moles of H2 gas to yield two moles of N. H three in its gaseous state. Alright, so the K. P. Here At 25°C is given to us equal to 1.4 times 10 to the -6. So the question for K. P is equal to the partial pressures of our N. H. Three, Raise the power of two and then denominator Is going to be our starting material. So the partial pressures of N2 times the partial pressure of age two raised to the power of three because we have a coefficient of three in our chemical equation here and then this raised to a power of two because of this coefficient here. All right, So, we're gonna go ahead and solve for this numerator here. So, we want to basically isolate this first. So, I can do this as well. Let's see we can divide or multiply here and here. They said they were multiplying both sides by this denominator here, this leaves us with the partial pressure of NH three raised to a power of two equaling two K. P times the partial pressure of N two times the partial pressure of H two raised to the power of three. They basically want to square root this to get rid of this to here. And that leaves us with the isolation of the partial pressure of NH three. Like we wanted. Then again, just here, just K. P. Multiplied by the partial pressure of N two, multiplied by the partial pressure of H two raised to the power of three is all going to be square rooted or to raise the power of one half. So now just using this equation here and putting the numerical values in. So my K. P. Is 1.4 times 10 to the negative six I pressure pressure of N two is 0. A. T. M. And my partial pressure of H two is 4.8026 times 10 to the negative seven A. T. M. All raised to the power of three. And now finally all of these values are going to be square rooted or raised to power of half. So once you put this into the calculator, we get the value of 3.4108 times 10 to the negative 13 units being an A. T. M. So for our volume, We know again that we have 11 times 10 to the six meter cubed. We were one of the first convert, this is into deaths meters so one D. M. For every 10 to the negative one m to go ahead and cube this all. Then we're gonna convert this into a leader. So for every one D. M. We have one leader. Let's see here that the meter cubicle cancel as well as our D. M. Here. So get the volume to be 1.1 times 10 to the leaders. So the ideal gas law equation as we know is pivoted so that PV equals to N R. T. We'll go ahead and sell for N the moles. So N will equal to So basically we're dividing both sides of this equation here by R. And T. So you get N equaling two P. V over R. T. So basically I'm gonna go ahead and use this equation now this new equation to go ahead and put in our values. So for our pressure here, that's 1.1 I apologize. We're doing pressure first. So that's 3.4108 times 10 to the negative 13 A. T. M. And then our volume is 1.1 times 10 to the 10 leaders. And for our our that's our gas law constant, that is 0.8 to one. Units being leaders times A. T. M. Her mole times kelvin. And then our temperature here is to 98.15 Calvin's were given the temperature actually to be 25°. We want to add 275 0.15 to go ahead and convert it into Kelvin's. Alright, so for our unit cancelation here we can see that A. T. M. Will cancel. Leaders will cancel and kelvin's will cancel leaving us with just the units of moles. So then once you put everything into the calculator you can see here that end is equal to 1.5327 times 10 to the negative four moles of N age three. But we're not just we're not quite done yet. We still need to convert this into molecules and that's how we get our final answer. So how can do this is by using avocados number. So starting off with the moles that we just calculated which is one 0.5327 times 10 to the negative four moles of N age three, just gonna go ahead and include avocados number. So for every one mole we have 6. times 10 to the molecules. So you can see here that the moles will cancel, leaving us with the final answer to be 2. times 10 to the Molecules of NH three. So this right here is going to be my final answer for this problem.

Ozone is unstable with respect to decomposition to ordi-nary oxygen: 2 O31g2 ∆ 3 O21g2 Kp = 1.3 * 1057 How many O3 molecules are present at equilibrium in 10 mil-lion cubic meters of air at 25 °C and 720 mm Hg pressure?

Verified Solution

Key Concepts

Equilibrium Constant (Kp)

Ideal Gas Law

Stoichiometry