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Was being formed at 0.27 Mls: In the following reaction, nitrogen 4 NH3 + 3 02 L 2 Nz + 6 HzOCalculate the rate of water formation.Calculate how fast oxygen was bei...

Question

Was being formed at 0.27 Mls: In the following reaction, nitrogen 4 NH3 + 3 02 L 2 Nz + 6 HzOCalculate the rate of water formation.Calculate how fast oxygen was being consumed.

was being formed at 0.27 Mls: In the following reaction, nitrogen 4 NH3 + 3 02 L 2 Nz + 6 HzO Calculate the rate of water formation. Calculate how fast oxygen was being consumed.



Answers

In the first step of the Ostwald process for the synthesis of nitric acid, ammonia is converted to nitric oxide by the high-temperature reaction
$$4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$
(a) How is the rate of consumption of $\mathrm{O}_{2}$ related to the rate of consumption of $\mathrm{NH}_{3} ?$
(b) How are the rates of formation of NO and $\mathrm{H}_{2} \mathrm{O}$ related to the rate of consumption of $\mathrm{NH}_{3} ?$

Just the react short dealing with and were given information felt rates based on different issues concentrations and so that let us figure out the rate law. So first we want to figure out the order end to So we're looking for next to experiments where the concentration of H two is constant. And so what will you will do is we'll choose experiments number one and two, concentration of end to goes up by a factor of two times two. Um, and when you see GOP is 0.3 becoming 0.6 And so let's see how the rate changes. Well, the rate if we take the rate of reaction to so rate two over rate one equals four. And so basically the rate increased by a factor of four. And so this means if we doubled this and we increased by a factor of four the order with respect and to his second order, do the same thing for age, too. We're looking for a reaction or 22 experiments where the concentration and two is constant and so we're going to use experiments one and three. The concentration of age to between these two is doubled, going from 0.12 point 02 and the rate? Well, if you divide rate of Experiment three overrated experiment one that actions be ate about eight. Exactly So really, we're saying, is that the Rays increasing by a factor of eight and so to to the third power equals eight. And so this is third order with respect Teoh H two. And so putting that together into the rate law, the raid equals first K. And now let's go back up. And to was with respect to second order. So constructs of and two squared for second order times a shoe to the third order. So age two to the third power, that is our rate. To calculate the rate constant, I'll just use Experiment one. So the rate was equal to 4.21 times 10 to the negative. Fifth equals okay, conscious of and two is 20.3 flared. Conservation age too, is 0.1 will Cubitt, and you'll get an answer for K, which is equal to 467 77 And since we have to, this is 1/5 order reaction. I wish I guess I answered the question later, we would have the units cancel out. And so we'll have leaders to the fourth over. Moles to the fourth. And the rate was given to us in minutes. So over minutes, um, you can check that. This is crap of this. So this will cancel out all units. So there's five moles per leaders here, so that cancel all four of them. It leaves one for the rate if you're not comfortable when I just did it. It's staying humble with, uh, what is the order? With respect to H, I'll go back to this reaction this over its third. Where, um And what is the overall reaction rate? Reaction order two plus three equals fifth order. I'm be honest. You This is really on, Tom, you know so many reactions case.

The rate law defines the relation between the concentration off the reactor and on how rapidly the reaction proceeds. The red low is expressed as late is a bust. Okay? It did the power and and we to the power and it here eminent are the power toe wish. The concentration office vicious has raised in the rich law expression. Andi also case the great constant. The given direction is and not toe. Oh, sorry. The given reaction is and two or five plus h two, which produces toe, toe edge and no. Three guests. Yeah. Here the reaction is for a struggle in each reactant Does the rate low Can we written? Has scared and toe okay. And two or five on a toe. The reaction is second order overall on first order in each reactant. So that is supposed to get and toe five and two here on two or five. And go four or five. This 0.132 Mm hmm. Andi I toe is 2. 30 and and also rated the first of 4.55 and to turn to the par minus four and for a minute, so okay can be calculated as um, Jade Bi on two or five and 205 and a door. So after putting all the values, it becomes 1.50 into it under the bar, minus five and in verse and many diverse. So this is that it constant alta?

So this problem is telling us that at a particular moment during the reaction, molecular hydrogen is reacting at a rate of 0.82 polar per second per second. And that's molecular hydrogen. And so we can look at our coefficients and that those provisions, we're gonna tell us how the reaction rates are related. So we didn't see that the co vision for end to is just one and the co vision for each to his three, meaning that the reaction rate for end to is gonna be 1/3 the reaction rate for H two. So I'm just going to 0.82 divided by three. That's gonna give me, um, Quincy 0.0 ah, to 73 um, and 33333 repeating. And I'll just round it there. And so that's gonna be Moeller per second for end to. And then we can see that the reaction rate for NH three is twice the reaction rate for end to. And so, um, we could just multiply this now by two to get the reaction rate for NH three. And we can just use our coefficients so I multiply that by two it's gonna get me 20.5 for six, seven more per second, and that's gonna be for NH three.


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