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S first Mt 1844 A gas reaction 2A B is second order in A and goes to completion in a reaction vessel of constant volume and temper- 2 ature with a half-life of 1 h....

Question

S first Mt 1844 A gas reaction 2A B is second order in A and goes to completion in a reaction vessel of constant volume and temper- 2 ature with a half-life of 1 h. If the initial pressure of A is 1 bar, what are the partial pressures of A and B, and what is the total CAJ pressure at 1 h,at 2 h, and at equilibrium?

s first Mt 1844 A gas reaction 2A B is second order in A and goes to completion in a reaction vessel of constant volume and temper- 2 ature with a half-life of 1 h. If the initial pressure of A is 1 bar, what are the partial pressures of A and B, and what is the total CAJ pressure at 1 h,at 2 h, and at equilibrium?



Answers

A gas reaction $2 \mathrm{A}=\mathrm{B}$ is second order in $\mathrm{A}$ and goes to completion in a reaction vessel of constant volume and temperature with a half-life of $1 \mathrm{h}$. If the initial pressure of $\mathrm{A}$ is 1 bar, what are the partial pressures of $A$ and $B,$ and what is the total pressure at $1 \mathrm{h},$ at $2 \mathrm{h},$ and at equilibrium?

So, given this equation, for the first part of this question, we want to find what KP is and define KP. We want to do like we usually dio for finding the equilibrium constant in reverse, a partial pressure. You do partial pressure of the products over partial pressure of the reactant raised toe. How many moles we have to hear. So we were given partial pressure values and the problems we know this is 0.6 is your 0.6 squared, which means that this is equivalent to 1.7. So we know that KP equals 1.7 for this be is a little bit more intensive than a So we're seeing an increase in pressure from 1.2 a. T. M. 21 0.5 80 m. Since the pressure is increasing, we know that means the volume is decreasing. So first we want to do a ratio between the two to find out just how much I would change. So this ratio this 1.25 which means the new volume this'll a little bit more obvious since little arrow we're looking at 0.8 of the initial volume. It's because of that. We know that the change in the volume of initial is equivalent to 0.2 so we can find the pressure, the partial pressure of A at this specific point as being our initial partial pressure pause. We're gonna dio two times who we have to hear in our story geometry of her balance equation. We're gonna multiply that by the change we saw. If they multiply that by the initial partial pressure again and that gives us zero point eight four a. T. M. And we'll do the same thing for be almost so they had the same initial partial pressure. However we do not. Some multiply this by two because there is just a one here in the story geometry of the balanced chemical equation. So we'll dio 0.2 times 0.6, which gives us zero point seven to a t. M. But we're not done at this point because this is not at equilibrium and we need the partial pressure at equilibrium. So any time we hear that we need to do something at equilibrium, we immediately think see if we can dio in ice table an initial change that at equilibrium so we can use the values we just found. So we know here we have 0.84 for the reactant since, and 0.72 for the products. This is going to be a change of minus two X. We have a two here and a positive change of just X on the product side. So that means at equilibrium we have zero point 84 minus two X For the products, we have 0.72 plus x. So since we found the KP value in part A, we can use that. We're gonna plug this into the KP equation so we can solve for X. So just like usual for equilibrium, constant expressions were going to do products over react IDs. We're gonna raise that to power of two because we have a two here. And if you calculate this out, X is equivalent to 0.0 775 That's all going around at this point. So now that we have X, you're able to find P a equilibrium first, this personal pressure a at equilibrium. So we take what we found, which is initial at this new volume and new pressure and they were going to subtract to times the X value that we just found. Which means the partial pressure of A at equilibrium is 0.69 are then looking at the partial pressure of be This is a little bit different. We're gonna since it is on the products. But we can do is take the total pressure that we got when we changed for Part B and they were going to subtract what we found to be the partial pressure A which is 0.69 And we can do this because we know that says there are only two substances in this simple equation that the total is comprised of the product in the reacted. So it's we already found the pressure pressure of the reacted because they attract that for the total pressure to find the partial pressure of the product, which is zero point 81 a. T. M. This is a T M as well

We're given the equilibrium partial pressures for both gases A and B based on this reaction and in part A. We want to use those equilibrium partial pressures to calculate the value of K p. We know that we conform the equilibrium expression for KP based on that overall reaction to be equilibrium. Partial pressure of B divided by the equilibrium, partial pressure of a squared. We're told of both of those or 0.60 atmospheres. Three new 0.6 divided by 0.6 squared. And that gives us our final answer for part A. The value of K P to be about 1.7 and now in part B. We want to determine what the equilibrium partial pressures are for both of these gases if the total pressure is now 1.5 atmospheres at equilibrium. So once we reach equal equilibrium, we know that the equilibrium expression for KP has to come out to a ratio of 1.7. So we know that KP is equal to 1.7, which is equal to equilibrium. Partial pressure, gas be divided by equilibrium, partial pressure of gas, a squared. We also knew that the total equilibrium pressure has been increased to 1.5 atmospheres, and that's the summation of the two individual partial pressures. Now, if we rewrite that second equation to ice elite, PB wouldn't say that P B is equal to 1.5 minus p a. And we can plug in that expression for a PB into the numerator of this expression in the equilibrium expression. And so when we do that, that value of K P 1.7 is equal to P B, which we substitute 1.5 minus p a by two by p a squared. And when we rearrange that you so to put in the form of a quadratic equation, we should be left with 1.7 p a squared plus p a minus 1.5 equals zero. Now we use either the quadratic formula or a calculator to solve for the two different values for P A that we should get. Since this is a quadratic equation in the first value that we have is a negative 1.29 We know that partial pressures cannot be negative, so we neglect that answer and the positive answer zero point 69 atmospheres that ends up being the equilibrium, partial pressure of gas A. And then we know that equilibrium partial pressure of gas be just 1.5 minus that value of 0.69 so PB is equal to zero 0.81 atmospheres.

For our equilibrium. Uh, got beach to s gas in equilibrium with each to gas and s gas, The store ever KP expression, which will be the partial pressure of each two. Partial pressure s divided by the part of pressure H two s and we've got these values at equilibrium. Let's substitute in. The partial pressure of each two s is going to be 0.20 atmospheres. We're gonna leave out the units. Partial pressure of H two is 0.45 and the partial pressure S is 0.30 Let's work out a k p or equilibrium constants. This works out to be 0.0 68

This is the mountains chemical equation. Initially, we have 98.4 tours of the nitric oxide and 41.3 tour of the roaming. And is your, um, pressure of the N O V e r at equilibrium, some amount of the animal and the BR used up to make an o b e r On the total pressure is 110.5 tour. Um so if we try to calculate the total pressure, we have to add all these terms up. In addition, we can solve for X. When we saw for X, we get 29.2 tour. Then we were asked to find the KP, which is to present it by this equation on. Once we put in the value of X, we get the value seven points, you know, 46 Ah, we are told in the second part of the problem that, um, anyone br to ours. You pointing it wasjust respected me on some amount. If it is, you stop to come to any equilibrium. So the concentrations are like a broom can be calculated by putting in the values, um, in K panic waiting it the sound ones, you know, force except you Calculated in the previous fun on on solving for X begins in 1083 And putting that into the creation will give us you open. 134 at most years for nitric oxide 0.217 And mortars were brawling, roaming. And you even want 66 at motions for an obeah.


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