Problem

Convert the values of $K_{c}$ to values of $K_{P}…

04:26

Problem 25 Medium Difficulty

Convert the values of $K_{c}$ to values of $K_{P}$ or the values of $K_{P}$ to values of $K_{c}$
$\begin{array}{ll}{\text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)} & {K_{c}=0.50 \mathrm{at} 400^{\circ} \mathrm{C}} \\ {\text { (b) } \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HII}(g)} & {K_{c}=50.2 \text { at } 448^{\circ} \mathrm{C}}\end{array}$
(c) $\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+10 \mathrm{H}_{2} \mathrm{O}(g) \quad K_{P}=4.08 \times 10^{-25} \mathrm{at} 25^{\circ} \mathrm{C}$
(d) $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g) \quad K_{P}=0.122$ at $50^{\circ} \mathrm{C}$

Answer

a) $1.6 \times 10^{-4}$
b) 50.2
c) $5.34 \times 10^{-39}$
d) $4.60 \times 10^{-3}$

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Chemistry 102

Chemistry

Chapter 13

Fundamental Equilibrium Concepts

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Video Transcript

So here we're given a series of equations and their equilibrium. Constance. A morass to convert. There's equilibrium constant, too, the opposite expression. So what I mean by that is if you're given the Casey, we want to convert that K p. And if you're given KP, we want to convert that Casey. So the first thing that we need to remember as we're doing this is the relationship between KP and Casey, where K P is equal to Casey comes the product of Artie raised the power of Delta End in this case, Delta. And it's gonna be a change in the number of moles of gas from the left side of the equation to the right. So we look at this for a problem. One. We have a mole, a ammonia or of nitrogen interacting with three moles of hydrogen. And that's an equilibrium with two moles of ammonia. So if we calculate adults end for this side of the for this equation, we have to minus four. It was negative, too. So we're going from four bowls of gas on the left. Two moles of gas on the right are changing and is negative too. So we're just gonna plug that value into this equation. We'll have K p is equal to the given Casey 0.5. We're animals by that by Artie. So our remember is the guy's constant Brennan atmospheres, and here were given the temperature at 400 degrees. We need to keep these in Calvin's. So we're going to write that a 6 73 come in. Where is that? To the negative, too. That means that the KP for this problem is going to be 1.6 times 10 to the negative forth. So your general steps here are first to determine you're changing end and then plug it into this equation and you wind up with your final answer for this section, which is K P equals 1.6 times 10 to the negative forth. So if we move on to Part B, we have one more hydrogen interacting with one mole of iodine, and they're an equilibrium with two moles of hydrochloric acid. So we go to calculator adults end here it zero what? That tells us we revisit this KP to Casey relationship. It's since Delta n zero. Anything raised 20 powers one. So this product of Artie. I'm going to be one. And what that means is that K P equals Casey. So in this case, we know that the KP is going to be exactly equal toe are given Casey, which is 50 point to sew because this one's of being one we don't actually have to worry about the temperature here is we don't need to do that calculation anyway, moving on to part. See, we're looking at, ah, hydrated salt on the equilibrium between that hydrated salt and water vapor above it. So we have no interacting species in our equilibrium equation on the left, and they're going to be an equilibrium with Ian hydrated salt and 10 molecules of water. Which means that our adult end for this is gonna be 10. The only difference between this problem and what we did for the 1st 2 is the fact that we're given R K p to start out. So it didn't solve for Casey. So we're just going toe reorganize his equation so that we can directly solve for Casey. So we're gonna have Casey is equal to the Given KP just 4.8 times 10 to the negative 25th. We will divide that by the product of 0.8 to 1 won't fly by our temperature again. We need to convert this 25 degrees Celsius to Calvin. So we have 298 Kelvin, and that is going to raise the 10th. Which means that our Casey, for part C is going to be 5.34 times 10 to the negative 39. Pretty unfavourable equilibrium there and then for the last one were simply looking at the evaporation of water. So we have liquid water on the left, which doesn't, uh, isn't gonna be involved in our equilibrium. Expression is an equilibrium with one mole of water on the right, which means that our delta end it's gonna be one again, were given KP. So we will have Casey. Equals are giving KP 0.1 to 2, divided by 0.8 to 1 times 50 degrees. Converted to Kelvin, which is gonna be 3 23 Race of the first power means you just keep it how it is. And that means that our, uh, Casey you r k C is equal to 4.6. So times 10 to the negative third. So are all of these, uh, conversions between K. P and K. C? You're just going to remember this general relationship between KP and Casey and then look at your reaction equation to see the change in moles of participating species from the left to the right of the equation, I just find the product straight through.

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