In this series of videos, we're gonna learn about ice charts what they are. And more importantly, when exactly do we use them here, we're going to say that in ice which stands for initial change equilibrium chart is a table used to simplify the calculations of equilibrium reactions. Here, we're gonna say they're used when we're missing more than one equilibrium amount for the compounds in a balanced equilibrium reaction. Now, here we're going to say the units of an ice chart are either in atmospheres because of the KP equilibrium constant or in molarity because of the KC equilibrium constant. So just remember equilibrium um equations can be complicated. Ice charts are just a way of us organizing the information before us so that we can get to our answer. Also remember that we use ice charts whenever we're missing more than one equilibrium amount for compounds within a balanced chemical reaction. And the units themselves have to be either atmospheres or molarity because of KP or KC.
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example
ICE Charts Example
Video duration:
8m
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The equation for the formation of hydrogen iodide from H two and I two is given as one mole of hydrogen gas reacts with one mole of iodine gas to produce two moles of hydrogen iodide gas. Here, we're saying the value of KC for the reaction is 0.0218 at 25 °C. Now we're asked what is the equilibrium concentration of hydrogen iodide in a sealed reaction vessel if the initial concentrations of H two and I two are both 0.10 molar. And initially there is no hydrogen iodide present. All right. So there's a lot of information being thrown at us. So first thing we're gonna do is step one is we're going to set up an ice chart if missing more than one equilibrium amount for the compounds in the balanced equation. Well, from the question itself, it tells us what we have. Initially, we have initially 0.10 molar of our reactants, we have nothing of our product. So we're missing three equilibrium amounts because initial amount and equilibrium amount are not the same thing. So that means we definitely have to set up an ice truck. Now, here it says using the initial row place the amount given for any compound. So remember, an ice chart stands for initial change equilibrium. And we're told that initially, we have 0.10 molar of both of our reactants and we have zero of our product step three. And this is important, we lose reactants to make products. So using the change row place A minus X for the reactants since we're losing it and we don't know how much we're losing and A plus X for the products because we're making them and we're not sure how much of it we're making. So minus X minus X plus X. Now we're going to say uh that the coefficient of the product must be placed before the X variable. So here, this isn't just plus X, there's a two here as a coefficient. So this is plus two X, then we're going to say the equilibrium row using the equilibrium role, set up the equilibrium constant expression and salt. So what we're gonna say here if we set up the equilibrium constant expression, remember KC equals products over reactants. So KC we're told is 0.0218 that equals so products would be H I squared over H two times I two, we plug in the value. So 0.0218 equals two X squared divided by and both of these are the same equation. So when we bring down everything, it's gonna be this times this on the bottom. So it's just 0.10 minus X squared at this point we're trying to solve for X. But there are some things we can keep in mind once we've set up the equilibrium constant expression, what we could do is we could check if a shortcut can be utilized to avoid the quadratic formula because sometimes we're gonna have to utilize the quadratic formula in order to solve for X. Here, we're gonna say if, if both shortcuts fail, then it must be used the quadratic formula. All right. So what are these shortcuts that I'm speaking of? Well, here, if we take a look, we have two of them, we have shortcut one and shortcut two shortcut one is called the square root method. Here's the equilibrium expression that I set up up above. And if we look, luckily the top and the bottom uh on the products on the right side, both of them are squared because of that, I can take the square root of both sides. So taking the square root of both sides here, I take a square root of this side and then I would take the square root of this side. That'll help me simplify my reaction or my expression. So let's take the square root of 0.0218. So that gives me 0.1476 taking the square root here, both of these squares cancel out. So this just becomes two X over 0.10 minus X. So shortcut one worked. So we could just go with that. But let's say it didn't work. If shortcut one doesn't work, then we go to shortcut two. This one is called the 500 approximation method. Here, we're gonna say when the ratio of initial concentration to K is greater than 500 you can ignore the minus X within your equilibrium expression. So let's say that we had as initial concentration not for this example, but for another one, let's say it was 0.55 molar and the equilibrium constant was 2.5 times 10 to negative four. When I punch this in, I'd get 2000, 200. A number that's greater than 500 which would mean that I could just drop this minus X avoid the quadratic formula and solve for X. Now, if you had to do the quadratic formula, remember the quadratic formula here it is minus B plus or minus the square root of B squared minus four ac over two A. Again, luckily, we don't have to worry about that year because we took the screw of both sides and we can avoid the quadratic. So coming over here, this is 0.1476 equals two X over 0.10 minus X, you're going to cross multiply these here. So when I do that, I'm gonna get 0.01476 minus 0.1476 X equals two X add 0.1476 to both sides. When I do that, I'm gonna get point 01476 equals 2.1476. X. We need to isolate X here. Divide by 2.1476 on both sides here. So when we do that, we're gonna get as our answer, we're gonna get here. X equals 0.00688 for X we solve for X. But is that our answer? Well, what is the question asking us to find? Well, the question is asking us to find the equilibrium concentration of H I equilibrium. Look at the equilibrium line at equilibrium. What is H I equal to? Well, H I is equal to two X. So we're not done. You come down here, we're gonna say here H I at equilibrium equals two X. Take the X that we found and plug it in. So it'll be two times 0.00688. So when we do that, we're gonna get as our concentration at the end for hydrogen iodide as 0.01376 molar. So this would be our final answer.
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Problem
Problem
At a given temperature the gas phase reaction: N2 (g) + O2 (g) ⇄ 2 NO (g) has an equilibrium constant of 4.18 x 10-7. What will be the concentration of NO at equilibrium if 2.00 moles of nitrogen and 6.00 moles oxygen are allowed to come to equilibrium in a 2.0 L flask?
A
1.1 × 10−3 M
B
3.1 × 10−7 M
C
1.2 × 10−6 M
D
6.2 × 10−7 M
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Problem
Problem
Consider the following reaction:
COBr2 (g) ⇌ CO (g) + Br2 (g)
A reaction mixture initially contains 0.15 atm COBr2. Determine the equilibrium concentration of CO if Kp for the reaction at 25°C is 4.08.
A
0.782 atm
B
0.145 atm
C
0.005 atm
D
4.225 atm
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concept
Calculating Equilibrium Constant
Video duration:
39s
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Now, ice charts can help us determine equilibrium amounts. But they can also be used to organize our information in order to calculate our equilibrium constant. Now recall that the equilibrium constant K is a ratio of product to reactant amounts at equilibrium. And we're gonna say when given, only when only given one equilibrium amount, the equilibrium constant K can be determined. So click on to the next video and let's take a look at a particular question where we have one given amount for an equilibrium concentration and how we can use that to help us solve for equilibrium constant K.
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example
ICE Charts Example 2
Video duration:
4m
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We're told that an important reaction in the formation of acid rain is two moles of sulfur dioxide reacts with one mole of oxygen gas to produce two moles of sulfur trioxide gas. Now, initially 0.30 molar of sulfur dioxide and 0.20 molar of oxygen are mixed and allowed to react in an evacuated flask at 340 °C. When an equilibrium is established, the equilibrium amount of sulfur trioxide was found to be 0.0150 molar. Here we need to calculate the equilibrium constant KC for the reaction at 340 °C, right. So what we're gonna do here is we're gonna follow steps 1 to 4 in order to determine the equilibrium constant expression for the reaction. So here, remember ice stands for initial change equilibrium. Initially, they're giving us amounts of in the question we're told we have 0.30 molar, we're given 0.20 molar and we're not giving anything of the product. So initially, it's zero. Remember here, we're gonna lose reactants to make products. Since this is a two, this is minus two X minus X, we're making product because it's a two this is plus two X, bring down everything 0.30 minus two, X 0.20 minus X and then plus two X. Now we're gonna say, let's see. They're telling us here for step five. Before we get there, we need to set up our equilibrium constant expression, which is KC equals products over reactants. So, so three squared divided by so two squared times 02. Now, if we take a look here, it says step five, make the given equilibrium amount equal to the value in the equilibrium row of the compound of sulfur. X plug the value of X for X into the equilibrium constant expression to solve for K. All right. So we're told here that at equilibrium, so three equals two X. But we're also told in the question that at equilibrium it's equal equilibrium amount is this number. So it equals two X and it equals that number. Therefore, they equal each other. And if they're equal to each other, we can solve for X. So divide both sides by two. So X equals 0.0075 because I know that now I can take this X value that I just found and plug it back in here in here and in here and find out what each of their amounts are at equilibrium. So let's do that. We're gonna say we have concentration of SS 02 which equals 0.30 minus two X. So 0.30 minus T uh two times 0.0075 that equals 0.285 we have 02 equals 0.20 minus X. So 0.20 minus 0.0075. So 0.1925 and then we already know that. So three at equilibrium equals two X, but we already know that's 0.0150. So we have all their equilibrium amounts. So now I just plug it into the formula or this expression, we're gonna have 0.0150 squared divided by. So two, which is 0.285 which is also squared. And then we have 0.1925. When we plug that in, that gives me 0.01439 for my equilibrium constant. If we look, let's see 0.30 0.20 both have two sig figs and then this number here has three. Remember we're going with the least number of sig figs. So my KCFN will be 0.014. Remember your equilibrium constants have no units. So this will be our final answer. 0.014 at the end.
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Problem
Problem
At a certain temperature, 0.810 mol NO is placed in a 5.00 L container. At equilibrium, 0.075 mol N2 is present. Calculate Kc.
2 NO (g) ⇌ N2 (g) + O2 (g)
A
0.013
B
0.026
C
0.66
D
0.11
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Problem
Problem
When 0.600 atm of NO2 was allowed to come to equilibrium the total pressure was 0.875 atm. Calculate the Kp of the reaction.