ICE Charts - Online Tutor, Practice Problems & Exam Prep

ICE charts, standing for Initial, Change, and Equilibrium, are essential tools in chemistry for simplifying the calculations of equilibrium reactions. They are particularly useful when multiple equilibrium amounts for compounds in a balanced chemical reaction are unknown. The units used in ICE charts are atmospheres for the equilibrium constant K_p or molarity for K_c. These charts not only assist in determining equilibrium amounts but also in organizing data to calculate the equilibrium constant, K, which is the ratio of product to reactant concentrations at equilibrium. When only one equilibrium amount is provided, ICE charts can be pivotal in solving for K, streamlining the process of understanding chemical equilibria.

Calculating Equilibrium Amount

In this series of videos, we're going to 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 going to 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 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.

ICE Charts Example

The equation for the formation of hydrogen iodide from H₂ and I₂ is given as one mole of hydrogen gas reacts with one mole of iodine gas to produce 2 moles of hydrogen iodide gas. Here we're saying the value of K_C for the reaction is 0.218 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 H₂ and I₂ are both 0.10 molar and initially there is no hydrogen iodide present.

Alright, so there's a lot of information being thrown at us. So first thing we're going to 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 3 equilibrium amounts because initial amount and equilibrium amount are not the same thing. So that means we definitely have to set up an ice chart.

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 lace, A -, X for the reactants since we're losing it and we don't know how much we're losing, and A + X for the products because we're making them and we're not sure how much of it we're making, O -, X -, X + X. Now we're going to say that the coefficient of the product must be placed before the X variable. So here this isn't just plus X. There's a 2 here as a coefficient, so this is +2 X.

Then we're going to say the equilibrium row. Using the equilibrium row, set up the equilibrium constant expression and saw. So what we're going to say here if we set up the equilibrium confident expression. Remember K_C equals products over reactants. So K_C, we're told is 0.218 that equals, so products would be HI squared over H₂ times I₂. We plug in the value so 0.0218 equals 2X squared divided by and both of these are the same equations, so when we bring down everything, it's going to be this times this on the bottom O It's just 10 -, X ^2. At this point we're trying to solve for X, but there are some things we can keep in mind.

Once we've set U, 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 going to have to utilize the quadratic formula in order to solve for X. Here we're going to say 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 on the product 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 sqrt 0.0218. So that gives me 0.1476. Taking the square root here, both of these squares cancel out, so this just becomes 2X over. Oh, point 10 -, X. So shortcut one worked, so we can 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 going to say when the ratio of initial concentration at 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 * 10 to the -4. When I punched this in, I'd get 2200, 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 have to do the quadratic formula, remember the quadratic formula? Here it is minus b ± sqrt b ^2 - 4 AC over 2A. Again, luckily we don't have to worry about that here because we took the square root of both sides and we can avoid the quadratic.

So coming over here, this is 0.1476 equals 2X over oh, point 10 -, X. You're going to cross multiply these here, so when I do that I'm going to get 0.01476 -, 0.1476 X equals 2X. Add 1476 to both sides. When I do that, I'm going to get oh 0.01476 = 2.1476 X. We need to isolate X here, divide by 2.1476 on both sides here. So when we do that, we're going to get as our answer. We're going to get. Here X equals 0.0688 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 HI. Equilibrium. Look at the equilibrium line. At equilibrium, what is HI equal to? Well, HI is equal to 2X. So we're not done. You come down here, we're going to say here, hi at equilibrium equals 2X. Take the X that we found and plug it in so it will be two times 0.00688. So when we do that, we're going to get as our concentration at the end for hydrogen iodide as 0.01376 molar. So this would be our final answer.

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

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

Calculating Equilibrium Constant

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 going to 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.

ICE Charts Example 2

Here we're told that an important reaction in the formation of acid rain is 2 moles of sulfur dioxide reacts with 1 mole of oxygen gas to produce 2 moles of sulfur trioxide gas. 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 degrees Celsius. When equilibrium is established, the equilibrium amount of sulfur trioxide was found to be 0.0150 Molar. We need to calculate the equilibrium constant k_c for the reaction at 340 degrees Celsius. So, we'll follow steps 1 to 4 to determine the equilibrium constant expression for the reaction. Remember, ICE stands for Initial, Change, and Equilibrium.

Initially, we have 0.30 Molar sulfur dioxide, 0.20 Molar oxygen, and no products, so initially, it's 0. During the reaction, the changes for the reactants are -2x for sulfur dioxide and -x for oxygen, and for the product sulfur trioxide, it's +2x, due to their stoichiometric coefficients. Therefore, at equilibrium, the concentrations are:

• Sulfur dioxide: 0.30 - 2x
• Oxygen: 0.20 - x
• Sulfur trioxide: 0 + 2x = 0.0150

We set up our equilibrium constant expression k_c, which is based on the products over reactants:

k_c = [SO 3 2 [SO 2 2 ⁢ [O 2 ]

Given the equilibrium amount of sulfur trioxide, we set the calculated 2x equal to the known equilibrium concentration which is 0.0150 Molar. Solving for x:

x = 0.0075

Using this value, we can calculate the equilibrium concentrations:

• Sulfur dioxide: 0.30 - 2(0.0075) = 0.285 M
• Oxygen: 0.20 - 0.0075 = 0.1925 M
• Sulfur trioxide: 2x = 0.0150 M (already given)

Substituting these values into the equilibrium expression:

0.0150 2 ( 0.285 2 )( 0.1925 ) = 0.01439

Rounding to match the least number of significant figures given in the reactants (2 significant figures), k_c = 0.014 (dimensionless).

This will be our final answer, indicating the equilibrium constant at 340 degrees Celsius for this reaction in the production of sulfur trioxide.

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 K_c. 

2 NO (g) ⇌ N₂ (g) + O₂ (g)

• A. 0.013
• B. 0.026
• C. 0.66
• D. 0.11

When 0.600 atm of NO­­₂ was allowed to come to equilibrium the total pressure was 0.875 atm. Calculate the K_p­ of the reaction. 

2 NO₂ (g) ⇌ 2 NO (g) + O₂ (g)

• A. 1.66
• B. 3.02
• C. 60.5
• D. 33.3

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How to do ICE charts?

An ICE chart is a tool used in chemistry to keep track of the changes in concentration for each species in a chemical reaction. ICE stands for Initial, Change, and Equilibrium. Here's how to create and use an ICE chart:

1. Write down the balanced chemical equation for the reaction.
2. Below the equation, create a chart with three rows labeled "I" for initial concentrations, "C" for change in concentrations, and "E" for equilibrium concentrations.
3. Fill in the initial concentrations of reactants and products. If a product is not present initially, its concentration is zero.
4. Use stoichiometry to determine the changes in concentration. These changes are often represented by 'x', which is added or subtracted from the initial concentrations to find the equilibrium concentrations. The changes should reflect the molar ratios from the balanced equation.
5. Apply the changes to the initial concentrations to calculate the equilibrium concentrations. For reactants, the change will be a subtraction, and for products, it will be an addition.
6. If you are given the equilibrium concentration for one species, you can calculate 'x' and then find the equilibrium concentrations for all other species.

Remember, the direction of the change (whether you add or subtract 'x') depends on whether the reaction is proceeding forward or in reverse.

How can ICE charts be solved?

An ICE chart, which stands for Initial, Change, and Equilibrium, is a tool used to solve equilibrium problems in chemistry. Here's how to solve an ICE chart step by step:

1. Identify the balanced chemical equation for the reaction and write it at the top of your chart.
2. Initial Concentrations: Write down the initial concentrations of reactants and products. If a substance is not present initially, write 0.
3. Change: Determine the change in concentration for each substance as the system moves towards equilibrium. Use variables like 'x' to represent this change. If the reaction consumes a reactant, the change will be negative (-x), and if it produces a product, the change will be positive (+x).
4. Equilibrium Concentrations: Add the initial concentration and change together to find the equilibrium concentrations for all substances.
5. Write the equilibrium expression for the reaction using the equilibrium concentrations.
6. Solve for 'x': Use the equilibrium constant (K) given for the reaction to solve for 'x'. This may involve some algebraic manipulation.
7. Check: Plug the value of 'x' back into the equilibrium concentrations to ensure they make sense chemically (e.g. no negative concentrations).
8. Answer: Now that you have the value of 'x', you can determine the equilibrium concentrations of all species involved.

How do you use ICE charts?

An ICE chart (which stands for Initial, Change, Equilibrium) is a tool used to keep track of the concentrations of reactants and products during a chemical reaction, especially when dealing with equilibrium reactions.

Here's how to use an ICE chart:

1. Identify the reaction: Write down the balanced chemical equation for the reaction you're analyzing.
2. Initial concentrations: List the initial concentrations (or pressures) of the reactants and products before any reaction has occurred under the "I" column. If a product is not present initially, its concentration is zero.
3. Change in concentrations: Under the "C" column, indicate the changes in concentrations (or pressures) as the reaction progresses. Use variables like 'x' to represent these changes. If the reaction consumes reactants, the change will be negative; if it produces more products, the change will be positive.
4. Equilibrium concentrations: Under the "E" column, express the equilibrium concentrations (or pressures) of the reactants and products. This is done by adding the initial concentrations and the changes together.
5. Apply the equilibrium constant: Use the equilibrium constant expression for the reaction to relate the changes in concentrations to the known equilibrium constant value. Solve for the unknown variable.
6. Solve for concentrations: Once you have the value of the change (x)