Just as important in knowing when to use an ICE Chart is knowing when **not** to use one.

**Tip #1 Only time to use an ICE Chart**

**An ICE Chart should only be used when we are missing more than one equilibrium amount for the compounds in our balanced chemical equation. **

**EXAMPLE:** At a certain temperature, the equilibrium constant, K, for the following chemical reaction is 2.25 x 10^{–6}: N_{2 }(g) + O_{2 }(g) ⇌ 2 NO (g). If the **equilibrium mixture** contains **0.110 M N**** _{2}** and

**0.250 M O**

**, what is the final concentration of NO?**

_{2}*HINT: **Remember if more than one compound is missing an equilibrium amount then an ICE Chart will be needed. Since only NO is missing a final (equilibrium) concentration then an ICE Chart isn’t needed. *

Since no ICE Chart is needed for this example, what do we do? All we need to do is set up a normal equilibrium expression and solve for our one missing variable.

**STEP 1:** Isolate the given variables from the question.

We know the equilibrium concentrations of N_{2} and O_{2} since we are told we have an **equilibrium mixture**. With only one compound, NO, missing an equilibrium amount an ICE Chart isn’t needed.

**STEP 2:** Set up an equilibrium expression with the equilibrium constant, K, and the compounds from the balanced chemical reaction.

**STEP 3: **Plug in the given values for K, N_{2} and O_{2} to solve for the equilibrium concentration of NO.

Multiply both sides by 0.110 M and 0.250 M to isolate [NO]** ^{2}**.

Multiplying the concentrations with the equilibrium constant isolates [NO]^{2} as 6.19 x 10^{–8}.

Taking the square root of both sides of the equation gives the equilibrium concentration of NO.

**Tip #2 Units of an ICE Chart**

**If an ICE Chart is used then the units can be in molarity if K _{c} is given or in atm if K_{p} is given. **

**EXAMPLE of Kp: **The equation for the formation of hydrogen iodide from H_{2} and I_{2} is given as H_{2 }(g) + I_{2 }(g) ⇌ 2 HI (g). The value of K_{p} for the reaction is 71 at 710.0°C. What is the equilibrium partial pressure of HI in a sealed reaction vessel at 710.0°C if the **initial partial pressures** of H_{2} and I_{2} are both **0.100 atm** and initially there is no HI present?

*HINT: **We are given two initial partial pressures of 0.100 atm, but remember we need equilibrium partial pressures. Since we only have the initial partial pressures and no equilibrium partial pressures we will need to use an ICE Chart. *

**STEP 1:** Isolate the given variables from the question.

With only **initial partial pressures** and no equilibrium partial pressures it’s time to use an ICE Chart.

**STEP 2: **Set up the ICE Chart with the initial amounts of each compound.

As a chemical reaction progresses our reactants will gradually break down in order to form products. This means the change in reactants will be represented by “ – x” and because of the coefficient in front of HI we will have “+**2**x” for the HI product.

Bringing down the information from the Initial line and Change line of the ICE Chart will give us the information for the Equilibrium line.

**STEP 3: **Set up the equilibrium expression with the equilibrium constant, K_{p}.

Since we are using the units of atm we use P in our equilibrium expression.

**STEP 4: **Plug in the given value for K_{p} and the equilibrium values from the ICE Chart for H_{2}, I_{2} and HI to determine your equilibrium expression.

Notice how both the top and bottom parts of the equilibrium expression are squared. This will help make solving for the x variable much easier.

**STEP 5: **Solve for the equilibrium partial pressure of HI.

**SHORTCUT:** *When both the top and bottom of the equilibrium expression are squared we can take the square root of both sides to avoid using the quadratic formula. *

After taking the square root of both sides of the equation we obtain:

Multiply both sides by 0.10 – x.

On the left side of the equation distribute 8.426.

Add 8.426x to both sides of the equation.

Divide both sides by 10.426 in order to isolate the *x* variable.

Take the answer for the *x* variable and plug it into the expression for HI to find its equilibrium partial pressure.

Now let’s take a look at an example where Kc is now given.

**EXAMPLE of Kc: **At a certain temperature, the K_{c} of the reaction below is 1.5 x10^{−11}. If **0.10 mol N _{2} **and

**0.20 mol O**

_{2}_{ }are reacted together in a

**1.00 L**container, how much NO is present at equilibrium?

N_{2} (g) + O_{2} (g) ⇌ 2 NO (g)

*HINT: **Did they give us the equilibrium amount for any of the compounds? If you answered, “no” , then you’d be correct. They only say that the compounds react together, but never mention those are their equilibrium amounts. So again we are missing the equilibrium amount for more than one compound which means it’s time for an ICE Chart. *

**STEP 1:** Isolate the given variables from the question.

Since the question uses K_{c }the desired units must be in terms of molarity. Remember that molarity is equal to moles divided by liters.

**STEP 2: **Set up the ICE Chart based on what we learned from the previous example.

Remember we ** lose** reactants to

**products so we have**

__make__**–x**for the reactants and +

**2**x for the product.

**STEP 3: **Set up the equilibrium expression with the equilibrium constant, K_{c}.

Since we are using the units for molarity we use brackets to represent their concentrations.

**STEP 4: **Plug in the given values for K_{c} and the equilibrium values from the ICE Chart for N_{2}, O_{2} and NO.

**STEP 5: **Solve the equilibrium concentration for NO.

Since both the top and bottom parts of the equilibrium expression are not squared we can’t utilize our first Shortcut.

**SHORTCUT:** *The 5% approximation method allows us to check if the – x variables on the bottom portion of our equilibrium expression can be ignored.*

To do the 5% approximation method we divide the **initial concentration** by the **equilibrium constant**. If this ratio gives a value greater than 500 then we can ignore the **– x** variables on the bottom portion of the equilibrium expression.

Dividing the **initial concentration** of **0.10 M** by the **equilibrium constant** of **1.5 x 10 ^{-11}** gives an answer of 6.7 x 10

^{9}. Since 6.7 x 10

^{9}is greater than 500 we can remove the

**– x**variables from the bottom portion of the equilibrium expression.

Multiply both sides by 0.10 and 0.20.

After multiplying all the values on the left side we obtain 3.0 x 10^{-13}.

Take the square root of both sides to isolate 2x.

Divide both sides by 2 to determine the value of the x variable.

Finally, take the answer for the *x* variable and plug it into the expression for NO to find its equilibrium concentration.

ICE Charts are useful in illustrating how the amounts of reactants and products change as they reach equilibrium. Follow the examples and techniques we’ve done and you’ll be solving with and without ICE Charts like a champ.