Supplemental Guide: ICE Charts & Equilibrium

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When an equilibrium reaction is initiated (reactants are added together in a closed system) or when an already existing equilibrium is disturbed, it can be challenging to determine exactly what the effect will be on the equilibrium system. One way to make this easier is with an ICE chart. ICE stands for Initial, Change, and Equilibrium. The general form of an ICE chart is as follows:

ICE img 1-1.png

Below are some example problems, along with demonstrations of how an ICE chart would be used in each of those cases.

Example 1:

For the equilibrium, $LaTeX: Br_2\left(g\right)+Cl_2\left(g\right)\longrightarrow2BrCl\left(g\right)$ $Br_2\left(g\right)+Cl_2\left(g\right)\longrightarrow2BrCl\left(g\right)$ at 400K, K_c = 7.0.

If 0.30 mol of Br₂ and 0.30 mol of Cl₂ are introduced into a 1.0 L container, what will be the equilibrium concentrations of Br₂, Cl₂, and BrCl?

1^st step: Calculate concentrations or pressures.

$LaTeX: \left[Br_2\right]=\frac{0.30\:mol\:Br_2}{1.0\:L}=0.30\:M\:Br_2$ $\left[Br_2\right]=\frac{0.30\:mol\:Br_2}{1.0\:L}=0.30\:M\:Br_2$ $LaTeX: \left[Cl_2\right]=\frac{0.30\:mol\:Cl_2}{1.0\:L}=0.30\:M\:Cl_2$ $\left[Cl_2\right]=\frac{0.30\:mol\:Cl_2}{1.0\:L}=0.30\:M\:Cl_2$

2^nd step: Rewrite equation and set up the ICE chart. Indicate unknown changes as “+ax” or “-ax”, where “a” is the coefficient of the reactant or product whose change is being determined.

ICE img 2.png

(Notice that the reactants have a change of “-x” whereas the products have a change of “+2x”. Whenever there is an initial zero value for a product, all products will have a positive change and all reactants will have a negative change. If there is ever an initial zero value for a reactant, then all reactants will have a positive change and the products will have the negative changes).

3^rd step: Write an equilibrium constant expression and then plug in the appropriate values.

$LaTeX: K_c=\frac{\left[BrCl\right]^2}{\left[Br_2\right]\left[Cl_2\right]}$ $K_c=\frac{\left[BrCl\right]^2}{\left[Br_2\right]\left[Cl_2\right]}$ , plug in all values to get $LaTeX: 7.0=\frac{\left(2x\right)^2}{\left(0.30-x\right)\left(0.30-x\right)}$ $7.0=\frac{\left(2x\right)^2}{\left(0.30-x\right)\left(0.30-x\right)}$ , which can be simplified to $LaTeX: 7.0=\frac{\left(2x\right)^2}{\left(0.30-x\right)^2}$ $7.0=\frac{\left(2x\right)^2}{\left(0.30-x\right)^2}$

4^th step: Solve for x using algebra, and use x to calculate the equilibrium concentrations.

$LaTeX: \sqrt{7.0}=\sqrt{\frac{\left(2x\right)^2}{\left(0.30-x\right)^2}}$ $\sqrt{7.0}=\sqrt{\frac{\left(2x\right)^2}{\left(0.30-x\right)^2}}$ , to give you $LaTeX: 2.646=\frac{2x}{0.30-x}$ $2.646=\frac{2x}{0.30-x}$ .

Multiply both sides by (0.30-x) to give LaTeX: 0.794-2.646x=2x $0.794-2.646x=2x$

Add 2.646x to both sides to get LaTeX: 0.794=4.646x $0.794=4.646x$ . Therefore, $LaTeX: x=0.171\:M$ $x=0.171\:M$

[Br₂] = [Cl₂] = (0.30-x) = 0.13 M [BrCl] = 2x = 0.34 M

Example 2:

At 100^oC, the following equilibrium is established: $LaTeX: SO_2Cl_2\left(g\right)\longleftrightarrow SO_2\left(g\right)+Cl_2\left(g\right)$ $SO_2Cl_2\left(g\right)\longleftrightarrow SO_2\left(g\right)+Cl_2\left(g\right)$

and the equilibrium concentrations of SO₂Cl₂, SO₂, and Cl₂ determined to be 0.108 M, 0.052 M, and 0.162 M, respectively. If the concentration of SO₂ is suddenly increased to 0.115 M, what will the concentrations of all three gases be after equilibrium is reestablished?

1^st step: Calculate K_c. Note that you do not need an ICE chart for this part since all equilibrium concentrations are initially given.

$LaTeX: K_c=\frac{\left[SO_2\right]\left[Cl_2\right]}{\left[SO_2Cl_2\right]}=\frac{\left(0.052\right)\left(0.162\right)}{\left(0.108\right)}=0.078$ $K_c=\frac{\left[SO_2\right]\left[Cl_2\right]}{\left[SO_2Cl_2\right]}=\frac{\left(0.052\right)\left(0.162\right)}{\left(0.108\right)}=0.078$

(Following Le Châtelier’s Principle, since product is ADDED to the equilibrium, equilibrium must shift to the left in order to REMOVE product. This is why there is a negative “Change” for the products and a positive change for the reactants in this case. If SO₂Cl₂ had been added instead, equilibrium would have shifted in the opposite direction.)

3^rd step: Write an equilibrium constant expression and then plug in the appropriate values.

$LaTeX: K_c=\frac{\left[SO_2\right]\left[Cl_2\right]}{\left[SO_2Cl_2\right]}$

$K_c=\frac{\left[SO_2\right]\left[Cl_2\right]}{\left[SO_2Cl_2\right]}$ , plug in all values to get $LaTeX: \frac{\left(0.115-x\right)\left(0.162-x\right)}{\left(0.108+x\right)}=0.078$

$\frac{\left(0.115-x\right)\left(0.162-x\right)}{\left(0.108+x\right)}=0.078$

which can be simplified and rearranged algebraically into a quadratic expression:

LaTeX: x^2-0.355x+0.010206=0 $x^2-0.355x+0.010206=0$

4^th step: Solve for x using the quadratic equation, and use x to calculate the equilibrium concentrations.

$LaTeX: x=\frac{-\left(-0.355\pm\sqrt{\left(-0.355\right)^2-4\left(1\right)\left(0.010206\right)}\right)}{2\left(1\right)}=0.323\:and\:0.0316$ $x=\frac{-\left(-0.355\pm\sqrt{\left(-0.355\right)^2-4\left(1\right)\left(0.010206\right)}\right)}{2\left(1\right)}=0.323\:and\:0.0316$

but 0.323 is invalid because it would make the product concentrations negative, and there is no such thing as a negative concentration. Therefore x = 0.0316 M

[SOCl₂] = (0.108 + x) = 0.140 M

[SO₂] = (0.115 - x) = 0.083 M

[Cl₂] = (0.162 - x) = 0.130 M

Problems that use ICE charts are many and varied, and the way an ICE chart will be used is different depending on the problem. Generally though, you will want to use an ICE chart whenever you are trying to determine the effect of disturbances to equilibrium or when you are trying to determine unknown quantities at equilibrium.