What does ice table stand for?

Initial, Change, Equilibrium An ICE (Initial, Change, Equilibrium) table is simple matrix formalism that used to simplify the calculations in reversible equilibrium reactions (e.g., weak acids and weak bases or complex ion formation).

chem.libretexts.org - ICE Tables - Chemistry LibreTexts

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An ICE (Initial, Change, Equilibrium) table is simple matrix formalism that used to simplify the calculations in reversible equilibrium reactions (e.g., weak acids and weak bases or complex ion formation).

Introduction

ICE tables are composed of the concentrations of molecules in solution in different stages of a reaction, and are usually used to calculate the K, or equilibrium constant expression, of a reaction (in some instances, K may be given, and one or more of the concentrations in the table will be the unknown to be solved for). ICE tables automatically set up and organize the variables and constants needed when calculating the unknown. ICE is a simple acronym for the titles of the first column of the table. I stands for initial concentration. This row contains the initial concentrations of products and reactants. stands for concentration. This row contains the initial concentrations of products and reactants. C stands for the change in concentration. This is the concentration change required for the reaction to reach equilibrium. It is the difference between the equilibrium and initial rows. The concentrations in this row are, unlike the other rows, expressed with either an appropriate positive (+) or negative (-) sign and a variable; this is because this row represents an increase or decrease (or no change) in concentration. stands for the in concentration. This is the concentration change required for the reaction to reach equilibrium. It is the difference between the equilibrium and initial rows. The concentrations in this row are, unlike the other rows, expressed with either an appropriate positive (+) or negative (-) sign and a variable; this is because this row represents an increase or decrease (or no change) in concentration. E is for the concentration when the reaction is at equilibrium. This is the summation of the initial and change rows. Once this row is completed, its contents can be plugged into the equilibrium constant equation to solve for (K_c). The procedure for filling out an ICE table is best illustrated through example. Example 1 Use an ICE table to determine (K_c) for the following balanced general reaction: [ ce{ 2X(g) <=> 3Y(g) + 4Z(g)} onumber] where the capital letters represent the products and reactants. This equation will be placed horizontally above the table, with each product and reactant having a separate column. A sample consisting of 0.500 mol of x is placed into a system with a volume of 0.750 liters. This statement implies that there are no initial amounts of Y and Z. For the I row of the Y and Z columns, 0.000 mol will be entered.

Notice that the initial composition is given in moles. The amounts can either be converted to concentrations before putting them into the ICE table or after the equilibrium amounts have been calculated. This example uses moles for the ICE table, and calculates concentrations later. At equilibrium, the amount of sample x is known to be 0.350 mol. For the equilibrium row of X, 0.350 mol will be entered. Desired Unknown [ K_c = ?

onumber ] Solution The equilibrium constant expression is expressed as products over reactants, each raised to the power of their respective stoichiometric coefficients: [ K_c = dfrac{[Y]^3[Z]^4}{[X]^2} onumber ] The equilibrium concentrations of Y and Z are unknown, but they can be calculated using the ICE table. STEP 1: Fill in the given amounts Reaction: 2X 3Y 4Z Initial amounts 0.500 mol 0.000 mol 0.000 mol Change in amount ? ? ? Equilibrium amount 0.350 mol ? ? This is the first step in setting up the ICE table. As mentioned above, the ICE mnemonic is vertical and the equation heads the table horizontally, giving the rows and columns of the table, respectively. The numerical amounts were given. Any amount not directly given is unknown. STEP 2: Fill in the amount of change for each compound Reaction 2X 3Y 4Z Initial amounts 0.500 mol 0.000 mol 0.000 mol Change in amount -0.150 mol +0.225 mol +0.300 mol Equilibrium amounts 0.350 mol ? ? Notice that the equilibrium in this equation is shifted to the right, meaning that some amount of reactant will be taken away and some amount of product will be added (for the Change row). The change in amount ((x)) can be calculated using algebra: [ Equilibrium ; Amount = Initial ; Amount + Change ; in ; Amount onumber ] Solving for the Change in the amount of (2x) gives: [ 0.350 ; mol - 0.500 ; mol = -0.150 ; mol

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onumber ] The change in reactants and the balanced equation of the reaction is known, so the change in products can be calculated. The stoichiometric coefficients indicate that for every 2 mol of x reacted, 3 mol of Y and 4 mol of Z are produced. The relationship is as follows: [ egin{eqnarray} Change ; in ; Product &=& -left(dfrac{ ext{Stoichiometric Coefficient of Product}}{ ext{Stoichiometric Coefficient of Reactant}} ight)( ext{Change in Reactant}) \ Change ; in ; Y &=& -left(dfrac{3}{2} ight)(-0.150 ; mol) \ &= +0.225 ; mol end{eqnarray} onumber ] Try obtaining the change in Z with this method (the answer is already in the ICE table). STEP 3: Solve for the equilibrium amounts Reaction 2X 3Y 4Z Initial amounts 0.500 mol 0.000 mol 0.000 mol Change in amounts -0.150 mol +0.225 mol +0.300 mol Equilibrium amounts 0.350 mol 0.225 mol 0.300 mol If the initial amounts of Y and/or Z were nonzero, then they would be added together with the change in amounts to determine equilibrium amounts. However, because there was no initial amount for the two products, the equilibrium amount is simply equal to the change: [egin{eqnarray} Equilibrium ; Amount &=& Initial ; Amount + Change ; in ; Amount \ Equilibrium ; Amount ; of ; Y &=& 0.000 ; mol; + 0.225 ; mol \ &=& +0.225 ; mol end{eqnarray} onumber ] Use the same method to find the equilibrium amount of Z. Convert the equilibrium amounts to concentrations. Recall that the volume of the system is 0.750 liters. [[Equilibrium ; Concentration ; of ; Substance] = dfrac{Amount ; of ; Substance}{Volume ; of ; System} onumber ] [ [X] = dfrac{0.350 ; mol}{0.750 ; L} = 0.467 ; M onumber ] [ [Y] = dfrac{0.225 ; mol}{0.750 ; L} = 0.300 ; M onumber ] [ [Z] = dfrac{0.300 ; mol}{0.750 ; L} = 0.400 ; M onumber ] Use the concentration values to solve the (K_c) equation: [ egin{eqnarray} K_c &=& dfrac{[Y]^3[Z]^4}{[X]^2} \ &=& dfrac{[0.300]^3[0.400]^4}{[0.467]^2} \ K_c &=& 3.17 imes 10^{-3} end{eqnarray}

onumber ]

Example 2: Using an ICE Table with Concentrations n this example an ICE table is used to find the equilibrium concentration of the reactants and products. (This example will be less in depth than the previous example, but the same concepts are applied.) These calculations are often carried out for weak acid titrations. Find the concentration of A- for the generic acid dissociation reaction: [ ce{HA(aq) + H_2O(l) <=> A^{-}(aq) + H_3O^{+}(aq)} onumber ] with ([HA (aq)]_{initial} = 0.150 M) and (K_a = 1.6 imes 10^{-2}) Solution This equation describes a weak acid reaction in solution with water. The acid (HA) dissociates into its conjugate base ((A^-)) and protons (H 3 O+). Notice that water is a liquid, so its concentration is not relevant to these calculations. STEP 1: Fill in the given concentrations Reaction: HA A- H 3 O+ I 0.150 M 0.000 M 0.000 M C ? ? ? E ? ? ? The contents of the leftmost column column are shortened for convenience. STEP 2: Calculate the change concentrations by using a variable 'x' Reaction: HA A- H 3 O+ I 0.150 M 0.000 M 0.000 M C -x M +x M +x M E ? ? ? The change in concentration is unknown, so the variable x is used to denote the change. x is the same for both products and reactants because equal stoichiometric amounts of A- and H 3 O+ are generated when HA dissociates in water. STEP 3: Calculate the concentrations at equilibrium Reaction: HA A- H 3 O+ I 0.150 M 0.000 M 0.000 M C -x M +x M +x M E 0.150 - x M x M x M To find the equilibrium amounts the I row and the C row are added. Use these values and K a (the equilibrium constant for acids) to find the concentration x. STEP 4: Use the ICE table to calculate concentrations with (K_a) The expression for K a is written by dividing the concentrations of the products by the concentrations of the reactants. Plugging in the values at equilibrium into the equation for K a gives the following: [K_a = dfrac{x^2}{0.150-x} = 1.6 imes 10^{-2} onumber] To find the concentration x, rearrange this equation to its quadratic form, and then use the quadratic formula to find x: [egin{align*} (1.6 imes 10^{-2})({0.150-x}) &= {x^2} \[4pt] x^2+(1.6 imes 10^{-2})x-(0.150)(1.6 imes 10^{-2}) &= 0 end{align*}] This is the typical form for a quadratic equation: [Ax^{2}+Bx+C=0

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onumber ] where, in this case: (A = 1)

(B = 1.6 imes 10^{-2})

(C =( -0.150)( 1.6 imes 10^{-2}) = -2.4 imes 10^{-3} ) The quadratic formula gives two solutions (but only one physical solution) for x: [x = dfrac{-B+sqrt{B^2-4AC}}{2A}

onumber ] and [x = dfrac{-B-sqrt{B^2-4AC}}{2A}

onumber ] Intuition must be used in determining which solution is correct. If one gives a negative concentration, it can be eliminated, because negative concentrations are unphysical. The x value can be used to calculate the equilibrium concentrations of each product and reactant by plugging it into the elements in the E row of the ice table. [Solution: x = 0.0416, -0.0576. x = 0.0416 makes chemical sense and is therefore the correct answer.]

chem.libretexts.org - ICE Tables - Chemistry LibreTexts

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