Medium ICE Table Practice Questions

Concept Explanation

An ICE table is a systematic bookkeeping method used in chemistry to calculate the changing concentrations of reactants and products in a chemical reaction as it reaches equilibrium. The acronym ICE stands for Initial concentration, Change in concentration, and Equilibrium concentration. This tool is essential for solving problems involving the equilibrium constant ($K_c$ or $K_p$) when the system is not yet at a steady state. By setting up an ICE table, you can relate the stoichiometry of the balanced equation to the unknown changes in molarity, typically represented by the variable '$x$'.

To master this concept, you must first ensure your chemical equation is balanced. The \"Change\" row is dictated by the stoichiometric coefficients; for example, if a reactant has a coefficient of 2, its change is $-2x$. This method is frequently used in Ka and Kb calculations to find the pH of weak acids and bases. According to Wikipedia's guide on equilibrium, the ICE table simplifies complex algebraic relationships into a manageable grid. It is particularly useful when the equilibrium constant is small enough to allow for approximations, or when a quadratic formula is required for more precise results.

Solved Examples

Below are three worked examples demonstrating how to apply the ICE method to gaseous and aqueous equilibria.

1. Example 1: Finding Equilibrium Concentrations
   For the reaction $H_2(g) + I_2(g) \u21cc 2HI(g)$, the $K_c$ is 54.3 at 430\u00b0C. If you start with 0.500 M of $H_2$ and 0.500 M of $I_2$, what are the equilibrium concentrations?

   1. Set up the ICE table. Initial: $[H_2]=0.500, [I_2]=0.500, [HI]=0$.

   2. Change: $[H_2]=-x, [I_2]=-x, [HI]=+2x$.

   3. Equilibrium: $[H_2]=0.500-x, [I_2]=0.500-x, [HI]=2x$.

   4. Substitute into $K_c$ expression: $54.3 = (2x)^2 / (0.500 - x)^2$.

   5. Take the square root of both sides: $7.37 = 2x / (0.500 - x)$.

   6. Solve for $x$: $3.685 - 7.37x = 2x \u2192 9.37x = 3.685 \u2192 x = 0.393$ M.

   7. Final concentrations: $[H_2] = 0.107$ M, $[I_2] = 0.107$ M, $[HI] = 0.786$ M.

2. Example 2: Calculating Kc from Equilibrium Molarity
   A 1.00 L flask is filled with 2.00 mol of $NOCl$. At equilibrium, 0.66 mol of $NO$ is present. $2NOCl(g) \u21cc 2NO(g) + Cl_2(g)$. Find $K_c$.

   1. Initial: $[NOCl]=2.00, [NO]=0, [Cl_2]=0$.

   2. Change: Since $[NO]$ at equilibrium is 0.66, the change for $NO$ is $+0.66$. Stoichiometry says $NOCl$ change is $-0.66$ and $Cl_2$ change is $+0.33$.

   3. Equilibrium: $[NOCl] = 2.00 - 0.66 = 1.34$; $[NO] = 0.66$; $[Cl_2] = 0.33$.

   4. $K_c = [NO]^2[Cl_2] / [NOCl]^2 = (0.66)^2(0.33) / (1.34)^2 = 0.080$.

3. Example 3: Weak Acid Dissociation
   Calculate the concentration of $H^+$ in a 0.10 M solution of acetic acid ($K_a = 1.8 \u00d7 10^{-5}$).

   1. Equation: $CH_3COOH \u21cc H^+ + CH_3COO^-$.

   2. Initial: $[HA]=0.10, [H^+]=0, [A^-]=0$. Change: $-x, +x, +x$.

   3. $K_a = x^2 / (0.10 - x) \u2248 x^2 / 0.10$ (assuming $x$ is small).

   4. $x^2 = 1.8 \u00d7 10^{-6} \u2192 x = 1.34 \u00d7 10^{-3}$ M.

Practice Questions

Test your skills with these medium-level ICE table practice questions. You may need a calculator and knowledge of the quadratic formula for some problems.

1. The reaction $CO(g) + H_2O(g) \u21cc CO_2(g) + H_2(g)$ has $K_c = 4.0$. If 1.0 mol of each reactant is placed in a 1.0 L container, calculate the equilibrium concentration of $H_2$.

2. For the dissociation of phosphorus pentachloride, $PCl_5(g) \u21cc PCl_3(g) + Cl_2(g)$, $K_c = 0.042$. If the initial concentration of $PCl_5$ is 0.200 M, what is the equilibrium concentration of $Cl_2$?

3. In the reaction $N_2(g) + O_2(g) \u21cc 2NO(g)$, $K_c = 0.10$ at 2000 K. If the initial concentrations are $[N_2] = 0.40$ M and $[O_2] = 0.40$ M, find the concentration of $NO$ at equilibrium.

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1. A solution of 0.20 M nitrous acid ($HNO_2$) has a $K_a = 4.5 \u00d7 10^{-4}$. Use an ICE table to find the equilibrium concentration of $H^+$.

2. For the reaction $2A(g) \u21cc B(g) + C(g)$, $K_c = 25$. If the initial concentration of $A$ is 0.60 M, what are the equilibrium concentrations of $B$ and $C$?

3. A 2.0 L flask contains 1.0 mol of $H_2$ and 2.0 mol of $I_2$ at a temperature where $K_c = 50.5$. Calculate the molarity of $HI$ at equilibrium.

4. Consider the reaction $SO_2Cl_2(g) \u21cc SO_2(g) + Cl_2(g)$ with $K_c = 0.078$. If the starting concentration of $SO_2Cl_2$ is 0.125 M, find the percent dissociation.

5. A mixture of 0.500 mol $H_2$ and 0.500 mol $Br_2$ is placed in a 1.00 L flask. $H_2(g) + Br_2(g) \u21cc 2HBr(g)$, $K_c = 1.9 \u00d7 10^{19}$. What is the equilibrium concentration of $H_2$?

Answers & Explanations

1. Answer: 0.67 M.
   Setup: $K_c = x^2 / (1.0-x)^2 = 4.0$. Taking the square root: $x / (1.0-x) = 2.0$. Solving for $x$: $x = 2.0 - 2x \u2192 3x = 2.0 \u2192 x = 0.67$. Since $[H_2] = x$, the answer is 0.67 M.

2. Answer: 0.073 M.
   Setup: $0.042 = x^2 / (0.200 - x)$. This requires the quadratic formula: $x^2 + 0.042x - 0.0084 = 0$. Using $x = [-b \u00b1 \u221a(b^2 - 4ac)] / 2a$, we find $x = 0.073$.

3. Answer: 0.11 M.
   Setup: $0.10 = (2x)^2 / (0.40 - x)^2$. Square root: $0.316 = 2x / (0.40 - x)$. $0.1264 - 0.316x = 2x \u2192 2.316x = 0.1264 \u2192 x = 0.0546$. Concentration of $NO = 2x = 0.109$ M.

4. Answer: 0.0093 M.
   Setup: $K_a = x^2 / (0.20 - x) = 4.5 \u00d7 10^{-4}$. Using the approximation $0.20 - x \u2248 0.20$ gives $x = 0.0095$. However, since $x$ is more than 5% of 0.20, use the quadratic formula to get $0.0093$ M. For more on this, see our pH calculation guide.

5. Answer: $[B] = [C] = 0.273$ M.
   Setup: $25 = x \u00b7 x / (0.60 - 2x)^2$. Square root: $5 = x / (0.60 - 2x)$. $3.0 - 10x = x \u2192 11x = 3.0 \u2192 x = 0.273$.

6. Answer: 0.932 M.
   Initial concentrations: $[H_2]=0.50, [I_2]=1.0$. $50.5 = (2x)^2 / (0.50-x)(1.0-x)$. This leads to a quadratic equation: $46.5x^2 - 75.75x + 25.25 = 0$. Solving for $x$ gives $x = 0.466$. $[HI] = 2x = 0.932$ M.

7. Answer: 50.4%.
   Setup: $0.078 = x^2 / (0.125 - x)$. Quadratic: $x^2 + 0.078x - 0.00975 = 0 \u2192 x = 0.063$. Percent dissociation = $(0.063 / 0.125) \u00d7 100 = 50.4\%$.

8. Answer: $1.15 \u00d7 10^{-10}$ M.
   Since $K_c$ is massive, the reaction goes almost to completion. Assume all $0.500$ M of $H_2$ and $Br_2$ react to form $1.00$ M $HBr$. Then use a reverse ICE table or $K_c$ expression: $1.9 \u00d7 10^{19} = (1.00)^2 / x^2$. $x^2 = 5.26 \u00d7 10^{-20} \u2192 x = 2.29 \u00d7 10^{-10}$. Since $[H_2] = x/2$ (depending on setup) or simply $x$ if solving for remaining reactant, the value is extremely small.

Quick Quiz

Frequently Asked Questions

What is the most common mistake when setting up an ICE table?

The most frequent error is forgetting to include stoichiometric coefficients in the \"Change\" row, which leads to incorrect algebraic expressions and wrong final values. Always multiply the variable $x$ by the coefficient from the balanced chemical equation.

Can I use moles instead of molarity in an ICE table?

You can use moles only if the total volume of the container is 1.0 L; otherwise, you must use molarity (mol/L) or partial pressures ($K_p$). Using moles in a volume other than 1 L will result in an incorrect equilibrium constant calculation.

How do I know if I need the quadratic formula?

You need the quadratic formula when the equilibrium constant $K$ is relatively large compared to the initial concentration, making the \"$x$ is small\" approximation invalid. Generally, if $x$ is more than 5% of the initial concentration, the approximation is not acceptable.

Does an ICE table work for reactions that go to completion?

ICE tables are designed for reversible reactions that reach equilibrium, but they can be adapted for limiting reactant problems by assuming the change $x$ equals the entire initial concentration of the limiting species. For more practice on energy in reactions, check out enthalpy change practice questions.

What is the difference between $Q$ and $K$ in the context of ICE tables?

$Q$ is the reaction quotient calculated using initial or non-equilibrium concentrations, while $K$ is the constant at equilibrium. Comparing $Q$ to $K$ tells you which direction the reaction will shift (the sign of $x$ in the ICE table).

Where can I find more advanced equilibrium resources?

Higher-level chemistry students often refer to LibreTexts Chemistry or Khan Academy for in-depth tutorials on Le Chatelier's principle and complex equilibrium systems.

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