How to Calculate the Equilibrium Constant (K): Kc, Kp, and ICE Tables
Chemical equilibrium is the state where the rate of the forward reaction equals the rate of the reverse reaction. For students, however, it often feels like a maze of variables, algebra, and thermodynamic concepts. Whether you are dealing with gas pressures, molar concentrations, or free energy, the math tells the story of how a reaction behaves.
Mastering the Equilibrium Constant ($K$) is not just about memorizing formulas; it is about understanding how to quantify the balance of a chemical system. In this guide, we will break down the calculation of $K_c$ and $K_p$, determining reaction direction with $Q$, and solving complex equilibrium concentrations using ICE tables.
Defining the Equilibrium Constant (K): Kc vs. Kp
The equilibrium constant is a number that expresses the relationship between the amounts of products and reactants present at equilibrium at a specific temperature. The size of $K$ tells you the extent of the reaction:
- If $K \gg 1$: The reaction favors the products (lies to the right).
- If $K \ll 1$: The reaction favors the reactants (lies to the left).
Writing the Expression
According to the Law of Mass Action, for a general reversible reaction:
$$ aA + bB \rightleftharpoons cC + dD $$The equilibrium constant expression for concentration ($K_c$) is:
$$ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} $$Important Rule: Pure solids ($s$) and pure liquids ($l$) are excluded from the expression because their concentrations (densities) remain constant. Only gases ($g$) and aqueous solutions ($aq$) are included.
Illustrative Example: Balancing and Deriving K
Consider the reaction between fluorine gas and hydrogen bromide:
$$ F_2(g) + HBr(g) \rightleftharpoons HF(g) + Br_2(g) $$
Before writing the K expression, we must ensure the law of conservation of mass is satisfied by balancing the equation.
- Balance F: We have 2 F on the left, so we need 2 HF on the right.
- Balance H and Br: Now we have 2 H on the right, so we need 2 HBr on the left. This also balances the Br (2 on left, 2 on right).
Balanced Equation:
$$ F_2(g) + 2HBr(g) \rightleftharpoons 2HF(g) + Br_2(g) $$Now, we can write the expressions based on the stoichiometry:
- Concentration ($K_c$): $$ K_c = \frac{[HF]^2 [Br_2]}{[F_2] [HBr]^2} $$
- Partial Pressure ($K_p$): $$ K_p = \frac{(P_{HF})^2 (P_{Br_2})}{(P_{F_2}) (P_{HBr})^2} $$
The Relationship Between Kc and Kp
While $K_c$ uses molarity ($M$), $K_p$ uses partial pressures (usually atm or bar). They are linked by the ideal gas law:
$$ K_p = K_c (RT)^{\Delta n} $$Where:
- $R$ = 0.08206 L·atm/(mol·K)
- $T$ = Temperature in Kelvin
- $\Delta n$ = (moles of gaseous product) - (moles of gaseous reactant)
Pro-Tip: Reversing Reactions
If you reverse a reaction, the new equilibrium constant is the reciprocal of the original.
Example: If $I_2(g) \rightleftharpoons 2I(g)$ has a constant $K_{fwd}$, then the reverse reaction $2I(g) \rightleftharpoons I_2(g)$ has a constant $K_{rev} = \frac{1}{K_{fwd}}$.
The Reaction Quotient (Q) and Predicting Direction
How do you know if a system is at equilibrium or which way it needs to shift to get there? You calculate the Reaction Quotient ($Q$).
The formula for $Q$ is identical to $K$, but it uses the current concentrations or pressures, not necessarily the equilibrium ones.
- $Q < K$: The ratio of products to reactants is too small. The reaction shifts forward (right) to make more products.
- $Q > K$: The ratio of products is too high. The reaction shifts reverse (left) to make more reactants.
- $Q = K$: The system is at equilibrium.
Case Study: The Haber-Bosch Process
Consider the synthesis of ammonia, critical for global fertilizer production:
$$ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) $$At 298 K, the equilibrium constant $K_p$ is determined to be 749.6.
Suppose a reaction vessel currently contains the following partial pressures:
- $P_{NH_3} = 10 \text{ bar}$
- $P_{H_2} = 6 \text{ bar}$
- $P_{N_2} = 2 \text{ bar}$
Step 1: Calculate Q
$$ Q = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} $$ $$ Q = \frac{(10)^2}{(2)(6)^3} = \frac{100}{2 \cdot 216} = \frac{100}{432} \approx 0.23 $$Step 2: Compare Q to K
Since $0.23 (Q) \ll 749.6 (K)$, the system is far from equilibrium. To reach $K$, the numerator (products) must increase. Therefore, the reaction will proceed forward.
Mastering the ICE Table: Step-by-Step Examples
When you know initial concentrations but need to find the equilibrium constant (or vice versa), an ICE Table (Initial, Change, Equilibrium) is your best tool. It organizes the stoichiometry systematically.
Practice Problem: Decomposition of Hydrogen Sulfide
Problem: 6.0 moles of $H_2S$ are placed in a 2.0 L container. At equilibrium, 5.0 moles of $H_2$ are present. Calculate $K_c$ for the reaction:
$$ 2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g) $$Step 1: Convert to Molarity (Concentration)
ICE tables work best with Molarity ($mol/L$).
- $[H_2S]_{initial} = 6.0 \text{ mol} / 2.0 \text{ L} = 3.0 \text{ M}$
- $[H_2]_{initial} = 0 \text{ M}$
- $[H_2]_{equilibrium} = 5.0 \text{ mol} / 2.0 \text{ L} = 2.5 \text{ M}$
Step 2: Set up the ICE Table
Use stoichiometry to define $x$. Since 2 moles of $H_2S$ produce 2 moles of $H_2$, we use $2x$ for convenience.
| Species | $2H_2S$ | $\rightleftharpoons$ | $2H_2$ | $S_2$ |
|---|---|---|---|---|
| Initial (M) | 3.0 | 0 | 0 | |
| Change (M) | $-2x$ | $+2x$ | $+x$ | |
| Equilibrium (M) | $3.0 - 2x$ | $2x$ | $x$ |
Step 3: Solve for x
We know the equilibrium concentration of $H_2$ is 2.5 M. Looking at the table:
Step 4: Calculate Equilibrium Concentrations
- $[H_2S] = 3.0 - 2(1.25) = 3.0 - 2.5 = 0.5 \text{ M}$
- $[H_2] = 2.5 \text{ M}$
- $[S_2] = x = 1.25 \text{ M}$
Step 5: Calculate Kc
$$ K_c = \frac{[H_2]^2 [S_2]}{[H_2S]^2} = \frac{(2.5)^2 (1.25)}{(0.5)^2} $$ $$ K_c = \frac{6.25 \cdot 1.25}{0.25} = \frac{7.8125}{0.25} = 31.25 $$Depending on significant figures (2 sig figs in the prompt), the answer is 31.
Relating $\Delta G^{\circ}$ to K
Thermodynamics determines if a reaction is spontaneous. The link between the Standard Gibbs Free Energy Change ($\Delta G^\circ$) and the equilibrium constant is one of the most important equations in physical chemistry:
$$ \Delta G^\circ = -RT \ln K $$- If $\Delta G^\circ$ is negative, $\ln K$ is positive, so $K > 1$ (Product favored).
- If $\Delta G^\circ$ is positive, $\ln K$ is negative, so $K < 1$ (Reactant favored).
Step-by-Step Calculation: From Energy to Equilibrium
Problem: Calculate $K_{eq}$ at 298 K for the reaction $CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)$ given the following standard free energies of formation ($\Delta G^\circ_f$):
- $CO(g) = -137.0 \text{ kJ/mol}$
- $Cl_2(g) = 0 \text{ kJ/mol}$
- $COCl_2(g) = -205.0 \text{ kJ/mol}$
Step 1: Calculate $\Delta G^\circ_{rxn}$
$$ \Delta G^\circ_{rxn} = \sum \Delta G^\circ_f (\text{products}) - \sum \Delta G^\circ_f (\text{reactants}) $$
$$ \Delta G^\circ_{rxn} = (-205.0) - [(-137.0) + 0] $$
$$ \Delta G^\circ_{rxn} = -205.0 + 137.0 = -68.0 \text{ kJ/mol} $$
Step 2: Convert Units (Crucial!)
The gas constant $R$ is in Joules ($8.314 J/mol\cdot K$), but $\Delta G$ is usually in kJ. You must convert.
Step 3: Rearrange and Solve for K
$$ \ln K = \frac{-\Delta G^\circ}{RT} $$ $$ \ln K = \frac{-(-68,000)}{(8.314)(298)} = \frac{68,000}{2477.6} \approx 27.45 $$To find K, take the exponent ($e^x$):
$$ K = e^{27.45} \approx 8.4 \times 10^{11} $$This huge value confirms that with a highly negative $\Delta G$, the reaction proceeds almost entirely to completion.
Key Takeaways: Chemical Equilibrium
- K is constant for a specific reaction at a specific temperature. It does not change with initial concentrations.
- Q tells the direction. If $Q < K$, shift right. If $Q > K$, shift left.
- ICE Tables are essential for organizing data when solving for missing equilibrium variables.
- Watch your units. When converting between $\Delta G$ and $K$, always ensure energy units match the gas constant $R$ (Joules).