What is the difference between $K_c$ and $K_p$ in terms of the data required for calculation?

$K_c$ requires equilibrium concentrations derived from moles and the container volume ($mol \, dm^{-3}$), whereas $K_p$ requires partial pressures derived from mole fractions and the total system pressure.

When calculating $K_p$, how is the partial pressure of an individual gas determined?

It is calculated by finding the mole fraction of the gas (moles of gas divided by total gaseous moles) and multiplying that fraction by the total pressure of the system.

Why are pure solids like $CaCO_3(s)$ excluded from equilibrium constant expressions?

The concentration of a pure solid is related to its density, which remains constant regardless of how much solid is present. Therefore, it does not change as the reaction progresses and is incorporated into the constant value itself.

What happens to the value of $K_c$ if the concentration of a reactant is doubled at constant temperature?

The value of $K_c$ remains unchanged. While the equilibrium position will shift to oppose the change (Le Chatelier's Principle), the ratio defined by the equilibrium expression will return to the same constant value.

What is a common error when calculating the 'Change' row in an ICE table?

Failing to account for stoichiometry. For example, if the reaction is $A + 2B \rightleftharpoons C$, a decrease of $x$ in $A$ must result in a decrease of $2x$ in $B$.

How do you determine the units for a specific $K_c$ value?

Substitute the units of concentration ($mol \, dm^{-3}$) into the equilibrium expression in place of the species and simplify the resulting fraction based on the powers involved.

What is the effect of an increase in temperature on $K_c$ for an exothermic reaction?

For an exothermic reaction, increasing temperature decreases the value of $K_c$. This is because the system shifts in the endothermic (reverse) direction to absorb heat, reducing the product-to-reactant ratio.

What error occurs if you use initial moles instead of equilibrium moles in a $K_c$ expression?

The resulting value will be the Reaction Quotient ($Q$), not the Equilibrium Constant ($K$). $K$ only describes the system once the rates of the forward and reverse reactions have equalized and concentrations have stabilized.

In a $K_p$ calculation, if the total pressure is given in $kPa$, what will be the units of $K_p$ for the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$?

The units would be $kPa^{-2}$. This is derived from $\frac{kPa^2}{kPa \cdot kPa^3} = \frac{1}{kPa^2}$.

How does a catalyst affect the equilibrium constant $K_p$?

A catalyst has no effect on the value of $K_p$. It increases the rate of both the forward and reverse reactions equally, meaning the system reaches the same equilibrium state faster but the final ratio of pressures remains the same.

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5. Advanced Physical Chemistry

Equilbrium Constant Calculations

Equilibrium Constant Calculations

Summary

Equilibrium constant calculations quantify the extent of a reversible chemical reaction at a specific temperature. By using stoichiometric relationships and experimental data, chemists determine the ratio of products to reactants through concentration-based ( $K_c$ ) or pressure-based ( $K_p$ ) expressions, which remain constant unless the system's temperature is altered.

1. Definition & Core Concepts

2. Underlying Principles: The ICE Method

A diagram showing the structure of an ICE table for a generic reaction A to 2B.

3. Methods & Techniques: Calculating Kp

4. Key Distinctions

5. Exam Strategy & Tips

6. Common Pitfalls & Misconceptions

Forgetting Powers: A frequent error is failing to raise the concentration or pressure to the power of the stoichiometric coefficient (e.g., forgetting to square $[HI]$ in the decomposition of hydrogen iodide).
Using Initial Moles: Students often mistakenly use initial amounts instead of equilibrium amounts in the $K$ expression. The ICE table is essential to avoid this.
Catalyst Confusion: A common misconception is that a catalyst increases the value of $K$ . In reality, a catalyst only increases the rate at which equilibrium is reached; it has no effect on the final position or the constant itself.