Q.Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium? C2H6

Equilibrium Constant Calculation: From Intuition to Precision

Imagine you're at a party where people can move between two rooms. Some people prefer the kitchen (more snacks), others prefer the living room (better music). After a while, the number of people in each room stops changing — not because everyone froze, but because the rate of people leaving the kitchen equals the rate of people entering it. The system is in dynamic equilibrium.

Chemical reactions work the same way. A reversible reaction like

$A + B \rightleftharpoons C + D$ doesn't stop when it reaches equilibrium. Instead, the forward reaction (making C and D) and the reverse reaction (making A and B) happen at the same rate. The concentrations of A, B, C, and D become constant — not equal, but constant.

The equilibrium constant $K$ is a number that tells you where this balance lies. It answers the question: At equilibrium, which side of the reaction is favoured?

The Intuitive Idea

Think of a seesaw. If $K$ is very large (say $10^6$), the equilibrium sits heavily on the product side — almost all A and B have turned into C and D. If $K$ is very small (say $10^{-6}$), the opposite is true: hardly any product forms. If $K$ is around 1, both sides have comparable amounts.

So $K$ is a ratio — a comparison of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients.

The Precise Statement

For a general reversible reaction at a given temperature:

the equilibrium constant $K_c$ (for concentrations in mol/L) is:

where $[X]$ means the equilibrium concentration of species X in moles per litre.

What $K$ Actually Depends On

$K$ is constant at a given temperature. Change the temperature, and $K$ changes. But $K$ does not depend on:

• Initial concentrations
• Presence of a catalyst (catalysts speed up both directions equally)
• Pressure or volume changes (for $K_c$; $K_p$ for gases has its own rules)

This is a critical exam point: if a problem gives you initial concentrations and asks for $K$, you must first find equilibrium concentrations — you cannot plug in initial values.

A Worked Example

Problem:

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

At 500°C, equilibrium concentrations are:

$[N_2] = 0.50\ \text{M}$, $[H_2] = 1.50\ \text{M}$, $[NH_3] = 0.20\ \text{M}$

Calculate $K_c$.

Solution:

Write the expression:

Substitute:

Common Pitfalls

• Forgetting the exponents: A coefficient of 2 means square the concentration, not double it.
• Using initial concentrations: You must use equilibrium concentrations only. …