Documentation:CHBE Exam Wiki/2.6 - Equilibrium Constants

2.6 – Equilibrium Constant

2.6.0 – Learning Objectives

By the end of this section you should be able to:

1. Understand the concept of equilibrium.
2. Calculate equilibrium constants.

2.6.1 – Introduction

There are some reactions that are irreversible, such as combustion, where the reaction proceeds in only one direction. On the other hand, there are reactions that are reversible, such as the hydrolysis of ethylene to ethanol.

$C_{2}H_{4}+H_{2}O\rightleftharpoons C_{2}H_{5}OH$

2.6.2 – Equilibrium

In the reaction shown above, the reactants form products and the products form reactants. When you put the ethylene in water, the forward reaction occurs. Once ethanol is present, the reverse reaction begins. As the concentrations of the $C_{2}H_{5}$ and $H_{2}O$ decrease, the rate of the decreases until the point is reached where the forward and reverse reaction rate is the same. This forward and reverse reaction is an equilibrium.

2.6.3 – Calculcating the Equilibrium Constant

The equilibrium constant is a function of temperature. Let's take the water-gas shift reaction as an example.

$CO{(g)}+H_{2}O{(g)}\rightleftharpoons CO_{2}{(g)}+H_{2}{(g)}$

The equilibrium constant will be:

$K(T)={\frac {y_{CO_{2}}y_{H_{2}}}{y_{CO}y_{H_{2}O}}}$

This equation shows that if your equilibrium constant is large, then you will have more products than reactants which means the forward reaction is strong. The opposite is true for a small equilibrium constant.