You are aware that the melting of ice can be represented by the equation
and that the freezing of water can be represented by the reverse equation
Either of these unilateral reactions is written with the implication that the reactant on the left is completely converted to the product on the right. The situation in which both states of H 2O are in equilibrium is shown by the reversible reaction
where the two arrows mean that some H 2O molecules are participating in the forward (melting) reaction and other molecules are simultaneously participating in the reverse (freezing) reaction. Therefore, equilibrium is the stable situation resulting from two offsetting reactions. At a pressure of 1 atmosphere and a temperature of 0°C, both solid ice and liquid water are stable and will coexist. Notice that this equilibrium condition can be reached from either side; it can begin with either pure ice at –10°C or pure water at 20°C.
Whether you warm such ice or cool such water, the second phase will appear at 0°C.
The second example demonstrates another aspect of equilibrium by using the transformations between dinitrogen tetroxide and nitrogen dioxide.
N 2O 4 is a colorless gas, whereas NO 2 is a dark reddish‐brown gas. Their relative abundances are a function of temperature because N 2O 4 dominates at room temperature, and NO 2 dominates at higher temperatures. At any one temperature, both gases are present in a mixture, and the color of the mixture allows an estimation of the ratio of the two nitrogen oxides. If a glass vessel containing them is colorless or pale, N 2O 4 exceeds NO 2. Warming that container would cause the color to slowly darken as N 2O 4 is converted to NO 2 until the ratio of the two species is appropriate for the higher temperature. (See Figure 1.)
Figure 1. Temperature and the N2O4-NO2 mixture.