an excess of oxygen. This corresponds to second order in ozone and order (−1) with respect to oxygen. When a partial order is negative, the overall order is usually considered as undefined. In the above example, for instance, the reaction is not described as first order even though the sum of the partial orders is , because the rate equation is more complex than that of a simple first-order reaction. Opposed reactions A pair of forward and reverse reactions may occur simultaneously with comparable speeds. For example, A and B react into products P and Q and vice versa (a, b, p, and q are the stoichiometric coefficients): {\mathit{a}A} + {\mathit{b}B} <=> {\mathit{p}P} + {\mathit{q}Q} The reaction rate expression for the above reactions (assuming each one is elementary) can be written as: where: k1 is the rate coefficient for the reaction that consumes A and B; k−1 is the rate coefficient for the backwards reaction, which consumes P and Q and produces A and B. The constants k1 and k−1 are related to the equilibrium coefficient for the reaction (K) by the following relationship (set v=0 in balance): Simple example In a simple equilibrium between two species: A <=> P where the reaction starts with an initial concentration of reactant A, [A]0, and an initial concentration of 0 for product P at time t=0. Then the equilibrium constant K is expressed as: where and are the concentrations of A and P at equilibrium, respectively. The concentration of A at time t, , is related to the concentration of P at time t, , by the equilibrium reaction equation: [A]_\mathit{t} = [A]0 - [P]_\mathit{t} The term [P]0 is not present because, in this simple example, the initial concentration of P is 0. This applies even when time t is at infinity; i.e., equilibrium has been reached: [A]_\mathit{e} = [A]0 - [P]_\mathit{e} then it follows, by the definition of K, that and, therefore, These equations allow us to uncouple the system of differential equations, and allow us to solve for the concentration of A alone. The reaction equation was given previously as: For A <=> P this is simply The derivative is negative because this is the rate of the reaction going from A to P, and therefore the concentration of A is decreasing. To simplify notation, let x be , the concentration of A at time t. Let be the concentration