7.2.1 The equilibrium constant
An expression for the equilibrium constant of a reaction can be put together from the concentrations of the reactants and products at equilibrium. A concentration of a reactant or product is represented by enclosing its chemical formula in square brackets. Thus, the concentration of NO(g) is written [NO(g)].
To write down the equilibrium constant of a reaction, we start with the concentrations of the products. Each one is raised to the power of the number that precedes it in the reaction equation, and the corresponding terms for each product are then multiplied together.
Question 44
Do this now for the products of the equilibrium system 8.2.
Answer
The result is [N2(g)] × [CO2(g)]2, or, taking the multiplication sign as understood, [N2(g)][CO2(g)]2. In Equation 8.2, CO2(g) is preceded by a two, so its concentration is squared.
Question 45
Now repeat the operation for the reactants in Equation 8.2.
Answer
The result is [NO(g)]2[CO(g)]2; in Equation 8.2, both NO(g) and CO(g) are preceded by a two.
The equilibrium constant, K, is obtained by dividing the result for the products by the result for the reactants:
We have raised the possibility that Reaction 8.1 does not happen because the equilibrium position for equilibrium system 8.2 lies well over to the left. In other words, at equilibrium, the concentrations of NO(g) and CO(g) are very high, and those of N2 (g) and CO2(g) are so small as to be undetectable.
Question 46
If so, will K be large or small?
Answer
It will be very small because the large quantities ([NO(g)] and [CO(g)]) occur on the bottom of Equation 8.5, and the small quantities ([N2(g)] and [CO2(g)]) occur on the top.
The value of K can be determined experimentally. A typical temperature in a car exhaust system is 525 °C. At this temperature, K turns out to be 1040 mol−1 litre.
Question 47
Given this information, does the equilibrium position lie to the left of Equation 8.2?
Answer
No; K is immense, so at equilibrium, the concentrations of the products (which appear on top of the fraction in Equation 8.5) must be much greater than those of the reactants (which appear on the bottom). The equilibrium position for Reaction 8.2 at 525 °C therefore lies well over to the right.