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The equilibrium constant (K ${}_{c}$ ), relates to a chemical reaction at equilibrium. It can be calculated if the equilibrium concentration of each reactant and product in a reaction at equilibrium is known.
Consider the following generalised reaction which takes place in a closed container at a constant temperature :
$A+B\rightleftharpoons C+D$
We know from "Factors affecting reaction rates" that the rate of the forward reaction is directly proportional to the concentration of the reactants. In other words, as the concentration of the reactants increases, so does the rate of the forward reaction. This can be shown using the following equation:
Rate of forward reaction $\propto $ [A][B]
or
Rate of forward reaction = k ${}_{1}$ [A][B]
Similarly, the rate of the reverse reaction is directly proportional to the concentration of the products. This can be shown using the following equation:
Rate of reverse reaction $\propto $ [C][D]
or
Rate of reverse reaction = k ${}_{2}$ [C][D]
At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. This can be shown using the following equation:
or
or, if the constants k ${}_{1}$ and k ${}_{2}$ are simplified to a single constant, the equation becomes:
A more general form of the equation for a reaction at chemical equilibrium is:
$aA+bB\rightleftharpoons cC+dD$
where A and B are reactants, C and D are products and a, b, c, and d are the coefficients of the respective reactants and products. A more general formula for calculating the equilibrium constant is therefore:
It is important to note that if a reactant or a product in a chemical reaction is in either the liquid or solid phase, the concentration stays constant during the reaction. Therefore, these values can be left out of the equation to calculate K ${}_{c}$ . For example, in the following reaction:
$C\left(s\right)+{H}_{2}O\left(g\right)\rightleftharpoons CO\left(g\right)+{H}_{2}\left(g\right)$
The formula for K ${}_{c}$ has the concentration of the products in the numerator and the concentration of reactants in the denominator. So a high K ${}_{c}$ value means that the concentration of products is high and the reaction has a high yield. We can also say that the equilibrium lies far to the right. The opposite is true for a low K ${}_{c}$ value. A low K ${}_{c}$ value means that, at equilibrium, there are more reactants than products and therefore the yield is low. The equilibrium for the reaction lies far to the left.
When you are busy with calculations that involve the equilibrium constant, the following tips may help:
Reactant 1 | Reactant 2 | Product 1 | |
Start of reaction | |||
Used up | |||
Produced | |||
Equilibrium |
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