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THE MEANING OF Kc

Reactions with Kc values significantly greater than 1 (e.g., on the order of 104 or larger) approach completion, and their equilibria lie far to the right or the product side. In general, reactions that go to completion have exceptionally large or undefined Kc values. In contrast, reactions that have Kc values much less than 1 (e.g., on the order of 10-4 or smaller), as in the case of the Ka value of acetic acid discussed later in this chapter (see Section 2.3), do not undergo significant change. Their equilibria lie far to the left or the reactant side, and thus very little product forms. In the case of acids, the smaller the value of Ka, the weaker the acid. For reactions that have equilibrium constant values between 100 and 0.01, it can be said that significant concentrations of both reactants and products are present at equilibrium.

CALCULATIONS FOR Kc

A remarkable feature of the equilibrium constant expression is that Kc value for a given reaction at a given temperature are always constant and unique, regardless of which of the three approaches or pathways to equilibrium is followed:

  • Only reactants present initially.
  • Only products present initially.
  • Both reactants and products present initially in a random ratio.

The Kc expression format, together with knowledge of the approach or pathway to equilibrium, can prove very valuable in developing a strategy to solving equilibrium problems, as shown in Examples 2.3 and 2.4.

Example 2.3

The reaction below is carried out in a 5-L vessel at 600 K. It is one way to convert carbon monoxide to carbon dioxide.

At equilibrium, it is found that 0.0200 mol of CO, 0.0215 mol of H2O, 0.0700 mol of CO2, and 2.00 mol of H2 are present. For this reaction, calculate:

A. K

B. Kp

Solution

A. The reaction is balanced as it stands. All four substances are in the gaseous state and thus have concentrations that can be inserted into the Kc expression.

Since the molar amounts are given at equilibrium, each must be divided by the volume to convert it into molarity, and then substituted into the Kc expression. Thus,

B. To convert to Kp, the conversion formula given in equation 2.4 is used:

Note that in this case, since

In fact, Kp = Kc whenever the product side and reactant side contain identical numbers of total moles.

Example 2.4

Ammonia gas, NH3 (g), is introduced into a previously evacuated reaction vessel in such a way that its initial concentration is 0.500 M. The ammonia decomposes into nitrogen and hydrogen gases, according to the reaction given below, and eventually reaches equilibrium. The equilibrium concentration of nitrogen is found to be 0.116 M. Determine the value of Kc.

Solution

Knowing the approach or pathway to equilibrium is vital to solving this problem. It is clear in the statement that only reactant is present initially; therefore, all of the products are formed from the decomposition of NH3. The equilibrium concentration of N2 is given as 0.116 M. Because of the 3-to-1 stoichiometric ratio between H2 and N2, 3(0.116 M) = 0.348 M of H2 must have also formed and is present at equilibrium.

Also, because of the stoichiometric ratio between NH3 and N2, 2(0.116 M) = 0.232 M of NH3 must have decomposed. Thus, the concentration of NH3 remaining and present at equilibrium is 0.500 M - 0.232 M = 0.268 M. Therefore, the values for the three species at equilibrium are as follows:

Substitute these values in the Kc expression:

 
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