Equilibrium :  Gas-Phase Equilibrium

Overview of Basic Kinetics

Chemical Kinetics - The study of the rates of chemical reactions.

Rate of a Reaction - The rate at which the reactants are changed into the products of a reaction.  The change in concentration of one of the reactants (DX), during a given period of time (Dt)

Rate Law - A mathematical equation that describes the rate of a chemical reaction.  It relates how the rate of a chemical reaction depends on the concentrations of the reactants consumed in that reaction.

rate = k[O3]

Collision Theory - A model for gas-phase reactions which assumes that molecules must collide in order to react.  This helps to explain why the rate of a reaction is proportional to the reactants in a reaction.  As a reaction proceeds in the forward direction, reactants are consumed.  As the reactants are consumed, the rate of the reaction decreases because there are fewer and fewer molecules available to collide and react.

When the rates of the forward and reverse reactions are equal, the reaction will seem like it has stopped.  The reactants are being consumed at the same rate that they are being produced, and there seems to be no net change in the concentrations of the reactants and the products.  The reaction is at equilibrium.

Equilibrium

Equilibrium has a two-part definition

  1. The point at which there is no longer a changed in the concentrations of the reactants and the products of a chemical reaction.
  2. The point at which the rates of the forward and reverse reactions are equal.

As was done with the acid-base equilibrium reactions, it is assumed that reactions that strongly favor the products and are not reversible go to completion.  This is indicated by a regular forward arrow:

On the other hand, reactions that are reversible and reach equilibrium are written using a forward/backward arrow:

Let's follow the equilibrium reaction above.  When this system is at equilibrium, by definition, the rates of the forward and reverse reactions are equal.  The rate laws for the forward and reverse reactions are:

Since at equilibrium, the rates of the forward and reverse reactions are equal, we can combine the rate laws:

If we rearrange the equation to isolate the k constants on one side, we have:

Since the k values are constants, they can be combined to make a single equilibrium constant (Kc).  The resulting equation is equilibrium constant expression:

Equilibrium constant expression - The mathematical product of the products of the reaction divided by the mathematical product of the reactants of the reaction:

**When a reaction reaches equilibrium, the ratio of the concentrations of the products to the concentrations of the reactants as described by the equilibrium constant expression will always be the same (at constant temperature).

How to write equilibrium constant expressions

  • In an equilibrium reaction, even though the reaction proceeds in both directions, the reagents to the left of the arrow are assumed to be the reactants and the reagents to the right of the arrow are assumed to be the products.  This is just like any non-equilibrium reaction.
  • The PRODUCTS are written on TOP, in the numerator
  • The REACTANTS are written on the BOTTOM in the denominator
  • Each of the products and reactants are raised to an exponent equal to the respective coefficient of each reagent in the balanced reaction.

Sample equilibrium constant expressions:

1) 2 NO2(g) N2O4(g)

The product, N2O4, goes on the top, and the reactant, NO2, goes on the bottom. The NO2 is raised to the second power (squared) since it has a coefficient of 2:

2) 2 SO3(g) 2 SO2(g) + O2(g)

The products, SO2 and O2, go on the top, and the reactant, SO3, goes on the bottom.  The SO2 and SO3 are both raised to the second power (squared) because they both have coefficients of 2:

3) N2(g) + 3 H2(g) 2 NH3(g)

The product, NH3, goes on the top, and the reactants, N2 and H2 goes on the bottom.  The H2 is raised to the third power (cubed) since it has a coefficient of 3.  The NH3 is raised to the second power (squared) since it has a coefficient of 2:

Reaction Quotient (Qc) - The mathematical product of the concentrations of the products of a reaction divided by the mathematical product of the concentrations of the reactants of a reaction AT ANY MOMENT IN TIME.  That is, this is basically the equilibrium constant expression, but when the reaction does not necessarily have to be at equilibrium.

The reaction quotient can be used to determine in which direction a system must shift in order to reach equilibrium.  There are 3 possible situations:

  1. Qc < Kc - This occurs if there is too much reactant and not enough product.  In order to raise Qc to equilibrium (Kc), some of the reactants must be converted to products, thereby decreasing the denominator and increasing the numerator.  In other words, the system must SHIFT RIGHT.
  2. Qc = Kc - The system is at EQUILIBRIUM.
  3. Qc > Kc  - This occurs if there is too much product and not enough reactant.  In order to lower Qc to equilibrium (Kc), some of the products must be converted back to reactants, thereby decreasing the numerator and increasing the denominator.  In other words, the system must SHIFT LEFT.

The relative difference between Qc and Kc indicates also how far a reaction must go to reach equilibrium.  If the difference is large, the reaction must shift a lot.  If the difference is relatively small, then the reaction will only shift slightly.

Sample Gas-Equilibrium Calculations

The method of solving gas-equilibrium calculations is much like how weak acid calculations are solved.

Summary of How to Solve Gas-Equilibrium Calculations

  1. Check the difference between Qc and Kc.
    • If the difference is relatively small, proceed with the problem as follows.
    • If the difference is relatively large, the following method will most likely not work.  See below for how to solve equilibrium problems when the assumption fails.
  2. Set up a concentration chart.
    • Assign as the unknown amount of reactant that is converted to products.
      • Therefore the equilibrium concentration of the reactants is:  initial - X, where x is the respective coefficients of the reactants.
    • Xis also the value for the the concentration of the products formed, where x is the respective coefficients of the products
  3. Write the gas equilibrium constant expression (Kc).
  4. Substitute
  5. Assume that is negligible when subtracted.
  6. Solve for
  7. Check to make sure the assumption was valid.
    • If the assumption fails then we must solve for by first shifting the reaction completely to the right and coming back to equilibrium by shifting slightly left.

1)

Assume the initial concentrations of both N2 and H2 are 0.100 M.  Calculate the equilibrium concentrations of N2, H2, and NH3 at 500C if Kc at this temperature is 0.040.

Check the DIFFERENCE between Qc and Kc.  Qc is initially 0 since no products have been formed yet.  0 < 0.040, but not by a great margin, so it we can proceed with the problem.

Set up the concentration CHART:

    [N2]     [H2]     [NH3 ]
Initial:

0.100 M

0.100 M

0
Equilibrium: 0.100 -

0.100 - 3

2

Set up the EQUILIBRIUM CONSTANT EXPRESSION (Qc) and SUBSTITUTE in the values from the chart and the given:

ASSUME that and 3are negligible when subtracted from the initial concentrations of the reactants:

SOLVE for :

CHECK to make sure that less than 5% of the reactants were converted to products:

Since our assumption was valid, we plug back into the values on our chart to find the equilibrium concentrations:

What if the assumption fails?

The above method will generally work for reactions in which the difference between Qc and Kc is small.  That is, the assumption will check out to be valid in the end.  If, however, the difference between Qc and Kc is quite large, then we cannot assume that the reaction will only proceed in the forward direction by a little bit.

2)

The initial concentration of NO is 0.100 M and the initial concentration of O2 is 0.050 M.  Calculate the equilibrium concentrations of NO, O2 and NO2 if the reaction takes place at 200C and the Kc at this temperature is 3 x 106.

Check the DIFFERENCE between Qc and Kc.  Qc is initially 0 since no products have been formed yet.  0 is much less than 3 x 106, so we can assume that the reaction has to shift a lot to reach equilibrium.  We cannot use the same method used above.  Instead we are going to push the reaction all the way to the right so that now there is an intermediate stage where the concentration of the product is 0.100 M (by mole ratio) and the concentrations of the reactants are 0:

    [NO]     [O2]     [NO2]
Initial: 0.100 M 0.050 M

0
Intermediate:

0

0

 0.100 M

Now, from the intermediate state, we can set up a chart with the intermediate and equilibrium concentrations.  Remember that the coefficients in the balanced reaction are the same coefficients in front of delta according to the mole ratios:

    [NO]     [O2]     [NO2]
Intermediate:

0

0

 0.100 M
Equilibrium:

2

0.100 - 2

Now we can set up the EQUILIBRIUM CONSTANT EXPRESSION (Qc) and SUBSTITUTE in the values from the chart and the given:

ASSUME that 2is negligible when subtracted from the intermediate concentration of the product:

SOLVE for :

CHECK to make sure that less than 5% of the product was converted back to reactants:

Since our assumption was valid, we can find the equilibrium concentrations by plugging in to the values on the chart:

Summary of How to Solve Gas-Equilibrium Calculations
if Qc is much smaller than Kc

  1. Force the reaction completely to the right, creating an INTERMEDIATE stage.
  2. Set up a concentration chart with the initial and intermediate concentrations.
    • The intermediate concentration of the products are determined by the mole ratios of the balanced reaction.
    • Assign as the unknown amount of product converted into reactants.
      • Therefore the equilibrium concentration of the products is:  intermediate - X, where x is the respective coefficients of the products.
    • Xis also the value for the the concentration of the reactants formed, where x is the respective coefficients of the reactants.
  3. Write the gas equilibrium constant expression (Kc).
  4. Substitute
  5. Assume that is negligible when subtracted.
  6. Solve for
  7. Check to make sure the assumption was valid.

Gas Equilibrium in Terms of Partial Pressure

Another way to express the gas equilibrium constant expression is with partial pressures of the gases (Kp) in the reaction instead of the concentrations.

For this reaction, the Kc and Kp expressions are:

The relationship between Kc and Kp is described in the following equation:

NOTE:  When there are equal moles of gas reactants and gas products, n = 0. Any number raised to the 0th power is 1.

 Therefore Kc = Kp when the moles of gas products = the moles of gas reactants.

Partial Pressure - The fraction of the total pressure of a mixture of gases that is due to one component of the mixture.

Reaction Quotient (Qp) - The mathematical product of the partial pressures of the products of a reaction divided by the mathematical product of the partial pressures of the reactants of a reaction AT ANY MOMENT IN TIME.  This can be used to determine in which direction a reaction will shift to reach equilibrium by comparing the value of Qp to Kp.  Qp follows the same guidelines as Qc descrbibed above, just with partial pressures instead of concentrations.

Consider the reaction:

Calculate the partial pressures of each of the gases if the total pressure is 2.0 atom.

Calculate the mole fraction of each gas and multiply this by the total pressure to get the partial pressure of each gas  The mole fraction of a gas is the moles of the one gas divided by the total moles of gas in the system:

Note that the sum of the partial pressures of all the gases equals the total pressure:

Gas Equilibrium Calculations with Kp are solved exactly as Kc problems are solved, just using partial pressures instead of concentrations.

Next:  "LeChâtelier's Principle"