**James Richard Fromm**

The effect of adding reactants to any gas-phase system at chemical equilibrium is to increase the partial pressures of the products, according to Le Chatelier's principle. The equilibrium partial pressures of the reactants will be less than the sum of the original equilibrium partial pressures and the partial pressures added, because some of the added reactants will be removed by reaction, but they will still be greater than the equilibrium partial pressures of the reactants before the addition.

Example. At 25^{o}C, the value of *K*_{p} for the gas-phase
equilibrium:

H_{2}(g) + I_{2}(g) 2HI(g)

has the numeric value of 617.1. (This value is calculated from thermodynamic
properties of pure substances as an example in another section.) Compute the partial
pressure of HI at equilibrium in a situation where the equilibrium partial pressures of
hydrogen and iodine are each 100 kPa. Then compute the equilibrium partial pressures of I_{2}
and HI when the partial pressure of hydrogen is increased so that the equilibrium partial
pressure of hydrogen is 150 kPa.

The initial partial pressures are hydrogen, 100 kPa; iodine, 100 kPa; and HI, 2484 kPa
since the equilibrium constant is 617.1 = *p*^{2}(HI)/*p*(H_{2})*p*(I_{2})
= *p*^{2}(HI)/(100)^{2}

*p*^{2}(HI) = 6.171 x 10^{+6} kPa^{2}

The equilibrium partial pressures after the increase are hydrogen, 150 kPa, with the values of iodine and HI to be calculated. For every kPa decrease in partial pressure of iodine, x, the partial pressure of HI will increase by 2 kPa since two moles of HI are formed for each mole of iodine lost.

617.1 = (2484 + 2x)^{2}/(150)(100 - x)

This equation could be solved by the quadratic formula, but a simpler approach is possible. Even if all of the added hydrogen went to form HI, the value of x could be only 100 kPa; twice 100 kPa is small compared to 2484 kPa. We therefore write, as a good approximation, 617.1 = 24842/(15000 - 150x) which yields a value of 33.3 kPa for x. The partial pressure of iodine, then, decreases by 33.3 kPa to 66.6 kPa while the equilibrium partial pressure of hydrogen iodide increases by twice 33.3, or 66.6, kPa to 2550.6 kPa. Using the quadratic formula gives a value of x of 30.1 kPa, which corresponds to a partial pressure of iodine of 69.9 kPa and a partial pressure of HI of 2544.2 kPa.

Example. Using the same equilibrium and the same initial conditions as in the previous
example, compute the effect on the partial pressures of H_{2}, HI, and I_{2}
at equilibrium when the partial pressure of hydrogen is increased so that the partial
pressure of hydrogen would have been 150 kPa if no equilibrium reaction had taken place.

The initial partial pressures are again hydrogen, 100 kPa; iodine, 100 kPa; and HI, 2484 kPa. The equilibrium partial pressure of hydrogen will now be 150 - y kPa, that of iodine will be 100 - y kPa, and that of HI will be 2484 + 2y kPa.

617.1 + (2484 + 2y)^{2}/(150 - y)(100 - y)

Making the same approximation as in the previous example, 617.1 = 24842/(15000 - 250y +
y^{2})

However, solution by the quadratic formula would still be required in this case. One
could argue, however, that the amount of reaction y cannot be much larger than it was in
the previous example, if indeed it is larger at all. Then 250y must be much larger than y^{2},
so 617.1 = 24842/(15000 - 250y) and y has the value 20.0 kPa. The equilibrium partial
pressure of hydrogen is 130 kPa, that of iodine is 80 kPa, and that of hydrogen iodide is
2524 kPa. Use of the quadratic formula without either of the approximations used above
gives a value of y of 20.3 kPa, which corresponds to an equilibrium partial pressure of
129.7 kPa for hydrogen, 79.7 kPa for iodine, and 2524.7 kPa for hydrogen iodide. The
approximations cause very little difference in the result of the calculation.

When the value of an equilibrium constant is of moderate size, as the equilibrium constant for the gas-phase hydrogen iodide reaction is at other temperatures, a quadratic or higher equation and some algebra cannot be avoided in equilibrium calculations. When the value of the equilibrium constant is either very large or very small, the calculations often become much simpler because the reaction can realistically be assumed to go all the way to completion, if the value of the equilibrium constant is very large, or to go not at all, if the value of the equilibrium constant is very small. Many of the aqueous equilibria taken up in other sections can be treated in this way.

For equilibrium constants which have no nominal units, such as the constant describing
the equilibrium 2HI <--> H_{2} + I_{2}, a value of 100 or more may
be considered very large and a value of 0.01 or less may be considered very small. When
equilibrium constants do have nominal units, their magnitude and the meaning of "very
large" and "very small" depends upon the nominal units they have.

Example. At 25^{o}C, the gas-phase equilibrium constant for the formation of
ammonia by the equilibrium N_{2} + 3H_{2} <--> 2NH_{3} has
the value 5.807 x 10^{-5}, with the nominal units Pa^{-2}. (This value is
calculated from thermodynamic properties of pure substances as an
example in another section.) Compute the partial pressure of ammonia if the equilibrium
partial pressures of nitrogen and hydrogen are 150 kPa. Then compute the partial pressure
of ammonia if the initial partial pressures of nitrogen and hydrogen are 150 kPa and there
is no ammonia initially present.

The equilibrium partial pressure of nitrogen is 1.5 x 10^{+5} Pa, as is that of
hydrogen. The partial pressure of ammonia at equilibrium must be given by:

5.807 x 10^{-5} Pa^{-2} = *p*^{2}(NH_{3})/*p*(N_{2})*p*^{3}(H_{2})

5.807 x 10^{-5} Pa^{-2} = *p*^{2}(NH_{3})/(1.5
x 10^{+5})^{4}

*p*^{2}(NH_{3}) = 2.940 x 10^{+16} Pa^{2};
*p*(NH_{3}) = 1.715 x 10^{+8} Pa = 171,500 kPa

The initial partial pressure of nitrogen is 1.5 x 10^{+5} Pa, so the
equilibrium partial pressure of nitrogen will be 1.5 x 10^{+5} - x, where x is the
nitrogen partial pressure decrease due to the equilibrium reaction. The equilibrium
partial pressure of hydrogen will be 1.5 x 10^{+5} - 3x, since three moles of
hydrogen are required for each mole of nitrogen. The equilibrium partial pressure of
ammonia will be 2x, since all of the ammonia is formed by reaction.

5.807 x 10^{-5} Pa^{-2} = (2x)^{2}/(1.5 x 10^{+5}
- x)(1.5 x 10^{+5} - 3x)^{3}

Mathematical solution of a fourth-order equation such as this would be practical only
by the method of successive approximations. An approximate solution is much easier. Since
the equilibrium lies heavily on the ammonia side at 25^{o}C, nearly all of the
hydrogen, 1.5 x 10^{+5} Pa, will have reacted to form ammonia since hydrogen is
the limiting reactant. The partial pressure of nitrogen will be 1.5 x 10^{+5} -
0.5 x 10^{+5} Pa, since only one-third of the amount initially present can be
consumed in the reaction. The partial pressure of ammonia will be the maximum possible,
100000 Pa, which can be produced from that amount of nitrogen (each mole of N_{2}
yields two moles of NH_{3}). The equilibrium expression gives:

5.807 x 10^{-5} Pa^{-2} = (1 x 10^{+5} Pa)^{2}/(1.0
x 10^{+5} Pa)(y Pa)^{3}

where y is the partial pressure of hydrogen. This will be the hydrogen not lost by
reaction to form ammonia according to the equilibrium stoichiometry. Then y^{3} =
1.722 x 10^{+9} Pa^{3}. The equilibrium partial pressures are ammonia,
100000 Pa; hydrogen, 1199 Pa; and nitrogen, 100000 Pa. The partial pressure of hydrogen is
negligible compared to that of ammonia.

The effects of adding products to any gas-phase system at chemical equilibrium is to decrease the partial pressures of the reactants, according to Le Chatelier's principle. The pressure which is the sum of the added products and the original equilibrium products will be decreased, and the original equilibrium pressures of the reactants will be increased, once the new position of equilibrium has been attained.

Example. Let us first calculate the equilibrium partial pressure of NO_{2} in a
mixture of N_{2}O_{4} and NO_{2} when the equilibrium partial
pressure of N_{2}O_{4} is 100 kPa. If the partial pressure of NO_{2}
is then increased by addition of 10 kPa of NO_{2}, we can compute what the new
equilibrium partial pressures of NO_{2} and of N_{2}O_{4} would
be.

Since the equilibrium is:

N_{2}O_{4}(g) 2NO_{2}(g),

the original equilibrium partial pressure of NO_{2} will be given by:

*K*_{p} = 1.484 x 10^{+4} Pa = *p*^{2}(NO_{2})/*p*(N_{2}O_{4})

*p*^{2}(NO_{2}) = (1.484 x 10^{+4})(100000),
*p*(NO_{2}) = 38.52 kPa

The equilibrium lies so heavily on the side of N_{2}O_{4} that addition
of 10 kPa of NO_{2} would cause reaction to the extent of almost 5 kPa of N_{2}O_{4}.
The partial pressure of N_{2}O_{4} at equilibrium would then be about 105
kPa and that of NO_{2} will be given by:

*p*^{2}(NO_{2}) = (1.484 x 10^{+4})(105000),
*p*(NO_{2}) = 39.47 kPa

Use of a quadratic equation or successive approximations shows that the reaction goes
heavily, but not totally, toward N_{2}O_{4} and the actual equilibrium
partial pressure of NO_{2} is a bit greater than 39.47 kPa. The addition of
product NO_{2} will result in the formation of more reactant N_{2}O_{4},
with only a small part of the addition remaining as N_{2}O_{4}.

Copyright 1997 James R. Fromm