# 13.3 Shifting equilibria: le châtelier’s principle  (Page 2/10)

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${\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)⇌2\text{HI}\left(g\right)\phantom{\rule{5em}{0ex}}{K}_{c}=50.0\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}400\phantom{\rule{0.2em}{0ex}}\text{°C}$

The numeric values for this example have been determined experimentally. A mixture of gases at 400 °C with [H 2 ] = [I 2 ] = 0.221 M and [HI] = 1.563 M is at equilibrium; for this mixture, Q c = K c = 50.0. If H 2 is introduced into the system so quickly that its concentration doubles before it begins to react (new [H 2 ] = 0.442 M ), the reaction will shift so that a new equilibrium is reached, at which [H 2 ] = 0.374 M , [I 2 ] = 0.153 M , and [HI] = 1.692 M . This gives:

${Q}_{c}=\phantom{\rule{0.2em}{0ex}}\frac{{\left[\text{HI}\right]}^{2}}{\left[{\text{H}}_{2}\right]\phantom{\rule{0.2em}{0ex}}\left[{\text{I}}_{2}\right]}=\phantom{\rule{0.2em}{0ex}}\frac{{\left(1.692\right)}^{2}}{\left(0.374\right)\left(0.153\right)}\phantom{\rule{0.2em}{0ex}}=50.0={K}_{c}$

We have stressed this system by introducing additional H 2 . The stress is relieved when the reaction shifts to the right, using up some (but not all) of the excess H 2 , reducing the amount of uncombined I 2 , and forming additional HI.

## Effect of change in pressure on equilibrium

Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system. An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for K c ) or partial pressure (for K P ). Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium.

As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Châtelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure.

Consider what happens when we increase the pressure on a system in which NO, O 2 , and NO 2 are at equilibrium:

$2\text{NO}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌2{\text{NO}}_{2}\left(g\right)$

The formation of additional amounts of NO 2 decreases the total number of molecules in the system because each time two molecules of NO 2 form, a total of three molecules of NO and O 2 are consumed. This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a decrease in the pressure on the system favors decomposition of NO 2 into NO and O 2 , which tends to restore the pressure.

Now consider this reaction:

${\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)⇌2\text{NO}\left(g\right)$

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