Effect of Adding a Reactant or Product

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 14:46

If we have a system which is already in equilibrium, addition of an extra amount of one of the reactants or one of the products throws the system out of equilibrium. Either the forward or the reverse reaction will then occur in order to restore equilibrium conditions. We can easily tell which of these two possibilities will happen from Le Chatelier's principle. If we add more of one of the products, the system will adjust in order to offset the gain in concentrationA measure of the ratio of the quantity of a substance to the quantity of solvent, solution, or ore. Also, the process of making something more concentrated. of this component. The reverse reaction will occur to a limited extent so that some of the added product can be consumed. Conversely, if one of the reactants is added, the system will adjust by allowing the forward reaction to occur to some extent. In either case some of the added component will be consumed.

We see this principle in operation in the case of the decomposition of HI at high temperatures:

      2HI(g) \rightleftharpoons H2(g) + I2(g)

In Example 1 from Calculating the Extent of a Reaction we saw that if 1 mol HI is heated to 745 K in a 10-dm3 flask, some of the HI will decompose, producing an equilibrium mixtureA combination of two or more substances in which the substances retain their chemical identity. of composition

1 [HI] = 0.0780 mol dm–3; [I2] = 0.0110 mol dm–3; [H2] = 0.0110 mol dm–3

This is a genuine equilibrium mixture since it satisfies the equilibrium law

      K_{c}=\frac{\text{ }\!\![\!\!\text{ H}_{\text{2}}\text{ }\!\!]\!\!\text{  }\!\![\!\!\text{ I}_{\text{2}}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ HI }\!\!]\!\!\text{ }^{\text{2}}}=\frac{\text{(0}\text{.011 mol dm}^{-\text{3}}\text{)(0}\text{.011 mol dm}^{-\text{3}}\text{)}}{\text{(0}\text{.078 mol dm}^{-\text{3}}\text{)}^{\text{2}}\text{ }}=\text{0}\text{.020}=K_{c}

If an extra moleThat chemical amount of a substance containing the same number of units as 12 g of carbon-12. of H2 is added to this mixture, the concentrations become

2 cHI = 0.0780 mol dm–3; cI2 = 0.0110 mol dm–3; cH2 = 0.110 mol dm–3

The system is no longer in equilibrium (hence the lack of square brackets to denote equilibrium concentrations) as we can easily check from the equilibrium law

      \frac{c_{\text{H}_{\text{2}}}\text{ }\times \text{ }c_{\text{I}_{\text{2}}}}{c_{\text{HI}}^{\text{2}}}=\frac{\text{0}\text{.011 mol dm}^{-\text{3}}\text{ }\times \text{ 0}\text{.111 mol dm}^{-\text{3}}}{\text{(0}\text{.078 mol dm}^{-\text{3}}\text{)}^{\text{2}}}=\text{0}\text{.201}\ne K_{c}

The addition of H2 has increased the concentration of this component. Accordingly, Le Chatelier’s principle predicts that the system will achieve a new equilibrium in such a way as to reduce this concentration. The reverse reaction occurs to a limited extent. This not only reduces the concentration of H2 but the concentration of I2 as well. At the same time the concentration of HI is increased. The system finally ends up with the concentrations calculated in Example 2 from Calculating the Extent of a Reaction, namely,

      H2+ I2 → 2HI

Figure 1 Le Chatelier’s principle: effect of adding a component. At 745 K, HI is partially decomposed into H2 and I2: 2HI \rightleftharpoons H2 + I2. If extra hydrogen (gray) is added to the equilibrium mixture, the system responds in such a way as to reduce the concentration of H2. Some I2 reacts with the H2, and more HI is formed. The equilibrium is shifted to the left. Note, however, that some of the I2 has been consumed, and its concentration is smaller than before.

3 [HI] = 0.0963 mol dm–3; [I2] = 0.001 82 mol dm–3; [H2] = 0.1018 mol dm–3

This is again an equilibrium situation since it conforms to the equilibrium law

\frac{\text{ }\!\![\!\!\text{ H}_{\text{2}}\text{ }\!\!]\!\!\text{  }\!\![\!\!\text{ I}_{\text{2}}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ HI }\!\!]\!\!\text{ }^{\text{2}}}=\frac{\text{0}\text{.1018 mol dm}^{-\text{3}}\text{ }\times \text{ 0}\text{.001 82 mol dm}^{-\text{3}}}{\text{(0}\text{.0963 mol dm}^{-\text{3}}\text{)}^{\text{2}}\text{ }}=\text{0}\text{.02}=K_{c}

The way in which this system responds to the addition of H2 is also illustrated schematically in Fig. 1. The actual extent of the change is exaggerated in this figure for diagrammatic effect.

Le Chatelier's principle can also be applied to cases where one of the components is removed. In such a case the system responds by producing more of the component removed. Consider, for example, the ionizationA process in which an atom, molecule, or negative ion loses an electron; a process in which a covalent molecule reacts with a solvent to form positive and negative ions; for example, a weak acid reacting with water to form its conjugate base (an anion) and a hydrogen (hydronium) ion. of the weak diproticDescribes an acid that can donate two hydrogen ions (protons) to a base. acidIn Arrhenius theory, a substance that produces hydrogen ions (hydronium ions) in aqueous solution. In Bronsted-Lowry theory, a hydrogen-ion (proton) donor. In Lewis theory, a species that accepts a pair of electrons to form a covalent bond. H2S:

      H2S + 2H2O \rightleftharpoons 2H3O+ + S2–

Since H2S is a weak acidAn acid that ionizes only partially in a given solvent., very few S2– ions are produced, but a much larger concentration of S2–ions can be obtained by adding a strong baseA base that dissociates completely or ionizes completely in a particular solvent.. The base will consume most of the H3O+ ions. As a result, more H2S will react with H2O in order to make up the deficiency of H3O+, and more S2–ions will also be produced. This trick of removing one of the products in order to increase the concentration of another product is often used by chemists, and also by living systems.

EXAMPLE 1 When a mixture of 1 mol N2 and 3 mol H2 is brought to equilibrium over a catalystA substance that increases the rate of a chemical reaction but that undergoes no net change during the reaction. at 773 K (500°C) and 10 atmAbbreviation for atmosphere, a unit of pressure equal to 101.325 kPa or 760 mmHg. (1.01 MPa), the mixture reacts to form NH3 according to the equation

      N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)      ΔHm = – 94.3 kJ

The yield of NH3, however, is quite small; only about 2.5 percent of the reactants are converted. Suggest how this yield could be improved (a) by altering the pressureForce per unit area; in gases arising from the force exerted by collisions of gas molecules with the wall of the container.; (b) by altering the temperature; (c) by removing a component; (d) by finding a better catalyst.

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture.

a) Increasing the pressure will drive the reaction in the direction of fewer molecules. Since Δn = - 2, the forward reaction will be encouraged, increasing the yield of NH3.

b) Increasing the temperature will drive the reaction in an endothermicIn chemical thermodynamics, describes a process in which energy is transferred from the surroundings to the system as a result of a temperature difference. direction, in this case in the reverse direction. In order to increase the yield, therefore, we need to lower the temperature.

c) Removing the product NH3 will shift the reaction to the right. This is usually done by cooling the reaction mixture so that NH3(l) condenses out. Then more N2(g) and H2(g) are added, and the reaction mixture is recycled to a condition of sufficiently high temperature that the rate becomes appreciable.

d) While a better catalyst would speed up the attainment of equilibrium, it would not affect the position of equilibrium. It would therefore have no effect on the yield.

Note: As mentioned in Chaps. 3 and 12, NH3 is an important chemical because of its use in fertilizers. In the design of a Haber-process plant to manufacture ammonia, attempts are made to use as high a pressure and as low a temperature as possible. The pressure is usually of the order of 150 atm (15 MPa), while the temperature is not usually below 750 K. Although a lower temperature would give a higher yield, the reaction would go too slowly to be economical, at least with present-day catalysts. The discoverer of a better catalyst for this reaction would certainly become a millionaire over-night.