3.4 Le chateliers principle  (Page 4/4)

Industrial applications

The Haber process is a good example of an industrial process which uses the equilibrium principles that have been discussed. The equation for the process is as follows:

Since the reaction is exothermic , the forward reaction is favoured at low temperatures, and the reverse reaction at high temperatures. If the purpose of the Haber process is to produce ammonia, then the temperature must be maintained at a level that is low enough to ensure that the reaction continues in the forward direction.

The forward reaction is also favoured by high pressures because there are four moles of reactant for every two moles of product formed.

The K value for this reaction will be calculated as follows:

Applying equilibrium principles

Look at the values of k calculated for the Haber process reaction at different temperatures, and then answer the questions that follow:

1. What happens to the value of K as the temperature increases?
2. Which reaction is being favoured when the temperature is 300 degrees celsius?
3. According to this table, which temperature would be best if you wanted to produce as much ammonia as possible? Explain.

Summary

• The rate of a reaction describes how quickly reactants are used up, or how quickly products form. The units used are moles per second.
• A number of factors can affect the rate of a reaction. These include the nature of the reactants , the concentration of reactants, temperature of the reaction, the presence or absence of a catalyst and the surface area of the reactants.
• Collision theory provides one way of explaining why each of these factors can affect the rate of a reaction. For example, higher temperatures mean increased reaction rates because the reactant particles have more energy and are more likely to collide successfully with each other.
• Different methods can be used to measure the rate of a reaction . The method used will depend on the nature of the product. Reactions that produce gases can be measured by collecting the gas in a syringe. Reactions that produce a precipitate are also easy to measure because the precipitate is easily visible.
• For any reaction to occur, a minimum amount of energy is needed so that bonds in the reactants can break, and new bonds can form in the products. The minimum energy that is required is called the activation energy of a reaction.
• In reactions where the particles do not have enough energy to overcome this activation energy, one of two methods can be used to facilitate a reaction to take place: increase the temperature of the reaction or add a catalyst.
• Increasing the temperature of a reaction means that the average energy of the reactant particles increases and they are more likely to have enough energy to overcome the activation energy.
• A catalyst is used to lower the activation energy so that the reaction is more likely to take place. A catalyst does this by providing an alternative, lower energy pathway, for the reaction.
• A catalyst therefore speeds up a reaction but does not become part of the reaction in any way.
• Chemical equilibrium is the state of a reaction, where the concentrations of the reactants and the products have no net change over time. Usually this occurs when the rate of the forward reaction is the same as the rate of the reverse reaction.
• The equilibrium constant relates to reactions at equilibrium, and can be calculated using the following equation:
  $$K_c=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}$$
  where A and B are reactants, C and D are products and a, b, c, and d are the coefficients of the respective reactants and products.
• A high K ${}_c$ value means that the concentration of products at equilibrium is high and the reaction has a high yield. A low K ${}_c$ value means that the concentration of products at equilibrium is low and the reaction has a low yield.
• Le Chatelier's Principle states that if a chemical system at equilibrium experiences a change in concentration, temperature or total pressure the equilibrium will shift in order to minimise that change and to re-establish equilibrium. For example, if the pressure of a gaseous system at eqilibrium was increased, the equilibrium would shift to favour the reaction that produces the lowest quantity of the gas. If the temperature of the same system was to increase, the equilibrium would shift to favour the endothermic reaction. Similar principles apply for changes in concentration of the reactants or products in a reaction.
• The principles of equilibrium are very important in industrial applications such as the Haber process, so that productivity can be maximised.

Summary exercise

1. For each of the following questions, choose the one correct answer from the list provided.
   1. Consider the following reaction that has reached equilibrium after some time in a sealed 1 dm ${}^3$ flask: $PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$ ; $\Delta H$ is positive Which one of the following reaction conditions applied to the system would decrease the rate of the reverse reaction?
      1. increase the pressure
      2. increase the reaction temperature
      3. continually remove Cl ${}_2$ (g) from the flask
      4. addition of a suitable catalyst
      (IEB Paper 2, 2001)
   2. The following equilibrium constant expression is given for a particular reaction: $K_c={\left[H_{2}O\right]}^4{\left[C{O}_2\right]}^3/\left[C_3{H}_8\right]{\left[O_2\right]}^5$ For which one of the following reactions is the above expression of K ${}_c$ is correct?
      1. $C_3{H}_8\left(g\right)+5O_2\left(g\right)\rightleftharpoons 4H_{2}O\left(g\right)+3C{O}_2\left(g\right)$
      2. $4H_{2}O\left(g\right)+3C{O}_2\left(g\right)\rightleftharpoons C_3{H}_8\left(g\right)+5O_2\left(g\right)$
      3. $2C_3{H}_8\left(g\right)+7O_2\left(g\right)\rightleftharpoons 6CO\left(g\right)+8H_{2}O\left(g\right)$
      4. $C_3{H}_8\left(g\right)+5O_2\left(g\right)\rightleftharpoons 4H_{2}O\left(l\right)+3C{O}_2\left(g\right)$
      (IEB Paper 2, 2001)
2. 10 g of magnesium ribbon reacts with a 0.15 mol.dm ${}^{-3}$ solution of hydrochloric acid at a temperature of 25 ${}^0$ C.
   1. Write a balanced chemical equation for the reaction.
   2. State two ways of increasing the rate of production of H ${}_2$ (g).
   3. A table of the results is given below:

      Time elapsed (min)	Vol of H ${}_2$ (g) (cm ${}^3$ )
      0	0
      0.5	17
      1.0	25
      1.5	30
      2.0	33
      2.5	35
      3.0	35

      1. Plot a graph of volume versus time for these results.
      2. Explain the shape of the graph during the following two time intervals: t = 0 to t = 2.0 min and then t = 2.5 and t = 3.0 min by referring to the volume of H ${}_2$ (g) produced.
   (IEB Paper 2, 2001)
3. Cobalt chloride crystals are dissolved in a beaker containing ethanol and then a few drops of water are added. After a period of time, the reaction reaches equilibrium as follows: $CoC{l}_4^{2-}$ (blue) $+6H_{2}O\rightleftharpoons Co{\left(H_{2}O\right)}_6^{2+}$ (pink) $+4C{l}^{-}$ The solution, which is now just blue, is poured into three test tubes. State, in each case, what colour changes will be observed (if any) if the following are added in turn to each test tube:
   1. 1 cm ${}^3$ of distilled water
   2. A few crystals of sodium chloride
   3. The addition of dilute hydrochloric acid to the third test tube causes the solution to turn pink. Explain why this occurs.
   (IEB Paper 2, 2001)