Presentation is loading. Please wait.

Presentation is loading. Please wait.

Unit 3 Equilibrium. HIGHER CHEMISTRY REVISION. Unit 3 :- Equilibrium 1. If both potassium iodide solution, KI(aq), and liquid chloroform, CHCl 3 (l),

Similar presentations


Presentation on theme: "Unit 3 Equilibrium. HIGHER CHEMISTRY REVISION. Unit 3 :- Equilibrium 1. If both potassium iodide solution, KI(aq), and liquid chloroform, CHCl 3 (l),"— Presentation transcript:

1 Unit 3 Equilibrium

2 HIGHER CHEMISTRY REVISION. Unit 3 :- Equilibrium 1. If both potassium iodide solution, KI(aq), and liquid chloroform, CHCl 3 (l), are added to a test tube with some iodine, the iodine dissolves in both. Two layers are formed as shown in the diagram. An equilibrium is set up: I 2 in KI(aq)  I 2 in CHCl 3 (l) The iodine is always distributed between the two layers in the same ratio: Concentration of I 2 in KI(aq) 3 = Concentration of I 2 in CHCl 3 (l)1 (a) What is meant by the term equilibrium? (b) When more potassium iodide solution is added to the top layer the equilibrium is disturbed. What happens to restore the equilibrium? (c) 0.4 g of iodine is dissolved in 10 cm 3 of KI(aq) and 10 cm 3 of CHCl 3 (l). Calculate the concentration of iodine, in g l -1, contained in CHCl 3 (l). (b) Adding potassium iodide solution decreases the concentration of iodine in the top layer. Iodine moves from the bottom layer to the top layer until the equilibrium is re-established. (c) At equilibrium there is 0.2g of iodine in 10cm 3 of chloroform. = 20 g in 1 litre. (a) A reaction is in equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction.

3 2. Consider the following industrial processes. Contact Process 2SO 2 (g) + O 2 (g)  2SO 3 (g)  H -ve Haber Process N 2 (g) + 3H 2 (g)  2NH 3 (g)  H -ve (a)For each process, circle the reactant that can be described as a raw material. (b) Explain why increasing the temperature in both processes decreases the equilibrium yield of the products. (c) Suggest why the Contact Process is carried out at atmospheric pressure but the Haber Process is carried out at 400 atmospheres. (d) Under certain conditions, 200 kg of hydrogen react with excess nitrogen in the Haber Process to produce 650 kg of ammonia. Calculate the percentage yield of ammonia (a)O 2 in Contact process, N 2 in Haber Process. (b)Applying Le Chatelier’s Principle, an increase in temperature will favour the reaction which would counteract the change ie it favours the endothermic step. In both reaction it is the back reaction which is endothermic and so the yield of product will decrease. (c)The Contact Process gives a high yield at atmospheric pressure whereas a high yield is only obtained in the Haber Process at high pressure. (d)N 2 (g) + 3H 2 (g)  2NH 3 (g) 3 mol  2 molat 100% yield 6 g  34gat 100% yield So 200 kg  200 / 6 x 34 kg = kg % yield = 650 / x 100 = 57.35%

4 3. Identify the two statements which can be applied to the role of a catalyst in a reversible reaction. AIt decreases the enthalpy change for the reaction. BIt decreases the time taken for equilibrium to be established. CIt alters the position of the equilibrium. DIt lowers the activation energy of the backward reaction. EIt increases the rate of the forward reaction more than the backward reaction. Statements in boxes B and D 4. Steam reforming of coal produces a mixture of carbon monoxide and hydrogen. (a)What name is given to this mixture of carbon monoxide and hydrogen? (b)This mixture could be used to produce methane, as shown by the following equilibrium. CO(g) + 3H 2 (g)  CH 4 (g) + H 2 O(g)  H = –206 kJ mol -1. Give two reasons why the yield of methane can be increased by cooling the reaction mixture from 400°C to 80°C. (a)Synthesis gas. (b)Decreasing the temperature favours the exothermic process – in this case the forward reaction which will increase the yield of methane. By cooling to 80 o C the water is turned from a gas into a liquid. Removing the steam favours the forward reaction.

5 5. The balanced equation for a reaction at equilibrium is: a) For this reaction, the equilibrium constant, K, can be defined as: where [A] represents the concentration of A, etc and a represents the number of moles of A, etc. (i) Write down the expression for the equilibrium constant for the following equilibrium. N 2 (g) + 3H 2 (g)  2NH 3 (g) (ii) What will happen to the position of the equilibrium if the reaction is carried out over a catalyst? (b) In industry, the reaction of nitrogen with hydrogen to produce ammonia by the Haber Process does not attain equilibrium. Give one feature of the operating conditions which leads to the Haber Process not reaching equilibrium (a) (i) [NH 3 ] 2 [N 2 ] [H 2 ] 3 (ii) No change in equilibrium position. (b)Unreacted nitrogen and hydrogen are recycled and ammonia gas is liquefied.

6 6. If the conditions are kept constant, reversible reactions will attain a state of equilibrium. (a) Circle the correct words in the table to show what is true for reactions at equilibrium. Rate of forward reaction compared to rate of reverse reaction.faster/ slower / same Concentrations of reactants compared to concentrations of products.Usually different / always the same (b) The following equilibrium involves two compounds of phosphorus. PCl 3 (g) + 3NH 3 (g)  P(NH 2 ) 3 (g) + 3HCl(g) (i) An increase in temperature moves the equilibrium to the left. What does this indicate about the enthalpy change for the forward reaction? (ii) What effect, if any, will an increase in pressure have on the equilibrium? (a)See box above. (b) (i) The forward reaction is exothermic. (ii) No effect.


Download ppt "Unit 3 Equilibrium. HIGHER CHEMISTRY REVISION. Unit 3 :- Equilibrium 1. If both potassium iodide solution, KI(aq), and liquid chloroform, CHCl 3 (l),"

Similar presentations


Ads by Google