1. AS 2.5 - Physical Chemistry Formative Assessments in Energy, Rates of Reaction, Equilibrium and Acid/Base Reactions

2. Overall Outline

3. Success Criteria Can classify a reaction as exothermic or endothermic Can classify several reactions correctly as exothermic or endothermic Can classify reactions as exothermic or endothermic from either descriptions, equations or energy profiles

4. Energy Change of StateHeat in or out? Evaporating Melting Freezing Condensing Sublimating State whether heat is taken in or given out in the following state changes. in out

5. Energy State whether the following are exothermic or endothermic reactions. ExampleExothermic or Endothermic Clothes drying outside Dehydrating CuSO 4 Combustion Photosynthesis Endothermic Exothermic Achieved Merit

6. ProcessExothermic or Endothermic Burning Wood Some ammonium chloride was dissolved in water. The temperature of the water changed from 19 o C to 10 o C CH 4 (g) + H 2 O (g) CO (g) + 3H 2 (g)  r H = + 206 kJ mol -1 State whether the following are exothermic or endothermic reactions Exothermic Endothermic Achieved Merit Excellence

7. Success Criteria Can carry out a single step calculation Can carry out a multi-step calculation and give the correct answer Can carry out a multi-step calculation and give the correct answer with appropriate sig.figures and unit

8. Octane is a key component in petrol, and burns according to the following equation: C 8 H 18 ( l ) + 12 1/ 2 O 2 (g)  8 CO 2 (g) + 9 H 2 O( l ) Δ r H =  5500 kJ mol  1 1.00 litre of Octane contains 6.12 moles of the fuel. Calculate the energy released when 1.00 litre of fuel is burnt. Achieved Energy Released = n x Δ r H = 6.12 x 5500 = 33 660 kJ

9. Achieved Merit Using hydrogen gas (H 2 ) as a fuel for cars, rather than octane, is often viewed as better for the environment. Calculate the mass of H 2 required to produce the same amount of energy as 1.00 litre of octane ( 33 660 kJ ). State your answer to 3 significant figures. H 2 (g) + 1/ 2 O 2 (g)  H 2 O(g) Δ r H =  286 kJ mol  1 n (H 2 ) = 33 660 / 286 = 118 mol m (H 2 ) = 117.7 x 2 = 235.4 Excellence = 235 g

10. 29.6 g of sodium hydroxide was dissolved in water and excess hydrochloric acid was added. Using the temperature increase and the heat capacity of water, it was calculated that 43.5 kJ of heat was released. Determine the enthalpy change,  rH, for the following reaction: NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) mol NaOH = 29.6 g 40.0 g mol -1 = 0.740 mol 0.740 mol reacts to produce 43.5 kJ 1 mol reacts to produce 43.5 = 58.82 kJ 0.740   rH = -58.8 kJ mol -1 Excellence Merit Achieved

11. 29.6 g of sodium hydroxide was dissolved in water and excess hydrochloric acid was added. Using the temperature increase and the heat capacity of water, it was calculated that 43.5 kJ of heat was released. NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) ΔrH = -58.8 kJ mol -1 What mass of sodium hydroxide is required to produce 150 kJ of energy? n (NaOH) = 150 kJ 58.8 kJ mol -1 = 2.55 mol m (NaOH) = 2.55 mol x 40.0 g mol -1 = 102.3 m = 102 g Achieved Merit Excellence

12. Carbohydrates are an important source of energy in our diet. Two common carbohydrates are glucose (C 6 H 12 O 6 ) and sucrose (C 12 H 22 O 11 ). The equation below shows the combustion of glucose to form carbon dioxide and water. C 6 H 12 O 6 (s) + 6O 2 (g) → 6CO 2 (g) + 6H 2 O(ℓ) ∆ r H = –2820 kJ mol –1 Calculate the enthalpy change, ∆ r H, when 100 g of glucose reacts to form carbon dioxide and water. M(C 6 H 12 O 6 ) = 180 g mol –1 n (glucose) = 150 / 180 = 0.556 mol ∆ r H = 0.556 x -2820 = -1567 ∆ r H = -1567 kJ mol -1 Achieved Merit Excellence

13. The equation below shows the combustion of sucrose to form carbon dioxide and water. C 12 H 22 O 11 (s) + 12O 2 (g) → 12CO 2 (g) + 11H 2 O(ℓ) When 150 g of sucrose, C 12 H 22 O 11, undergoes combustion, 2478 kJ of energy is released. Calculate the enthalpy change when 1 mole of sucrose undergoes complete combustion. M(C 12 H 22 O 11 ) = 342 g mol –1 n (sucrose) = 150 / 342 = 0.439 mol ∆ r H = 1 x -2478 0.439 = - 5650 ∆ r H = -5650 kJ mol -1 Achieved Merit Excellence

14. Success Criteria Can describe what happens to the particles due to change in condition Can explain how the change in condition affects the frequency of collisions Can fully explain the change in rate due to the frequency of collisions and link to observation seen

15. The reaction between 20.0 mL of 0.500 mol L –1 hydrochloric acid and 20.0 mL of 0.250 mol L –1 sodium thiosulfate solution at room temperature (25°C) produces a precipitate of sulfur that makes the solution go cloudy after about 5 minutes. With reference to the collisions of particles, explain how and why the reaction is affected, if the reaction is carried out in a water bath at a temperature of 50°C. An increase in temperature means the particles have more kinetic energy. There will be an increase in the frequency of collisions. And when the particles collide there is more chance that they will reach the activation energy required for the reaction to take place. Therefore the frequency of successful collisions will increase and hence the reaction rate will increase, so lowering the time taken for the reaction to take place. Achieved Merit Excellence

16. Hydrogen peroxide decomposes at room temperature (25 o C) according to the following equation. 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) On addition of a very small amount of solid manganese dioxide, the rate at which the bubbles of gas are produced is increased so that rapid fizzing is observed. Further observation indicates that manganese dioxide remains after reaction has stopped. With reference to the collisions of particles, explain why the reaction rate has increased. MnO 2 is a catalyst for the reaction. The catalyst provides an alternative pathway of lower activation energy for the reaction. So molecules that previously have enough energy to react, can now reach the lowered activation energy. Therefore the successful collision rate is increased and the reaction rate is increased. So more rapid fizzing is observed. AchievedMerit Excellence

17. Hydrogen peroxide decomposes at room temperature (25 o C) according to the following equation. 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) Hydrogen peroxide is stored at a low temperature. Discuss this statement in terms of reaction rate. The low temperature means that the molecules have less energy. So there is a decrease in the frequency of collisions. Less energy means when the molecules collide, they have less chance of reaching the activation energy for the reaction. Therefore there are fewer successful collisions in the same time, so the reaction rate decreases and the rate of decomposition is decreased. AchievedMerit Excellence

18. Can place products on top and reactants on the bottom Can correctly write Kc expressions using square brackets and subscripts Can discuss the relationship shown by Kc Success Criteria

19. The following equilibrium system is established when thiocyanate ions (SCN - ) are added to iron (III) ions (Fe 3+ ). The resulting aqueous solution is a dark red colour. The equation representing the equilibrium system and the colours of each species involved are given below. Fe 3+ (aq) + SCN – (aq) FeSCN 2+ (aq) pale orangecolourless dark red Write the equilibrium constant expression for the above reaction. Kc = FeSCN 2+ Fe 3+. SCN - Achieved Merit Kc = [FeSCN 2+ ] [Fe 3+ ]. [SCN - ]

20. Ammonia is produced industrially according to the Haber Process as shown below: N 2 (g) + 3H 2 (g) 2NH 3 (g) Complete the equilibrium constant expression for the above reaction. Kc = NH 3 N 2. H 2 Kc = [NH 3 ] 2 [N 2 ]. [H 2 ] 3 Achieved Merit

21. At 25°C the value of K c is 1.70  10 7. Circle the species that would be present in the higher concentration in the equilibrium mixture at this temperature. Ag + (aq) or Ag(NH 3 ) 2 + (aq) Justify your choice. Ag + (aq) + 2NH 3 (aq) Ag(NH 3 ) 2 + (aq) Kc = Ag(NH 3 ) 2 + (aq) Ag + (aq). NH 3 (aq ) Kc = [ Ag(NH 3 ) 2 + (aq)] [Ag + (aq)]. [NH 3 (aq )] 2 Ag(NH 3 ) 2 + (aq) Kc is very large, so the concentration of product is high compared to that of reactants (as the product concentration is on top of the ratio). Achieved Merit Excellence

22. At 200°C the value of K c is 1.10  10 – 5. Circle the species that would be present in the higher concentration in the equilibrium mixture at this temperature. NO 2 (g) or NO(g) Justify your choice. 2NO 2 (g) 2NO(g) + O 2 (g) Kc = NO (g). O 2 (g) NO 2(g) Kc = [NO (g) ] 2. [O 2 (g) ] [NO 2(g) ] 2 NO 2 (g). Kc is very small, so concentration of products is low compared to that of reactants. So the product concentration is the top of the ratio. Achieved Merit Excellence

23. Success Criteria State observation seen Explain how the equilibrium shifts due to the change in condition Relates equilibrium theory to current situation to explain observation seen

24. The following equilibrium system is established when thiocyanate ions (SCN - ) are added to iron (III) ions (Fe 3+ ). The resulting aqueous solution is a dark red colour. Fe 3+ (aq) + SCN – (aq) FeSCN 2+ (aq) pale orangecolourless dark red When iron (III) ions (Fe 3+ ) are removed from the equilibrium mixture (by adding sodium fluoride), a colour change is observed. Describe the colour change you would expect to see and explain why it occurs. The red colour lightens and becomes more orange. Removal of the Fe 3+ causes the equilibrium position to shift towards the reactants in order to minimise the change. This replaces some of the Fe 3+ that has been removed. Therefore there is less FeSCN 2+ present which results in a lighter colour.

25. Ammonia is produced industrially according to the Haber Process as shown below: N 2 (g) + 3H 2 (g) 2NH 3 (g) The pressure of the system at equilibrium is increased (by decreasing the total volume of the system). Describe the effect of this change on the amount of NH 3 in the system. Explain your answer. The amount of NH 3 will increase. Increasing the pressure of the system causes a shift to the right. This is to decrease the amount of pressure by forming fewer gas moles of gas.

26. The following reaction is exothermic: 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) Both N 2 O 5 and O 2 are colourless gases and NO 2 is a brown gas. A mixture of these gases exists at equilibrium and is observed as a brown colour. The mixture of gases is heated (at constant pressure). Describe the expected observation and explain why this occurs. The brown colour will lighten. When the mixture is heated the endothermic reaction is favoured. In this case, this is the reverse reaction. So the amount of NO 2 gas is decreased.

27. The pressure is increased, by decreasing the volume of the container. Describe the expected observation and explain why this occurs. The following reaction is exothermic: 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) Both N 2 O 5 and O 2 are colourless gases and NO 2 is a brown gas. A mixture of these gases exists at equilibrium and is observed as a brown colour. The brown colour will become lighter. As the pressure is increased the formation of fewer moles of gas is favoured. This favours the reverse reaction as the ratio is 5:2 moles of gas. Therefore the amount of brown gas will decrease.

28. An equilibrium system involving different species of cobalt(II) is shown in the equation below. [CoCl 4 ] 2– (aq) + 6H 2 O(ℓ) [Co(H 2 O) 6 ] 2+ (aq) + 4Cl – (aq) [CoCl 4 ] 2– (aq) is blue and [Co(H 2 O) 6 ] 2+ (aq) is pink. At room temperature (25°C) the equilibrium mixture is pink. Describe the expected observation when solid sodium chloride (NaCl) is added to the equilibrium mixture. Explain your answer. The colour of the solution will turn blue. Adding NaCl will increase the concentration of the Cl 1- ions. The equilibrium shift to decrease the concentration of the chloride ion. In this case, it will move in favour of the reactants so more blue [CoCl 4 ] 2– is formed.

29. An equilibrium system involving different species of cobalt(II) is shown in the equation below. [CoCl 4 ] 2– (aq) + 6H 2 O(ℓ) [Co(H 2 O) 6 ] 2+ (aq) + 4Cl – (aq) [CoCl 4 ] 2– (aq) is blue and [Co(H 2 O) 6 ] 2+ (aq) is pink. At room temperature (25°C) the equilibrium mixture is pink. The enthalpy change (∆ r H) for this reaction as written above, has a negative value. State the ion that would be present in the higher concentration when the equilibrium mixture is heated. Explain your answer. [CoCl 4 ] 2– would be in present in the higher concentration. As the temperature increases, the equilibrium will shift to reduce the temperature increase by moving in the endothermic direction. As the reaction is exothermic, the equilibrium will move in the reverse direction creating more [CoCl 4 ] 2–.

30. Success Criteria Can identify a conjugate acid-base pair Can identify several conjugate acid-base pairs Can identify species that can act as acids or bases and their conjugate pair

31. Chickens make egg shell, CaCO 3, using carbon dioxide gas from the air. The carbon dioxide forms carbonic acid (H 2 CO 3 ), which then reacts to form the carbonate ions (CO 3 2– ) needed to make egg shell. Two equations showing part of this process are given below. Equation 1:H 2 CO 3 (aq) + H 2 O( l ) HCO 3  (aq) + H 3 O + (aq) Equation 2:HCO 3  (aq) + H 2 O( l ) CO 3 2  (aq) + H 3 O + (aq) Identify three conjugate acid-base pairs in the equations above. H 2 CO 3 / HCO 3  HCO 3    CO 3 2  H 3 O + / H 2 O

32. Chickens make egg shell, CaCO 3, using carbon dioxide gas from the air. The carbon dioxide forms carbonic acid (H 2 CO 3 ), which then reacts to form the carbonate ions (CO 3 2– ) needed to make egg shell. Two equations showing part of this process are given below. Equation 1:H 2 CO 3 (aq) + H 2 O( l ) HCO 3  (aq) + H 3 O + (aq) Equation 2:HCO 3  (aq) + H 2 O( l ) CO 3 2  (aq) + H 3 O + (aq) HCO 3  can act as both an acid and a base. Specify which equation above (1 or 2) shows HCO 3  acting as an acid. Give a reason for your answer. Equation two HCO 3 1- is donating a proton / H 1+

33. Complete the table below to show the conjugate acid-base pairs. Conjugate acid Conjugate base NH 4 + H 2 PO 4 – Cl – HSO 4 – NH 3 HPO 4 2- HCl H 2 SO 4

34. Which ion below can act as both an acid and a base. CH 3 COO – HCO 3 – Justify your choice. HCO 3 – It can donate an H + or it can accept an H +

35. Success Criteria Describe properties of weak or strong acids (conductivity, pH and reactions with metals and carbonates) Explain how properties seen are linked to the strength of either weak or strong acids Fully explain how properties seen are linked to the strength of weak and strong acids

36. Two acids of the same concentration, hydrochloric acid (HCl) and propanoic acid (CH 3 CH 2 COOH), have properties as shown below: PropertyHydrochloric acid (0.100 mol L  1 ) Propanoic acid (0.100 mol L  1 ) Relative conductivity of solution HighLow pH of solution1.002.93 Explain the differences in the conductivity and pH of the two acids. In your explanation include reference to the species present in each solution. HCl has a low pH and a high conductivity as it is a strong acid. This means it completely dissociates into its ions, so there is a high [H 3 O + ], which results in a low pH. The high concentration of ions overall results in the high conductivity. Propanoic acid is a weak acid. It only partially dissociates in water, resulting in a low [H 3 O + ]. This causes a high pH and the low overall ion concentration results in low conductivity.

37. The concentration and pH of three acids, HA, HB and HC, are shown in the table below. acidconcentration (mol L –1 ) pH HA0.1001.00 HB0.1002.50 HC0.001003.00 A small piece of magnesium is added to a 20 mL sample of each of the acids. State which acid that would be expected to react most rapidly with the magnesium. Explain why this acid will react the fastest. HA would react the most rapidly with the magnesium. This is because HA has the lowest pH, which means it is the strongest acid. It will completely dissociate into its ions, producing a high [H 3 O + ]. This will result in more particles being available and a faster reaction rate.

38. The table below shows the pH of two acids, HA and HB, each with the same concentration. AcidpH HA1.00 HB4.00 When these acids react with magnesium metal, hydrogen gas (H 2 ) Is produced. Discuss the reactions with both acids, HA and HB, with magnesium metal when the same volume of each acid is used. HA and HB will react with the magnesium metal and produce the same amount of hydrogen gas. Although HA will produce it faster. This is because HA is a strong acid as it has a low pH. This means it fully dissociates into its ions and so has a high [H 3 O + ]. Therefore there are more particles present to react. HB is a weak acid as it has a high pH. This means it only partially dissociates into its ions, so has a low [H 3 O + ]. Therefore there are less particles present to react with initially, so a slower rate of reaction is seen.

39. Success Criteria Can use Kw or pH expressions to calculate an unknown Can use Kw or pH expressions to solve several unknowns Can use Kw and pH expressions to solve unknowns

40. Complete the following table showing hydronium ion concentration, hydroxide ion concentration and pH for some solutions. K w = 1.00  10 – 14 Solution[H 3 O + ][OH – ]pH 10.0350 210.8 35.66 × 10 – 6 1.58 x 10 -11 1.77 x 10 -9 2.86 x 10 -13 6.31 x 10 -4 1.46 8.75

41. Complete the table below to show the hydronium ion concentration, hydroxide ion concentration, and pH for the three solutions shown. K w = 1.00 × 10 –14 Solution [H 3 O + ] mol L –1 [OH – ] mol L –1 pH hydrochloric acid (HCl) 0.0720 sodium hypochlorite (NaOCl) 11.4 hypochlorous acid (HOCl) 2.24 × 10 –11 1.39 x 10 -13 1.14 3.98 x 10 -12 2.51 x 10 -3 4.46 x 10 -4 3.35

42. If a solution of sodium hydrogen carbonate has a pH of 9.20, calculate the concentration of hydroxide ions,  OH  , present in the solution. State your answer to 3 significant figures. pH = -log [H 3 O + ] [H 3 O + ] = inv log -9.20 = 6.31 x 10 -10 mol L -1 [OH - ] = 1 x 10 -14 6.31 x 10 -10 = 1.58 x 10 -5 mol L -1

43. Success Criteria Identifying the species as either an acid or a base, due to its reaction with water Identifying the species as either acid or base using equations Linking the products of the equation to the increase or decrease in pH

44. A solution of sodium ethanoate (NaCH 3 COO) is tested and found to have a pH of 8.50. Discuss why the pH of the solution is greater than 7. Include appropriate equation(s) in your answer. Sodium ethanoate solution contains both Na 1+ and CH 3 COO 1- ions. Ethanoate ions react with water to accept H 1+ since ethanoic acid is a weak acid. So [OH 1- ] is increased. Therefore the [OH 1- ] > [H 3 O 1+ ], which results in a pH greater than 7. CH 3 COO 1- + H 2 O CH 3 COOH + OH 1-

45. A solution of sodium hypochlorite, NaOCl, is basic. Discuss the above statement, including appropriate chemical equation(s) in your answer. The hypochlorite ion is basic. Therefore it accepts a proton from water to form hydroxide ions. The [OH 1- ] is now greater than the [H 3 O 1+ ], producing a basic solution. OCl 1- + H 2 O HOCl + OH 1-