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AS 2.5 - Physical Chemistry

AS 2.5 - Physical Chemistry. Formative Assessments in Energy, Rates of Reaction, Equilibrium and Acid/Base Reactions. Overall Outline. Success Criteria. Energy. State whether heat is taken in or given out in the following state changes. in. in. out. out. in. Energy.

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AS 2.5 - Physical Chemistry

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

  2. Overall Outline

  3. Success Criteria

  4. Energy State whether heat is taken in or given out in the following state changes. in in out out in

  5. Energy State whether the following are exothermic or endothermic reactions. Merit Achieved Endothermic Endothermic Exothermic Endothermic

  6. State whether the following are exothermic or endothermic reactions Merit Excellence Achieved Exothermic Endothermic Endothermic Exothermic

  7. Success Criteria

  8. Octane is a key component in petrol, and burns according to the following equation: C8H18(l) + 121/2 O2(g)  8 CO2 (g) + 9 H2O(l) ΔrH = 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. Energy Released = n x ΔrH = 6.12 x 5500 = 33 660 kJ Achieved

  9. Excellence • = 235 g Using hydrogen gas (H2) as a fuel for cars, rather than octane, is often viewed as better for the environment. Calculate the mass of H2required to produce the same amount of energy as 1.00 litre of octane ( 33 660 kJ ). State your answer to 3 significant figures. H2(g) + 1/2 O2(g)  H2O(g) ΔrH = 286 kJ mol1 n (H2) = 33 660 / 286 = 118 mol Achieved • m (H2) = 117.7 x 2 • = 235.4 Merit

  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 Achieved 0.740 mol reacts to produce 43.5 kJ 1 mol reacts to produce 43.5 = 58.82 kJ 0.740 Merit  rH = -58.8 kJ mol-1 Excellence

  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 Achieved Merit m (NaOH) = 2.55 mol x 40.0 g mol-1 = 102.3 Excellence m = 102 g

  12. Carbohydrates are an important source of energy in our diet. Two common carbohydrates are glucose (C6H12O6) and sucrose (C12H22O11). The equation below shows the combustion of glucose to form carbon dioxide and water. C6H12O6(s) + 6O2(g) →6CO2(g) + 6H2O(ℓ) ∆rH = –2820 kJ mol–1 Calculate the enthalpy change, ∆rH, when 100 g of glucose reacts to form carbon dioxide and water. M(C6H12O6) = 180 g mol–1 n (glucose) = 150 / 180 = 0.556 mol Achieved ∆rH = 0.556 x -2820 = -1567 Merit Excellence ∆rH = -1567 kJ mol-1

  13. The equation below shows the combustion of sucrose to form carbon dioxide and water. C12H22O11(s) + 12O2(g)→12CO2(g) + 11H2O(ℓ) When 150 g of sucrose, C12H22O11, undergoes combustion, 2478 kJ of energy is released. Calculate the enthalpy change when 1 mole of sucrose undergoes complete combustion. M(C12H22O11 )= 342 g mol–1 n (sucrose) = 150 / 342 = 0.439 mol ∆rH=1 x -2478 0.439 = - 5650 Achieved Merit Excellence ∆rH = -5650 kJ mol-1

  14. Success Criteria

  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. Excellence Merit Achieved

  16. Hydrogen peroxide decomposes at room temperature (25oC) according to the following equation. 2H2O2(aq) 2H2O(l) + O2(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. MnO2 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. Excellence Achieved Merit

  17. Hydrogen peroxide decomposes at room temperature (25oC) according to the following equation. 2H2O2(aq) 2H2O(l) + O2(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. Excellence Achieved Merit

  18. Success Criteria

  19. The following equilibrium system is established when thiocyanate ions (SCN-) are added to iron (III) ions (Fe3+). 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. • Fe3+ (aq) + SCN– (aq) FeSCN2+ (aq) • pale orange colourless dark red • Write the equilibrium constant expression for the above reaction. Achieved Kc = [FeSCN2+ ] [Fe3+] . [SCN-] Kc = FeSCN2+ Fe3+ . SCN- Merit

  20. Ammonia is produced industrially according to the • Haber Process as shown below: • N2 (g) + 3H2 (g) 2NH3 (g) • Complete the equilibrium constant expression for the above reaction. Achieved Kc = [NH3]2 [N2] . [H2]3 Kc = NH3 N2 . H2 Merit

  21. Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) At 25°C the value of Kc is 1.70  107. Circle the species that would be present in the higher concentration in the equilibrium mixture at this temperature. Ag+(aq) or Ag(NH3)2+(aq) Justify your choice. Achieved Kc = [Ag(NH3)2+(aq)] [Ag+(aq)] . [NH3(aq)]2 Kc = Ag(NH3)2+(aq) Ag+(aq) . NH3(aq) Merit Excellence Ag(NH3)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).

  22. 2NO2(g) 2NO(g) + O2(g) At 200°C the value of Kc is 1.10  10–5. Circle the species that would be present in the higher concentration in the equilibrium mixture at this temperature. NO2(g) or NO(g) Justify your choice. Achieved Merit Kc = [NO (g)]2 . [O2 (g)] [NO2(g)]2 Kc = NO (g) . O2 (g) NO2(g) Excellence NO2(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.

  23. Success Criteria

  24. The following equilibrium system is established when thiocyanate ions (SCN-) are added to iron (III) ions (Fe3+). The resulting aqueous solution is a dark red colour. • Fe3+ (aq) + SCN– (aq) FeSCN2+ (aq) • pale orange colourless dark red • When iron (III) ions (Fe3+) 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 Fe3+ causes the equilibrium position to shift towards the reactants in order to minimise the change. This replaces some of the Fe3+ that has been removed. Therefore there is less FeSCN2+ present which results in a lighter colour.

  25. Ammonia is produced industrially according to the Haber Process as shown below: • N2 (g) + 3H2 (g) 2NH3 (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 NH3 in the system. Explain your answer. The amount of NH3 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: • 2N2O5(g) 4NO2(g) + O2(g) • Both N2O5 and O2 are colourless gases and NO2 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 NO2 gas is decreased.

  27. The following reaction is exothermic: • 2N2O5(g) 4NO2(g) + O2(g) • Both N2O5 and O2 are colourless gases and NO2 is a brown gas. A mixture of these gases exists at equilibrium and is observed as a brown colour. The pressure is increased, by decreasing the volume of the container. Describe the expected observation and explain why this occurs. 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. • [CoCl4]2–(aq) + 6H2O(ℓ) [Co(H2O)6]2+(aq) + 4Cl–(aq) • [CoCl4]2–(aq) is blue and [Co(H2O)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 Cl1- 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 [CoCl4]2– is formed.

  29. An equilibrium system involving different species of cobalt(II) is shown • in the equation below. • [CoCl4]2–(aq) + 6H2O(ℓ) [Co(H2O)6]2+(aq) + 4Cl–(aq) • [CoCl4]2–(aq) is blue and [Co(H2O)6]2+(aq) is pink. • At room temperature (25°C) the equilibrium mixture is pink. The enthalpy change (∆rH) 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. [CoCl4]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 [CoCl4]2– .

  30. Success Criteria

  31. Chickens make egg shell, CaCO3, using carbon dioxide gas from the air. The carbon dioxide forms carbonic acid (H2CO3), which then reacts to form the carbonate ions (CO32–) needed to make egg shell. Two equations showing part of this process are given below. Equation 1: H2CO3(aq) + H2O(l) HCO3(aq) + H3O+(aq) Equation 2: HCO3(aq) + H2O(l) CO32(aq) + H3O+(aq) Identify three conjugate acid-base pairs in the equations above. H2CO3 / HCO3 HCO3CO32 H3O+ / H2O

  32. Chickens make egg shell, CaCO3, using carbon dioxide gas from the air. The carbon dioxide forms carbonic acid (H2CO3), which then reacts to form the carbonate ions (CO32–) needed to make egg shell. Two equations showing part of this process are given below. Equation 1: H2CO3(aq) + H2O(l) HCO3(aq) + H3O+(aq) Equation 2: HCO3(aq) + H2O(l) CO32(aq) + H3O+(aq) HCO3 can act as both an acid and a base. Specify which equation above (1 or 2) shows HCO3 acting as an acid. Give a reason for your answer. Equation two HCO31- is donating a proton / H1+

  33. Complete the table below to show the conjugate acid-base pairs. NH3 HPO42- HCl H2SO4

  34. Which ion below can act as both an acid and a base. CH3COO– HCO3– Justify your choice. HCO3– It can donate an H + or it can accept an H +

  35. Success Criteria

  36. Two acids of the same concentration, hydrochloric acid (HCl) and propanoic acid (CH3CH2COOH), have properties as shown below: 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 [H3O+], 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 [H3O+]. 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. A small pieceof 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 [H3O+]. 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. When these acids react with magnesium metal, hydrogen gas (H2) 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 [H3O+]. 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 [H3O+]. Therefore there are less particles present to react with initially, so a slower rate of reaction is seen.

  39. Success Criteria

  40. Complete the following table showing hydronium ion concentration, hydroxide ion concentration and pH for some solutions. Kw = 1.00  10–14 2.86 x 10-13 1.46 1.58 x 10-11 6.31 x 10-4 1.77 x 10-9 8.75

  41. Complete the table below to show the hydronium ion concentration, • hydroxide ion concentration, and pH for the three solutions shown. • Kw = 1.00 × 10–14 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 [H3O+] [H3O+] = 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

  44. A solution of sodium ethanoate (NaCH3COO) 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 Na1+ and CH3COO1- ions. Ethanoate ions react with water to accept H1+ since ethanoic acid is a weak acid. CH3COO1- + H2O CH3COOH + OH1- So [OH1-] is increased. Therefore the [OH1-] > [H3O1+], which results in a pH greater than 7.

  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. OCl1- + H2O HOCl + OH1- The [OH1-] is now greater than the [H3O1+], producing a basic solution.

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