Transition Metals Test PRi Flashcards
Q1a) Identify a suitable reagent, give the formula for copper containing species and write equation for reaction of [Cu(H2O)6]2+ ⟶ Pale Blue ppt.
Reagent – Ammonia
Species – [Cu(H2O)4(OH)2]
Equation – [Cu(H2O)6]2+ + 2NH3 ⟶ [Cu(H2O)3(OH)3] + 2NH4+
Q1b) Reagent, species and equation for Reaction 2
Reagent – Ammonia in excess
Species – [Cu(H2O)2(NH3)4]2+
Equation – [Cu(H2O)4(OH)2] + 4NH3 ⟶ [Cu(H2O)2(NH3)4]2+ + 2H2O + 2OH-
Q1c) Reagent, species and equation for Reaction 3
Reagent – Na2CO3
Species – [Cu(H2O)6]2+
Equation – [Cu(H2O)6]2+ + CO3(2-) ⟶ CuCO3 + 6H2O
Q1d) Reagent, species and equation for Reaction 4
Reagent – HCl (Conc/xs)
Species – [Cu(H2O)6]2+
Equation – [Cu(H2O)6]2+ + 4Cl- ⟶ [CuCl4]2- + 6H2O
Q2a) Explain why the reaction between I- ions and S2O82- ions is slow
They are both negatively charged
-ve ions repel one another
Q2b) Other than having variable oxidation states, explain why Fe2+ ions are good catalysts for this reaction
Positive ions attract -ve ions in catalysed process
Q2c) Write a half-equation for the reduction of S2O8(2-) ions to SO4(2-) ions
S2O8(2-) + 2e- ⟶ 2SO4(2-)
Q2d) Construct an overall equation for the reaction between S2O8(2-) ions and I- ions
S2O8(2-) + 2I- ⟶ 2SO4(2-) + I2
Q3a) Calculate the entropy change for the reaction between ammonia and oxygen
∆S = S(s products) - Σ(s reactants)
((4x211) + (6x189)) - ((4x193) + (5x205)) = 181 J/K/mol
Q3b) Calculate a value for the Gibbs free-energy change (∆G), in kJ/mol, for the reaction between ammonia and oxygen at 600ºC
∆G = ∆H - T∆S 600ºC = 873K ∆G = -905 - 873(181x10-3) = -1063 kJ/mol
Q3c) The reaction between ammonia and oxygen was carried out at a higher temperature.
Explain how this change affects the value of ∆G for this reaction
1) ∆G becomes more negative
2) Entropy change/∆S is +ve/T∆S gets bigger
Q3d) Platinum acts as a heterogeneous catalyst in the reaction between ammonia and oxygen. It provides an alternative reaction route with a lower activation energy.
Describe the stages of this alternative route
1) Reactants absorbed onto Platinum surface/Pt provides surface/active site
2) Reaction on surface. Bond breaking/forming occurs on surface
3) Desorption of the product or words to that effect (wtte)
Q3e) Deduce the change in oxidation state of Nitrogen, when NH3 is oxidised to NO
Oxidation state changes from -3 to +2 (or +5 overall)
Q3f) When ammonia reacts with oxygen, Nitrous Oxide (N2O) can be produced instead of NO
Give an equation for this reaction
2NH3 + 2O2 ⟶ N2O + 3H2O
Q4a) Explain, in terms of electrons, why the complexes are different colours. (You are NOT required to explain why the observed colours are red-violet and green)
1) In each of P & Q, oxidation state of Cr = +3
2) In each of P & Q, electron config is same/3d3
3) Ligands are different
4) Different energies of electrons/different orbital energies
5) Different wavelengths of light absorbed
6) Different wavelengths of light transmitted/reflected