The equation for reaction 1 is shown. reaction 1 X → 2Y Reaction 1 is first order with respect to the concentration of X. The half-life of the reaction, t₁⁄₂, is 900 s at 20 °C.

1. (a)(i)A solution of X with a concentration of 0.180 moldm⁻³ is prepared at 20 °C. Calculate the average rate of reaction 1 over the first 1800 s. average rate of reaction 1 =[2]
2. (a)(ii)Complete the rate equation for reaction 1. rate =[1]
3. (a)(iii)Show that the rate constant, k, is 7.70 × 10⁻⁴ s⁻¹ at 20 °C.[1]
4. (a)(iv)Calculate the initial rate of reaction 1 when the concentration of X is 0.150 mol dm⁻³. Include units. rate = units[2]
5. (b)(i)Platinum is a transition element. Explain why transition elements behave as catalysts.[1]
6. (b)(ii)Name the metal catalyst in the Haber process and explain why it is a heterogeneous catalyst. metal[1]
7. (b)(iii)Platinum acts as a heterogeneous catalyst in the removal of nitrogen dioxide, NO₂, from the exhaust gases of car engines. Describe the mode of action of a platinum catalyst in this process.[2]
8. (b)(iv)NO₂ acts as a homogeneous catalyst in the oxidation of atmospheric sulfur dioxide, SO₂. Write equations for the two reactions that occur. equation 1 equation 2[1]
9. (c)SO₂ dissolves in water, forming H₂SO₃. H₂SO₃ can be oxidised under acidic conditions. The relevant electrode reaction and its E° value are shown. SO₄²⁻ + 4H⁺ + 2e⁻ ⇌ H₂SO₃ + H₂O E° = +0.17 V Four more half-equations for reactions occurring under acidic conditions, and their E° values, are shown. H₃BO₃ + 3H⁺ + 3e⁻ ⇌ B + 3H₂O E° = –0.73 V BiO⁺ + 2H⁺ + 3e⁻ ⇌ Bi + H₂O E° = +0.28 V S + 2H⁺ + 2e⁻ ⇌ H₂S E° = +0.14 V Sb + 3H⁺ + 3e⁻ ⇌ SbH₃ E° = –0.51 V Select the oxidising agent that could oxidise H₂SO₃ to SO₄²⁻ ions under acidic conditions. Write an equation, and give the E°cell value, for the reaction that occurs. oxidising agent equation E°cell = V[3]