The aquation of tris(1,10-phenanthroline)iron(ii) in acid solution takes place according to the equation: fe(phen)32+ + 3 h3o+ + 3 h2o → fe(

Question

The aquation of tris(1,10-phenanthroline)iron(ii) in acid solution takes place according to the equation: fe(phen)32+ + 3 h3o+ + 3 h2o → fe(h2o)62+ + 3 phenh+. if the activation energy, ea, is 126 kj/mol and the rate constant at 30°c is 9.8 × 10-3 min-1, what is the rate constant at 35°c? the aquation of tris(1,10-phenanthroline)iron(ii) in acid solution takes place according to the equation: fe(phen)32+ + 3 h3o+ + 3 h2o → fe(h2o)62+ + 3 phenh+. if the activation energy, ea, is 126 kj/mol and the rate constant at 30°c is 9.8 × 10-3 min-1, what is the rate constant at 35°c? 4.4 × 10-3 min-1 4.5 × 101 min-1 2.3 × 102 min-1 2.2 × 10-2 min-1

in progress 0
Maria 1 week 2022-11-24T15:52:06+00:00 1 Answer 0 views 0

Answers ( )

    0
    2022-11-24T15:53:14+00:00

    Answer:

    2.2 × 10⁻² min⁻¹  

    Step-by-step explanation:

    We can use the Arrhenius equation  

    [tex]\ln(\frac{k_2 }{k_1}) = (\frac{E_{a} }{R})(\frac{1}{T_1} – \frac{1 }{T_2 })[/tex]

    Data:

    k₁ = 9.8 × 10⁻³ min⁻¹; k₂ = ?

    Eₐ = 126 kJ·mol⁻¹

    T₁ = 30 °C; T₂ = 35°C  

    Calculations:

    (a) Convert temperatures to kelvins.

    T₁ = (30 + 273.15) K = 303.15  K

    T₂ = (35 + 273.15) K = 308.15  K

    (b) Activation energy

    [tex]\ln(\frac{k_{2} }{9.8 \times 10^{-3}}) = (\frac{126 000 }{8.314})(\frac{ 1}{303.15} – \frac{1 }{308.15 })[/tex]

    [tex]\ln(\frac{k_{2} }{9.8\times10^{-3}}) = 15160\times 5.35 \times10^{-5}[/tex]

    [tex]\ln(\frac{k_{2} }{9.8 \times 10^{-3}}) = 0.811[/tex]

    [tex]\frac{k_{2} }{9.8\times 10^{-3}} = \text{e}^{0.811}[/tex]

    [tex]\frac{k_{2} }{9.8 \times 10^{-3}} = 2.25[/tex]

    k₂ = 9.8 × 10⁻³× 2.25

        = 0.022 min⁻¹

        = 2.2 × 10⁻² min⁻¹

Leave an answer

45:5+15-5:5+20+17*12-15 = ? ( )