## 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(

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

## Answers ( )

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