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,E_a,is 126 kJ/mol and the rate constant at 30°C is 9.8 × 10^-3 min^-1,what is the rate constant at 40°C?

Question

Quizplus · Accepted Answer

We can use the Arrhenius equation:

k2 = k1 * exp((Ea/R) * (1/T1 - 1/T2))

where k1 is the rate constant at temperature T1, k2 is the rate constant at temperature T2, Ea is the activation energy, R is the gas constant (8.314 J/mol·K), and T1 and T2 are the temperatures in kelvin.

Plugging in the given values:

k2 = (9.8 × 10-3 min-1) * exp((126000 J/mol / 8.314 J/mol·K) * (1/303 K - 1/313 K))

k2 = 4.8 × 10-2 min-1

Therefore, the rate constant at 40°C is approximately 4.8 × 10-2 min-1, which corresponds to answer choice B.

The Aquation of Tris(1,10-Phenanthroline)iron(II)in Acid Solution Takes Place According to the Equation