The aquation of tris (1,10-phenanthroline) iron(II) in acid solution takes place according to the equation: Fe(phen) ₃²⁺ + 3H₃O⁺ + 3H₂O → Fe(H₂O) ₆²⁺ + 3phenH⁺.
If the activation energy, E_a, is 126 kJ mol^-1 and the rate constant at 30 °C is 9.8 × 10^-3 min^-¹, what is the rate constant at 35°C?

Question

Quizplus · Accepted Answer

The rate constant at a different temperature can be calculated using the Arrhenius equation:

k2 = A * exp(-Ea/RT2)

where k2 is the rate constant at the new temperature, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T2 is the new temperature in Kelvin.

To solve for k2, we need to know A, which is not given in the problem. However, we can assume that A does not change significantly over this small temperature range, so we can use the rate constant at 30°C as an approximation for A:

A ≈ k1 = 9.8 × 10^-3 min^-1

Plugging in the given values for Ea and T2 = 35+273 = 308 K, we get:

k2 = (9.8 × 10^-3) * exp(-126000/(8.314*308)) ≈ 2.2 × 10^-2 min^-1

Therefore, the answer is B.

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