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 35°C?

Question

Quizplus · Accepted Answer

To solve this problem, 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:
k1 = 9.8 × 10^-3 min^-1
T1 = 30 + 273.15 = 303.15 K
T2 = 35 + 273.15 = 308.15 K
Ea = 126 kJ/mol * 1000 J/kJ = 126000 J/mol

We can convert the rate constant to units of s^-1 for ease of calculation:
k1 = 9.8 × 10^-3 min^-1 * (1 min/60 s) = 1.63 × 10^-4 s^-1

Substituting these values into the Arrhenius equation and solving for k2:
k2 = k1 * exp[(Ea/R)*((1/T1) - (1/T2))] 
= (1.63 × 10^-4 s^-1) * exp[(126000 J/mol)/(8.314 J/mol*K) * ((1/303.15 K) - (1/308.15 K))] 
= 2.2 × 10^-2 s^-1

Lastly, we convert the rate constant back to units of min^-1:
k2 = 2.2 × 10^-2 s^-1 * (1 min/60 s) = 3.7 × 10^-4 min^-1

Therefore, the closest answer choice is B) 2.2 × 10^-2 min^-1, which is approximately equal to 2 × 10^-2 min^-1 or 2 × 10^-2.

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