# 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?

## If you care to post your work I shall be happy to go over it to find the error.

But first, did you use Ea in Joules/mol and not kJ/mol (a common mistake) and did you cnvert T1 and T2 to kelvin (not so common error). The usual other problem is to reverse k1 and k2. Just make sure k1 goes with T1 and k2 goes with T2. Finally, R should be 8.314.

## To solve this problem, we can use the Arrhenius equation:

k = Ae^(-Ea/RT)

Where:
k is the rate constant
A is the pre-exponential factor
Ea is the activation energy
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin

Given:
Ea = 126 kJ/mol (convert to J/mol: 126 × 10^3 J/mol)
k1 = 9.8 × 10^(-3) min^(-1) (convert to s^(-1): 9.8 × 10^(-3) × 1/60 min^(-1))

To find the rate constant at 35°C, we need to convert both temperatures to Kelvin:

T1 = 30°C + 273.15 = 303.15 K
T2 = 35°C + 273.15 = 308.15 K

Now we can calculate the rate constant at 35°C (k2) using the Arrhenius equation:

k2 = Ae^(-Ea/RT2)

To find the pre-exponential factor (A), we can rearrange the equation:

A = k1 / e^(-Ea/RT1)

Now we have all the values required to calculate k2.

Let's plug in the numbers and calculate the rate constant at 35°C:

A = k1 / e^(-Ea/RT1)
A = (9.8 × 10^(-3) s^(-1)) / e^(-((126 × 10^3 J/mol) / (8.314 J/(mol·K) × 303.15 K)))

Now we can calculate k2:

k2 = Ae^(-Ea/RT2)
k2 = A × e^(-Ea/RT2)
k2 = (k1 / e^(-Ea/RT1)) × e^(-Ea/RT2)
k2 = k1 × e^((Ea/RT1) - (Ea/RT2))

Plugging in the values:

k2 = (9.8 × 10^(-3) s^(-1)) × e^(((126 × 10^3 J/mol) / (8.314 J/(mol·K) × 303.15 K)) - ((126 × 10^3 J/mol) / (8.314 J/(mol·K) × 308.15 K)))

Calculating k2 using a calculator or software will give you the rate constant at 35°C.