A given reaction has an activation energy of 24.52 kJ/mol. At 25�C the half-life is 4 minutes. At what temperature will the half-life
be reduced to 20 seconds?
i hate chem at UT!
To solve this question, we need to use the Arrhenius equation, which relates the rate constant of a reaction to temperature and activation energy.
The Arrhenius equation is given by:
k = A * exp(-Ea/RT)
Where:
- k is the rate constant,
- A is the pre-exponential factor or frequency factor,
- Ea is the activation energy,
- R is the ideal gas constant (8.314 J/(mol*K)),
- T is the absolute temperature in Kelvin.
We are given the half-life at two different temperatures: 4 minutes at 25°C and 20 seconds at an unknown temperature. We need to find this unknown temperature.
First, let's convert the given temperatures to Kelvin:
- 25°C + 273.15 = 298.15 K (temperature for the 4-minute half-life)
Let's assume the rate constant at 298.15 K is k1, and the unknown temperature is T2. The half-life can be expressed in terms of the rate constant as follows:
t1/2 = (0.693) / k1
t1/2 = (0.693) / (A * exp(-Ea/RT1))
where t1/2 is the half-life (4 minutes = 240 seconds), and t1 is the temperature (298.15 K).
Similarly, for T2:
t1/2' = (0.693) / k2
t1/2' = (0.693) / (A * exp(-Ea/RT2))
where t1/2' is the half-life (20 seconds), and T2 is the unknown temperature.
Now, let's solve for the unknown temperature T2.
Dividing the two equations above, we can eliminate the pre-exponential factor (A) and simplify:
(t1/2') / (t1/2) = exp((Ea/R) * ((1/T2) - (1/T1)))
Taking the natural logarithm (ln) of both sides:
ln((t1/2') / (t1/2)) = (Ea/R) * ((1/T2) - (1/T1))
Finally, solving for T2:
T2 = (1 / ((1/T1) + ((R/Ea) * ln((t1/2') / (t1/2)))))
Now, plug in the known values:
T1 = 298.15 K
t1/2 = 240 seconds
t1/2' = 20 seconds
Ea = 24.52 kJ/mol (convert to J/mol)
Using the given values, we can calculate the value of T2:
T2 = (1 / ((1/298.15) + ((8.314 J/(mol*K))/(24.52 kJ/mol) * ln((20 s) / (240 s)))))