The given reaction,
2N2O5 4NO2 + O2,
follows first-order kinetics and has an activation energy of 103 kJ mol-1. At 298 K, k = 0.0000354 s-1. What is the half-life (min) for this reaction at 338 K ? Round your answer to 3 significant figures.
k = 0.693/t1/2
Substitute for k and calculate t1/2.
?? and then what do you do?
.0000354 = 0.693/t(1/2)
t(1/2) = 19576.27119
the answer is suppose to be 2.38 min
which isomer Maleic acid or fumanic acid has the strongest intermolecular bonding? why?
To determine the half-life of a reaction, you need to use the rate constant (k) and the reaction temperature (T). The half-life can be calculated using the formula:
t(1/2) = (0.693 / k)
Given that the rate constant (k) is 0.0000354 s^-1 at 298 K and the activation energy is 103 kJ mol^-1, we can use the Arrhenius equation to find the new rate constant (k') at 338 K. The Arrhenius equation is given by:
k' = Ae^(-Ea / (RT))
where:
- A is the pre-exponential factor (assumed to be the same for both temperatures)
- Ea is the activation energy (103 kJ mol^-1)
- R is the gas constant (8.314 J K^-1 mol^-1)
- T is the temperature in Kelvin (298 K and 338 K)
First, let's convert the activation energy from kJ to J by multiplying it by 1000:
Ea = 103 kJ mol^-1 * 1000 J kJ^-1 = 103000 J mol^-1
Now, let's calculate the rate constant (k') at 338 K using the Arrhenius equation:
k' = Ae^(-Ea / (RT'))
Substituting the known values:
- A is assumed to be the same for both temperatures
- R is 8.314 J K^-1 mol^-1
- Ea is 103000 J mol^-1
- T' is 338 K
k' = 0.0000354 s^-1 * e^(-103000 J mol^-1 / (8.314 J K^-1 mol^-1 * 338 K))
Calculating this expression will give you the new rate constant (k') at 338 K.
Next, we can use the new rate constant (k') to calculate the half-life (t(1/2)) at 338 K using the formula mentioned earlier:
t(1/2) = (0.693 / k')
By substituting the value of k' into this equation, you can calculate the half-life for the reaction at 338 K.