A particular reaction has an activation energy of 51.6kJ/mol-rxn. How much faster is the reaction at 50 degrees C versus 25degrees C?
I do not know how to start this!
I think we've gone over this before. You start by substituting into the Arrhenius equation. Ea goes in in J/mol (not kJ/mol) and you change T to kelvin.
So the equation I use is ln k = a + (-ea/rt) correct?
I would use ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)
To determine how much faster the reaction is at 50 degrees Celsius compared to 25 degrees Celsius, you can use the Arrhenius equation. This equation relates the rate constant of a reaction to the activation energy and temperature.
The Arrhenius equation is given by:
k = Ae^(-Ea/RT)
where:
k is the rate constant
A is the pre-exponential factor (also known as the frequency factor or attempt frequency)
Ea is the activation energy
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
First, we need to convert the temperatures from Celsius to Kelvin. The conversion from Celsius to Kelvin is simple: K = °C + 273.15
So, for 50 degrees Celsius, the temperature in Kelvin is:
T1 = 50 + 273.15
And for 25 degrees Celsius, the temperature in Kelvin is:
T2 = 25 + 273.15
Next, plug these values into the Arrhenius equation to determine the rate constants at each temperature:
k1 = A * e^(-Ea/RT1)
k2 = A * e^(-Ea/RT2)
To determine how much faster the reaction is, we can calculate the ratio of the rate constants:
faster = k1 / k2
This gives us the relative increase in the rate constant between the two temperatures, which reflects how much faster the reaction is at 50 degrees Celsius compared to 25 degrees Celsius.