If he activation energy of a reaction is 4.86Kj, then waht is the percent increase in the rate constant of a raction when the temperature is increased from 45 to 75C
The reaction rate is assumed to be of the form
Rate = Constant * e^(-Ea/RT)
where T must be in Kelvin
Rate(75C)/Rate(45C) = e^-[Ea/348R - Ea/318R]
where R = molar gas constant
= 8.317 J/(mol*K)
and Ea = 4860 J/mol is the activation anergy
Rate(75C)/Rate(45C) = e^[(Ea/k)(1/318 - 1/348)]
= e^(584*2.71*10^-4)= e^.158 = 1.17
The percent rate increase is 17%
Check my math. I tend to be careless
I understand how to get the Constant but I don't get how you know that it is a 17% increase?
The constant does not matter. You can't even compute it, but you don't have to. They only ask for the PERCENT INCREASE in reaction rate.
The 17% increase comes from the 1.17 factor increase in the reaction rate
Thank you
To calculate the percent increase in the rate constant of a reaction when the temperature is increased, we need to use the Arrhenius equation that relates temperature and rate constant.
The Arrhenius equation is given as:
k = A * e^(-Ea / (R * T))
Where:
k = rate constant
A = pre-exponential factor
Ea = activation energy
R = gas constant (8.314 J/(mol*K))
T = temperature in Kelvin
First, we need to convert the activation energy from kilojoules to joules. We multiply it by 1000, since there are 1000 joules in a kilojoule:
Ea = 4.86 kJ * 1000 J/kJ = 4860 J
Next, we need to convert the temperatures from Celsius to Kelvin. To do this, we add 273.15 to each temperature:
T1 = 45 °C + 273.15 = 318.15 K
T2 = 75 °C + 273.15 = 348.15 K
Now, we can calculate the rate constant at each temperature. Let's assume that the pre-exponential factor is the same for both temperatures.
k1 = A * e^(-Ea / (R * T1))
k2 = A * e^(-Ea / (R * T2))
Finally, we can calculate the percent increase in the rate constant:
Percent increase = ((k2 - k1) / k1) * 100
Substitute the values of k1 and k2 obtained in the previous calculations to find the final answer.