At 19.0°C, what is the fraction of collisions with energy equal to or greater than an activation energy of 151.0 kJ/mol?
I got 1.06 which wasn't right
Never mind! I didn't change my R to kJ
Would you show how you arrived at the right answer?
f=e^(-Ea/RT)
f=e^(-151.0/(8.314e-3 x 292)
To find the fraction of collisions with energy equal to or greater than an activation energy, you need to use the Boltzmann distribution formula. The formula for the fraction of collisions (f) is given by:
f = e^(-Ea/RT)
Where:
- Ea is the activation energy in joules/mol
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
To solve this problem, we need to convert the temperature from Celsius to Kelvin by adding 273.15 K. Therefore, the temperature is 19.0°C + 273.15 = 292.15 K.
Now we can substitute the values into the equation:
f = e^(-151000 / (8.314 * 292.15))
To evaluate this expression, you need to use a calculator or a computer with exponentiation functionality. The result should give you the fraction of collisions with energy equal to or greater than the activation energy at 19.0°C.