The standard free energy of activation of one reaction is 94.3 kj/mol^-1 (22.54 kcal/mol^-1). The standard free energy of activation for another reaction is 76.4 kJ/mol^-1 (18.26 kcal/mol^-1). Assume temp of 298K and 1 M conc. By what factor is one faster than the other?
Im not sure if this is right but do you subtract the lower kcal value from the other and then multiply by 10? Thanks!
To determine the factor by which one reaction is faster than the other, you need to use the Arrhenius equation, which relates the rate constant (k) to the standard free energy of activation (ΔG‡). The equation is as follows:
k = Ae^(-ΔG‡/RT)
Where:
k = rate constant
A = pre-exponential factor (frequency factor)
ΔG‡ = standard free energy of activation
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
Using this equation, you can compare the rate constants of the two reactions and determine how much faster one reaction is compared to the other.
Let's calculate the factor by which one reaction is faster than the other:
Reaction 1:
ΔG‡1 = 94.3 kJ/mol^-1 = 94.3 × 10^3 J/mol^-1
R = 8.314 J/(mol·K)
T = 298 K
Reaction 2:
ΔG‡2 = 76.4 kJ/mol^-1 = 76.4 × 10^3 J/mol^-1
Now, using the Arrhenius equation:
k1 = A1e^(-ΔG‡1/RT)
k2 = A2e^(-ΔG‡2/RT)
Since the concentrations and temperature are the same for both reactions (1 M and 298 K), we can compare the rate constants directly:
k1/k2 = (A1/A2)e^((-ΔG‡1+ΔG‡2)/RT)
Let's calculate the factor by substituting the values:
k1/k2 = e^((ΔG‡2-ΔG‡1)/RT)
k1/k2 = e^((76.4 × 10^3 - 94.3 × 10^3)/(8.314 × 298))
Calculating the exponential term:
k1/k2 = e^-8.69
k1/k2 ≈ 0.00015
Therefore, one reaction is approximately 0.00015 times faster than the other. So, the second reaction is slower compared to the first one by a factor of 0.00015.
To determine the relative rate of two reactions based on their standard free energy of activation, you need to use the Arrhenius equation. The Arrhenius equation relates the rate constant of a reaction to its activation energy (standard free energy of activation) and temperature.
The Arrhenius equation is as follows:
k = A * e^(-Ea/RT)
Where:
k is the rate constant
A is the pre-exponential factor (a constant that depends on the nature of the reaction)
Ea is the activation energy (standard free energy of activation)
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
In this case, we have the standard free energy of activation in kJ/mol^-1. To convert it to J/mol^-1, multiply the value by 1000.
Now, let's calculate the ratio of the rate constants for the two reactions based on their standard free energy of activation.
For the first reaction, with an activation energy (Ea1) of 94.3 kJ/mol^-1:
Ea1 = 94.3 kJ/mol^-1 * 1000 J/mol^-1 = 94300 J/mol^-1
For the second reaction, with an activation energy (Ea2) of 76.4 kJ/mol^-1:
Ea2 = 76.4 kJ/mol^-1 * 1000 J/mol^-1 = 76400 J/mol^-1
Now we can substitute these values into the Arrhenius equation, along with the temperature (T = 298 K) and assume the pre-exponential factor (A) is the same for both reactions:
k1/k2 = (A * e^(-Ea1/RT)) / (A * e^(-Ea2/RT))
k1/k2 = e^((Ea2 - Ea1)/RT)
k1/k2 = e^((76400 - 94300)/(8.314 * 298))
Using a calculator, you can calculate the exponential value. The result gives you the factor by which the first reaction is faster than the second reaction.
Note: It's important to double-check and ensure that the units are consistent throughout the calculation to obtain accurate results.