Given that ∆Go’ = -15 kJ/mol and k1 = 1.5 X 10-6s-1, calculate k2.
To calculate k2, we need to use the relationship between ∆Go', k1, and k2. The relationship is as follows:
∆Go' = -RT ln(k2 / k1)
where ∆Go' is the standard Gibbs free energy change, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, k1 is the rate constant for the forward reaction, and k2 is the rate constant for the reverse reaction.
First, we need to convert the given standard Gibbs free energy change from kilojoules per mole (kJ/mol) to joules per mole (J/mol). Since 1 kJ = 1000 J, we have ∆Go' = -15,000 J/mol.
Next, we need to rearrange the equation to solve for k2:
k2 / k1 = e^(-∆Go' / RT)
Now, we can plug in the known values and solve for k2:
k2 / (1.5 x 10^(-6) s^(-1)) = e^(-(-15,000 J/mol) / (8.314 J/(mol·K) * T)
Simplifying further:
k2 = (1.5 x 10^(-6) s^(-1)) * e^(-15000 J/mol / (8.314 J/(mol·K) * T)
Since the temperature (T) is not given in the question, you need to know the temperature at which the reaction is taking place to obtain a numerical value for k2.