The activation energy of a certain reaction is 30.3 kJ/mol. At 20 C, the rate constant is 0.0130 s^{-1}. At what temperature would this reaction go twice as fast?
Can't you use Arrhenius' equation for this. Look it up. You have Ea. you have T1. You have k1 and k2. That leaves T2 as the only unknown.
Yes, you can definitely use Arrhenius' equation to solve this problem! Arrhenius' equation relates the rate constant (k) of a reaction to the temperature (T) and the activation energy (Ea) of the reaction. The equation is given as:
k = A * exp(-Ea / (R * T))
In this equation, k is the rate constant, A is the pre-exponential factor (also called the frequency factor), Ea is the activation energy, R is the gas constant (8.314 J/(mol*K)), and T is the temperature in Kelvin.
To solve the problem, we have the following information:
Ea = 30.3 kJ/mol (convert to Joules/mol for consistency)
T1 = 20 °C = 293.15 K
k1 = 0.0130 s^{-1} (given rate constant)
k2 = 2 * k1 (reaction goes twice as fast, so k2 = 2 * k1)
We can rearrange Arrhenius' equation to solve for T2:
k2 = A * exp(-Ea / (R * T2))
Divide the two equations to eliminate the A factor:
k2 / k1 = (exp(-Ea / (R * T2))) / (exp(-Ea / (R * T1)))
Since exp(-Ea / (R * T)) is the same for both sides, we can simplify it as:
k2 / k1 = exp(-Ea / (R * T2 + Ea / (R * T1)))
Taking natural logarithm on both sides of the equation:
ln(k2 / k1) = -Ea / (R * T2) + Ea / (R * T1)
Rearranging the equation to solve for T2:
ln(k2 / k1) + (Ea / (R * T1)) = -Ea / (R * T2)
T2 = (-Ea) / (R * (ln(k2 / k1) + (Ea / (R * T1))))
Now, we can substitute the given values into the equation and calculate T2:
Ea = 30.3 kJ/mol = 30,300 J/mol
T1 = 293.15 K
k1 = 0.0130 s^{-1}
k2 = 2 * k1 = 2 * 0.0130 s^{-1}
Plugging these values into the equation:
T2 = (-30,300 J/mol) / (8.314 J/(mol*K) * (ln(2 * 0.0130 s^{-1} / 0.0130 s^{-1}) + (30,300 J/mol) / (8.314 J/(mol*K) * 293.15 K)))
By evaluating this equation, you will find the temperature (in Kelvin) at which the reaction would go twice as fast.