A certain reaction has an activation energy of 58.80 kJ/mol. At what Kelvin temperature will the reaction proceed 6.50 times faster than it did at 303 K?
Use the Arrhenius equation.
Remember to convert kJ/mol to J/mol
Use k1 for k1 and 6.5*k1 for k2. That way k1 cancels and have just the ratio.
To find the Kelvin temperature at which the reaction will proceed 6.50 times faster, we can use the Arrhenius equation:
k2 = k1 * exp((Ea / R) * (1/T2 - 1/T1))
where:
k2 = rate constant at the new temperature (unknown)
k1 = rate constant at the initial temperature (known)
Ea = activation energy (known)
R = gas constant (8.314 J/(mol·K))
T2 = new temperature (unknown)
T1 = initial temperature (known)
First, let's convert the activation energy from kJ/mol to J/mol:
Ea = 58.80 kJ/mol * 1000 J/1 kJ = 58,800 J/mol
Next, we rearrange the Arrhenius equation to solve for T2:
(1/T2 - 1/T1) = (Ea / R) * ln(k2/k1)
Now, we have two equations:
Equation 1: k2 = 6.50 * k1
Equation 2: (1/T2 - 1/T1) = (Ea / R) * ln(k2/k1)
Since k2 is 6.50 times greater than k1, we can substitute 6.50 * k1 for k2 in Equation 2:
(1/T2 - 1/T1) = (Ea / R) * ln(6.50 * k1 / k1)
(1/T2 - 1/T1) = (Ea / R) * ln(6.50)
We rearrange this equation to solve for T2:
1/T2 = (Ea / R) * ln(6.50) + 1/T1
Now, substitute the known values:
1/T2 = (58,800 J/mol / (8.314 J/(mol·K))) * ln(6.50) + 1/303 K
Solving this equation will give us the reciprocal of T2. To find T2, take the reciprocal of both sides:
T2 = 1 / ( (58,800 J/mol / (8.314 J/(mol·K))) * ln(6.50) + 1/303 K )
Calculating this expression will give you the Kelvin temperature at which the reaction will proceed 6.50 times faster than it did at 303 K.