The activation energy of a certain reaction is 32.9 kJ/mol At 25 degrees C the rate constant is 0.0160 units s-1 units. At what temperature in degrees Celsius would this reaction go twice as fast?
Given that the initial rate constant is 0.0160 s-1 at an initial temperature of 25 degrees C , what would the rate constant be at a temperature of 130 degrees C for the same reaction described in Part A?
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I managed to be given the formula: ln(k2/0.0160) = (32900/0.08206)(1/298/15 - 1/403.15) but when I answered it, I got the wrong answer.
Note that there are two questions here. The answer I gave you yesterday was for the second one. Did you report that answer for the first or the second one? Next, you simply copied what I gave you. Show your work, including the answer, and I will try to find the error.
To solve the first part of the question, you need to use the Arrhenius equation, which describes the relationship between the rate constant of a chemical reaction and the temperature.
The Arrhenius equation is given by:
k2 = k1 * e^(-Ea/R * (1/T2 - 1/T1))
Where:
k1 is the initial rate constant at temperature T1 (25 degrees C in this case),
k2 is the rate constant at temperature T2,
Ea is the activation energy (given as 32.9 kJ/mol),
R is the ideal gas constant (8.314 J/(mol·K)),
T1 is the initial temperature in Kelvin (25 degrees C + 273.15),
and T2 is the temperature you want to find in Kelvin.
To find the temperature at which the reaction goes twice as fast, we can rearrange the equation:
k2 = 2k1
2k1 = k1 * e^(-Ea/R * (1/T2 - 1/T1))
Cancelling out k1:
2 = e^(-Ea/R * (1/T2 - 1/T1))
Now, we can solve for T2 by taking the natural logarithm of both sides:
ln(2) = -Ea/R * (1/T2 - 1/T1)
Rearranging the equation to solve for T2:
1/T2 = (ln(2) * R) / Ea + 1/T1
Finally, plugging in the values into the equation will give us the temperature in degrees Celsius.
For the second part of the question, you can use the same Arrhenius equation.
Again, plug in the given values into the equation and evaluate the rate constant at the new temperature of 130 degrees C.
Please note that in the calculation mentioned above, I have used the ideal gas constant (R) in units of J/(mol·K). If the given activation energy is in kJ/mol, it's important to make sure the units are consistent throughout the calculation.