For a given reaction, the rate constant doubles when the temperature is increased from 27 o C to 49 o C. What is the activation energy?
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Use the Arrhenius equation.
Use k1 = k1 or choose any convenient number (like 1).
Then k2 = 2k1 or twice the number you choose (use 2 if you used k1 = 1)
Remember to use K for temperature.
T1= 27+273.15=300.15K
T2=90+273.15=363.15 K
In(1/300.15)-(2/363.15)= 5.709???? Did I get it?
No. I suggest you look up the Arrhenius equation and use it. Where is Ea? Where is R? Which R will you use? You have T1 and T2 right but no where in the Arrhenius equation do you take ln (1/T1 - 1/T2)
To determine the activation energy, we can use the Arrhenius equation, which relates the rate constant (k) to the temperature (T) and the activation energy (Ea):
k = A * exp(-Ea/RT)
In this equation, A is the pre-exponential factor, R is the gas constant (8.314 J/(mol*K)), and T is the absolute temperature (in Kelvin).
First, let's convert the given temperatures to Kelvin. To convert from Celsius to Kelvin, we add 273.15 to the Celsius temperature:
T1 = 27 + 273.15 = 300.15 K (temperature at 27 o C)
T2 = 49 + 273.15 = 322.15 K (temperature at 49 o C)
Next, let's consider the ratio of the rate constants at the two temperatures:
k2/k1 = 2 (since the rate constant doubles)
Now, we can substitute these values into the Arrhenius equation:
2 = exp(-Ea/(8.314 * (1/T2 - 1/T1)))
Simplifying further, we get:
2 = exp(-Ea/(8.314 * (1/322.15 - 1/300.15)))
To solve for the activation energy (Ea) in this equation, we can take the natural logarithm (ln) of both sides:
ln(2) = -Ea/(8.314 * (1/322.15 - 1/300.15))
Rearranging the equation to solve for Ea, we have:
Ea = -8.314 * (1/322.15 - 1/300.15) * ln(2)
Calculating this expression will give us the value of the activation energy.