If a temperature increase from 11.0∘C to 22.0∘C doubles the rate constant for a reaction, what is the value of the activation barrier for the reaction?
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To find the value of the activation barrier for the reaction, we can use the Arrhenius equation:
k = A * exp(-Ea/RT)
where:
k is the rate constant
A is the pre-exponential factor
Ea is the activation energy
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
We are given that the temperature increases from 11.0°C to 22.0°C, so we need to convert these temperatures to Kelvin:
T1 = 11.0°C + 273.15 = 284.15 K (initial temperature)
T2 = 22.0°C + 273.15 = 295.15 K (final temperature)
With this information, we know that the rate constant, k, doubles when the temperature increases. In other words, k2 = 2 * k1.
Now let's set up the equation using the Arrhenius equation for both temperatures:
k1 = A * exp(-Ea/R * (1/T1))
k2 = A * exp(-Ea/R * (1/T2))
Since k2 = 2 * k1, we have:
2 * k1 = A * exp(-Ea/R * (1/T2))
Now, divide both sides of the equation by k1:
2 = exp(-Ea/R * (1/T2))
Next, take the natural logarithm of both sides:
ln(2) = -Ea/R * (1/T2)
Now we can solve for Ea:
Ea = -ln(2) * R * T2
Substituting the values:
Ea = -ln(2) * 8.314 J/(mol·K) * 295.15 K
Calculating this expression will give us the value of the activation barrier for the reaction.