A certain reaction has an activation energy of 37.30 kj/mol. At what Kelvin temperature will the reaction proceed 3.00 times then it did at 303 k?
Did you make a typo and omit the word FASTER in 3x.... than it did at 303K? Use the Arrhenius equation. No k1 is given but you can call it k1 and k2 becomes 3k1 (in which case k1 cancels) OR you can make up a number for k1, then multiply it by 3 to become k2.
No its not a typo but I did put the wrong activation energy. Its suppose to be a certain reaction has an activation energy of 37.70kj/mol. At what kelvin temperature will the reaction proceed 3.00 times then it did at 303k?
I was working the problem and got stuck on finishing the equation 1/T2=1/303k-1.099/4534.25k?
Yes, I think it is a typo for the sentence doesn't make sense as is. I think it should read, "At what Kelvin temperature will the reaction proceed 3.00 times FASTER THAN it did at 303 k?
ln(k2/k1) = Ea/R(1/T1-1/T2)
ln(3k1/k1) = [37700/8.314](1/303 - 1/T2)
To solve this problem, we can use the Arrhenius equation, which relates the rate constant (k) of a reaction to the activation energy (Ea) and the temperature (T) in Kelvin:
k = Ae^(-Ea/RT)
Where:
k = rate constant
A = pre-exponential factor or frequency factor
Ea = activation energy
R = ideal gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
We want to find the temperature (T) at which the reaction proceeds 3.00 times faster than at 303 K.
Step 1: Convert the activation energy to joules.
Ea = 37.30 kJ/mol × 1000 J/1 kJ = 37,300 J/mol
Step 2: Use the given information.
Let's assume the rate constant at 303 K is k1, and we want to find the temperature at which the rate constant is 3.00 times faster, which will be denoted as k2.
k2 = 3.00 × k1
Step 3: Set up the equation using the Arrhenius equation.
k2 = Ae^(-Ea/RT2)
k1 = Ae^(-Ea/RT1)
Divide the second equation by the first equation:
k2/k1 = [Ae^(-Ea/RT2)] / [Ae^(-Ea/RT1)]
Cancel out the A's:
k2/k1 = e^(-Ea/RT2) / e^(-Ea/RT1)
Since k2 = 3.00 × k1, the equation becomes:
3.00 = e^(-Ea/RT2) / e^(-Ea/RT1)
Step 4: Simplify the equation.
Take the natural logarithm (ln) of both sides to eliminate the exponential term:
ln(3.00) = ln [e^(-Ea/RT2) / e^(-Ea/RT1)]
Apply the logarithmic rule, ln(A/B) = ln(A) - ln(B):
ln(3.00) = [-Ea/RT2 - (-Ea/RT1)]
Simplify further:
ln(3.00) = -Ea/RT2 + Ea/RT1
ln(3.00) = Ea/RT1 - Ea/RT2
Step 5: Solve for T2.
Rearrange the equation to solve for T2:
Ea/RT2 = Ea/RT1 - ln(3.00)
Multiply both sides by RT2:
Ea = (Ea/RT1 - ln(3.00)) × RT2
Divide both sides by (Ea/RT1 - ln(3.00)):
T2 = Ea / [(Ea/RT1) - ln(3.00)]
Plug in the given values:
T1 = 303 K
Ea = 37,300 J/mol
Now, we can calculate T2 by substituting the values:
T2 = (37,300 J/mol) / [(37,300 J/mol / (8.314 J/(mol·K) × 303 K)) - ln(3.00)]
Using a calculator, evaluate the right-hand side of the equation, and you will get the value of T2 in Kelvin.