Q: 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?
My work:
ln(0.0160/0.032) = (32.9 x 10-3/8.314)(1/T2 - 1/298.15)
-0.6931 = 3,957.180(1/T2 - 0.00335)
-0.6932 = 3,957.180(1/T2) - 13.2565
13.187T2 = 3,957.180
T2 = 300.08 K = 26.98 degrees C
The answer above is not correct, and I would like to know where I went wrong. In addition, I need this answer for a follow-up question.
Where did you get 0.032 to plug in for K2?
Thank you for posting your work.
I didn't work the problem but it appears to me that the error is in Ea. That's 32.9 kJ/mol which is 32,900 J/mol and that's the number that goes in for Ea (not 32.9E-3). 0.032 is twice 0.0160 and that's ok. By the way, since the rate doubles you know the temperature is higher by approximately 10 C. Do you remember that from class; i.e., the rate (approximately) doubles for every 10 C rise in temperature.
DrBob222, when you divide 32,900 by 8.314, you still get 3,957.180.
To determine where you went wrong, let's go through the steps of solving the problem correctly:
1. Start with the Arrhenius equation:
ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)
Where:
k1 = the initial rate constant
k2 = the final rate constant
Ea = activation energy
R = gas constant (8.314 J/mol·K)
T1 = initial temperature (25 degrees C = 298.15 K)
T2 = final temperature (unknown)
2. Set up the equation using the given values:
ln(k2/0.0160) = (32.9 × 10^3 J/mol)/(8.314 J/mol·K) × (1/298.15 K - 1/T2)
3. Solve for ln(k2/0.0160):
Since we want the reaction to go twice as fast, k2 will be twice the initial rate constant:
ln(2/0.0160) = (32.9 × 10^3 J/mol)/(8.314 J/mol·K) × (1/298.15 K - 1/T2)
4. Simplify the equation:
ln(125) = (3.957180)(1/298.15 K - 1/T2)
5. Solve for 1/T2:
3.957180(1/298.15 K - 1/T2) = ln(125)
6. Rearrange the equation:
1/298.15 K - 1/T2 = ln(125)/3.957180
7. Solve for 1/T2:
1/T2 = 1/298.15 K - ln(125)/3.957180
8. Calculate T2:
T2 = 1/(1/298.15 K - ln(125)/3.957180)
Calculating this value gives T2 approximately equal to 492.84 K or 219.69 degrees C.
Therefore, the correct answer is approximately 219.69 degrees Celsius, not 26.98 degrees Celsius.
Apologies for the confusion caused by the incorrect answer in your previous attempt.