The following was calculated as described in the lab handout.
| T=20.1C | Data provided courtesy of B. Smith, R. Suburu, A. Avalos(additional data added to Arrhenius graph provided by A. West, F. Albadawi, and K. Fong | |||||
| g EtAc= | 0.865 | |||||
| mol EtAc= | 0.009817 | |||||
| [EtAc] @t=0 | 0.377587 | |||||
| mole HCl in each sample
(1.mL) (xM) |
0.000962 |
|
||||
| Time/min | mL NaOH | VM(NaOH)=total moles acid | [HAc] | [EtAc] @ time | =LN(EtAc) | =1/[EtAc] |
| 0 | 0.377587456 | -0.97395 | 2.648393064 | |||
| 10 | 5.3 | 0.0010865 | 0.124961538 | 0.252625918 | -1.37585 | 3.958422037 |
| 20 | 5.25 | 0.00107625 | 0.114711538 | 0.262875918 | -1.33607 | 3.804076115 |
| 30 | 5.3 | 0.0010865 | 0.124961538 | 0.252625918 | -1.37585 | 3.958422037 |
| 40 | 5.7 | 0.0011685 | 0.206961538 | 0.170625918 | -1.76828 | 5.860774336 |
| 50 | 5.55 | 0.00113775 | 0.176211538 | 0.201375918 | -1.60258 | 4.965837082 |
| 60 | 5.8 | 0.001189 | 0.227461538 | 0.150125918 | -1.89628 | 6.661075015 |
As discussed in class, sometimes it is hard to see from the graphs if a reaction
is 1st or 2nd order. It would be best to collect data over two half-lifes or more.
A first order half-life would be =LN(2)/k = 53 minutes. A second order half-life
would be =1/(k*[EtAc]) where this is the initial concentration. The 2nd order
half-life is predicted to be 460 minutes. Looking at the data above, we have
dropped to less than half the starting amount after 60 minutes, indicating this is
a first-order reaction. So, using their higher T data...
and using the temperatures and 1st order rate constants (slope = -k for 1st order) to calculate an activation energy and pre-exponential factor.
| Temp | T/K | 1/T | k(min^-1) |
| 19 | 292 | 0.00342466 | 0.0046 from West et al. |
| 20.1 | 293.1 | 0.0034118 | 0.013 |
| 42 | 315 | 0.0031746 | 0.653 |
Here slope = -Ea / R. So Ea = 22 kJ/mol.
The intercept = LN (A) = 9.0334, so exp(9.0334) = A = 8 X 103 min-1 .