# Using a modified Arrhenius plot to find activation energy

I have a question on how to determine activation energy from a plot using the Arrhenius equation. I am working as an assistant with general chemistry II students and their lab is the iodination of acetone. I am familiar with using a plot of $\ln(k)$ vs $1/T$ to get the slope of $E_\mathrm{a}/R$ to solve for $E_\mathrm{a}$. Their lab manual though tells then to plot $\ln(t)$ vs $1/T$ to determine $E_\mathrm{a}$. I was wondering if anyone could shed some light on the derivation of this equation.

• I have no clue but as the exponential decay function $N(t)=N_0~e^{-\lambda~t}$ looks very similar to Arrhenius function $k=A~e^{-\frac{E_a}{R~T}}$ there might be some way to either convert one into another or to simply replace the equal terms ... – pH13 - Yet another Philipp Jul 16 '15 at 14:41

The use of $t$ only makes sense if $t$ is defined as the time it takes for the reaction to progress a certain relative amount (and if we are talking about first-order reactions). For example if $t$ is the time it takes for half (or 90% or 99% or whatever, as long as it is specified) the reactant to be consumed, then the idea makes sense. In that case it would be more helpful and clear to say $t_{1/2}$ (or $t_{90\%}$ or whatever) instead of just $t$.

Assuming $t_{1/2}$ is what is meant by $t$:

### Arrhenius equation:

$k = A_0 e^{\frac{-E_a}{RT}}$

$k = A_0 e^{\frac{-E_a}{R}\frac{1}{T}}$

$\ln k = \ln A_0 - \frac{E_a}{R}\frac{1}{T}$

### First-order kinetics:

$C = C_0 e^{-k t}$

$\ln \frac{C_0/2}{C_0} = - k t_{1/2}$

$k = \frac{1}{t_{1/2}} \ln 2$

$\ln k = - \ln t_{1/2} + \ln \ln 2 = -\ln{t_{1/2}} + \ln \ln 2$

### Combining:

$\ln k = \ln A_0 - \frac{E_a}{R}\frac{1}{T} = -\ln{t_{1/2}} + \ln \ln 2$

$\ln t_{1/2} - \ln \ln 2 = \frac{E_a}{R} T^{-1} - \ln A_0$

$\ln t_{1/2} = \frac{E_a}{R} T^{-1} - \ln A_0 + \ln \ln 2$

So if $t$ means $t_{1/2}$, then a plot of $\ln t_{1/2}$ vs. inverse temperature has a slope of $+\frac{E_a}{R}$, instead of $-\frac{E_a}{R}$ if you plot $\ln k$ vs. inverse temperature.

N.B. If $t$ just means time, then the idea doesn't make any sense, because then instead of $\ln \ln 2$ you would have $\ln \ln \frac{C(t)}{C_0}$, which is a term that depends on time, so instead of being an explicit equation that is clearly a line, you would have an implicit equation, non-linear in $t$, that would need to be solved before the plot was made, and the result would not be linear and would be impossible to use to determine $E_a$.

• I believe that is the equation that they are trying to get at in this lab. Although it has no explanation as to what t really is. I assume it's t when 100% of the iodine has been consumed in the reaction. I don't know if it is easier to do the ln(t) vs 1/T or the ln(k) vs 1/T. Personally I preferred the normal Arrhenius equation. That's for helping me understand. Maybe I should ask them to clarify the lab manual. – JuliusDariusBelosarius Jul 16 '15 at 18:30
• First, if all you care about is the activation energy, it doesn't matter if it is $t_{1/2}$ or $t_{90\%}$ or whatever; the equation shows it only affects the intercept, not the slope. Second, it can't be $t_{100\%}$, as that would imply an infinite intercept! It must really be $t_{99.99\%}$ or similar, corresponding to the highest iodine concentration that the measurement method cannot distinguish from zero. – Curt F. Jul 16 '15 at 18:35