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. 




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 $E\_a$. 


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