# A chemist's rule of thumb is that 10^@ "C" of difference results in a two-fold change in the reaction rate. If the temperature is varied from 35^@ "C" to 45^@ "C" for the rxn of crystal violet with hydroxide, by what factor does the rate multiply?

Apr 1, 2017

You can use the Arrhenius equation.

$k = A {e}^{- {E}_{a} \text{/} R T}$

where $A$ is the frequency factor, ${E}_{a}$ is the activation energy, $R$ is the universal gas constant, and $T$ is the temperature in $\text{K}$.

For the same reaction at most temperatures and pressures, the activation energy is assumed to not change. It is certainly true that $R$ and $A$ don't change with temperature.

Thus, all that can vary are $k$ and $T$. Recall that the $i$th trial of any reaction can be expressed as a rate law:

${r}_{i} \left(t\right) = {k}_{i} {\left[A\right]}_{i}^{m} {\left[B\right]}_{i}^{n}$

When we assume the concentrations are held constant as well to observe the effects of temperature change, we thus have:

$\frac{{r}_{i} \left(t\right)}{{r}_{j} \left(t\right)} = \frac{{k}_{i} \cancel{{\left[A\right]}_{i}^{m} {\left[B\right]}_{i}^{n}}}{{k}_{j} \cancel{{\left[A\right]}_{i}^{m} {\left[B\right]}_{i}^{n}}}$

for trial $j \ne i$. Thus, the initial rates of reaction are directly proportional to the rate constants.

As a result, we can therefore find the ratio of the rate constants, which is also the ratio of the initial rates.

${k}_{i} / {k}_{j} = \left(\cancel{A} {e}^{- {E}_{a} \text{/"RT_i))/(cancel(A)e^(-E_a"/} R {T}_{j}}\right)$

$= {e}^{- {E}_{a} \text{/"RT_i)*e^(E_a"/} R {T}_{j}}$

$= {e}^{- {E}_{a} \text{/"RT_i + E_a"/} R {T}_{j}}$

= e^(-E_a/R(1/(T_i) - 1/(T_j))

$= \text{exp} \left[- {E}_{a} / R \left(\frac{1}{{T}_{i}} - \frac{1}{{T}_{j}}\right)\right]$

If we take ${T}_{i} = 45 + 273.15 = \text{318.15 K}$ and ${T}_{j} = 35 + 273.15 = \text{308.15 K}$, we thus say that ${k}_{i}$ is for the reaction at the higher temperature.

A typical experimental result for the activation energy for this reaction of ${\text{CV}}^{+}$ with ${\text{OH}}^{-}$ is on the order of 20 ~ 60 $\text{kJ/mol}$. Take ${E}_{a} \approx \text{50 kJ/mol}$. We get:

${k}_{i} / {k}_{j} = \textcolor{b l u e}{\frac{{r}_{i} \left(t\right)}{{r}_{j} \left(t\right)}}$

= "exp"[-"50000 J/mol"/("8.314472 J/mol"cdot"K")*(1/"318.15 K" - 1/"308.15 K")]

$= \textcolor{b l u e}{1.847}$

So it's not a bad estimate. It's somewhat close to $2$. If we chose a more accurate ${E}_{a}$ we may have gotten closer to $2$.