The temperature coefficient for a certain reaction is 2. If the temperature of the reaction was raised by "40 K", what is the approximate factor to which the rate is multiplied?

Apr 26, 2017

I have no idea what this is, so I googled it... The temperature coefficient on Wikipedia is defined as:

Q_(10) = ((r_2(t))/(r_1(t)))^(10^@ "C""/"(T_2 - T_1)

where ${r}_{i} \left(t\right)$ is the initial rate of reaction for the reaction at temperature ${T}_{i}$. Each temperature is in $\text{^@ "C}$ to ensure the units cancel out!

The temperature was increased by $\text{40 K}$, hence it is increased by ${40}^{\circ} \text{C}$, and we have:

$\log {Q}_{10} = \left({10}^{\circ} \text{C")/(40^@ "C}\right) \log \left(\frac{{r}_{2} \left(t\right)}{{r}_{1} \left(t\right)}\right)$

$\implies \log \left(2\right) = \frac{1}{4} \log \left(\frac{{r}_{2} \left(t\right)}{{r}_{1} \left(t\right)}\right)$

$\implies \textcolor{b l u e}{\frac{{r}_{2} \left(t\right)}{{r}_{1} \left(t\right)}} = {10}^{4 \log 2}$

$=$ $\textcolor{b l u e}{16}$

So, when the temperature coefficient is $2$, a ${40}^{\circ} \text{C}$ increase in the reaction temperature accomplishes a reaction that is 16 times faster.