Question

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?

Answer 1

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(t)` is the initial rate of reaction for the reaction at temperature `T_i`. Each temperature is in `""^@ "C"` to ensure the units cancel out!

The temperature was increased by `"40 K"`, hence it is increased by `40^@ "C"`, and we have:

`logQ_(10) = (10^@ "C")/(40^@ "C")log((r_2(t))/(r_1(t)))`

`=> log(2) = 1/4log((r_2(t))/(r_1(t)))`

`=> color(blue)((r_2(t))/(r_1(t))) = 10^(4log2)`

`=` `color(blue)(16)`

So, when the temperature coefficient is `2`, a `40^@ "C"` increase in the reaction temperature accomplishes a reaction that is 16 times faster.