# Question 66e6d

##### 1 Answer
Jun 27, 2016

The answer is (3) $27$

#### Explanation:

The idea here is that the temperature coefficient, which you'll sometimes see referred to as the ${Q}_{10}$ coefficient, relates the change in the rate of a reaction to a ${10}^{\circ} \text{C}$ increase in the temperature of the reaction.

Simply put, the ${Q}_{10}$ coefficient tells you how increasing the temperature of the reaction by ${10}^{\circ} \text{C}$ will affect its rate.

The equation that allows you to calculate the ${Q}_{10}$ coefficient looks like this

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {Q}_{10} = {\left({\text{rate"_2/"rate}}_{1}\right)}^{\frac{10}{{T}_{2} - {T}_{1}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

${T}_{1}$ - the initial absolute temperature of the reaction
${T}_{2}$ - the new absolute temperature of the reaction
${\text{rate}}_{1}$ - the rate at ${T}_{1}$
${\text{rate}}_{2}$ - the rate at ${T}_{2}$

Now, a ${Q}_{10}$ coefficient of $3$ means that with every ${10}^{\circ} \text{C}$ increase in temperature, the rate of the reaction increases by a factor of $3$.

It's important to notice that because you're dealing with change in temperature, you can skip the conversion to Kelvin.

That is the case because an increase by ${1}^{\circ} \text{C}$ is equal to an increase by $\text{1 K}$.

So in the case of a ${10}^{\circ} \text{C}$ increase, you would have

${T}_{2} - {T}_{1} = \text{10 K}$

and

Q_(10) = ("rate"_2/"rate"_1)^(10/10) = "rate"_2/"rate"_1

In your case,

3 = "rate"_2/"rate"_1 implies color(purple)(|bar(ul(color(white)(a/a)color(black)("rate"_2 = 3 xx "rate"_1)color(white)(a/a)|)))#

For your reaction, the temperature goes from ${35}^{\circ} \text{C}$ to ${65}^{\circ} \text{C}$. This is equivalent to

$\Delta T = {65}^{\circ} \text{C" - 35^@"C" = 30^@"C}$

So, if the rate of the reaction increases by factor of $3$ with every ${10}^{\circ} \text{C}$ increase in temperature, it follows that a

${30}^{\circ} \text{C" = 3 xx 10^@"C}$

increase will cause the rate to go up by

${\text{rate"_2 = overbrace(color(white)(aaa)3color(white)(aaa))^(color(darkgreen)("for a 10"^@"C increase")) xx overbrace(color(white)(aaa)3color(white)(aaa))^(color(darkgreen)("for a 10"^@"C increase")) xx overbrace(color(white)(aaa)3color(white)(aaa))^(color(darkgreen)("for a 10"^@"C increase")) xx "rate}}_{1}$

${\text{rate"_2 = 27 xx "rate}}_{1}$

Therefore, the rate of the reaction will increase by a factor of $27$.