# A first order reaction has an Arrhenius constant for the reaction is #9.9 xx 10^10 sec^ -1#, and an activation energy of #"62.0 kJ"#. At what temperature (in Celsius) is the rate constant equal to #0.434 sec^-1#?

##### 1 Answer

#### Answer:

#### Explanation:

Your tool of choice here will be the **Arrhenius equation**, which looks like this

#color(blue)(|bar(ul(color(white)(a/a)k = A * "exp"(-E_a/(RT))color(white)(a/a)|)))" "# , where

*rate constant* for a given reaction

*pre-exponential factor*, specific to a given reaction

*activation energy* of the reaction

**absolute temperature** at which the reaction takes place

Notice that he problem provides you with the pre-exponential factor, the activation energy, and the rate constant, and asks for the temperature *in degrees Celsius* at which the reaction has that particular rate constant.

Your strategy here will be to use the Arrhenius equation to solve for

#color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))#

to go from Kelvin to degrees Celsius.

So, rearrange the Arrhenius equation to isolate the exponential term on one side of the equation

#k/A = "exp"(-E_a/(RT))#

Take the natural log of both sides of the equation

#ln(k/A) = ln["exp"(-E_a/(RT))]#

This will be equivalent to

#ln(k/A) = -E_a/(RT)#

Rearrange to isolate

#T = -E_a/R * 1/ln(k/A)#

Plug in your values to get - **do not** forget to convert the activation energy from *kilojoules* to *joules*

#T = - (62.0 * 10^3color(red)(cancel(color(black)("J"))))/(8.314color(red)(cancel(color(black)("J")))color(red)(cancel(color(black)("mol"^(-1))))"K"^(-1)) * 1/(ln( (0.434 color(red)(cancel(color(black)("s"^(-1)))))/(9.9 * 10^(10)color(red)(cancel(color(black)("s"^(-1))))))#

#T = "285.14 K"#

Expressed in degrees Celsius, the temperature will be

#t[""^@"C"] = "285.14 K" - 273.15 = color(green)(|bar(ul(color(white)(a/a)12^@"C"color(white)(a/a)|)))#

The answer is rounded to two **sig figs**, the number of sig figs you have for the pre-exponential factor.